PODMESWAR HELP SECTOR: Chapter - 9 Structure in C, Computer Science, Class 10 /Class ten, ASSEB (SEBA), Prepared by Podmeswar through AI

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The below answers and discussions are directly prepared by copy-paste from www.google.com, Googole AI Overview, https://gemini.google.com/app , https://chatgpt.com/,etc,

These are only for understanding to imrpove the concept only; not for answers. We have not checked these clearly. Some mistakes may be happened. We are not responsible for any wrong information and answers. Please check yourself, yourselves and discuss with your expert teachers for the final approval or accepted the answers.

Structure in C

Example 9.1 A C program where multiple arrays are used to store information about students.

Answer:

#include<stdio.h>

int main()

{

int roll[20], age[20], studentCount;

printf( "\n How many students? " );

scanf( "%d", &studentCount );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Enter roll for student %d ", index );

scanf(" %d", &roll[index] );

printf( " \n Enter age for student %d ", index );

scanf( "%d", &age[index] );

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Roll for student %d is %d", index, roll[index] );

printf( " \n Age for student %d is %d", index, age[index] );

}

return 0;

}

Meaning & Output:

This C program collects and displays roll numbers and ages for a user-specified number of students (up to 20), using parallel arrays roll[20] and age[20]. It prompts for the student count, iterates to take input for each, and then prints the entered data.

Key Aspects:

· Input: Accepts an integer for studentCount and uses a for loop to scan data for each student.

· Storage: Stores up to 20 integers for roll numbers and 20 for ages.

· Output: Displays the stored data in a formatted list.

· Safety: The program lacks input validation, allowing users to enter more than 20 students, which could cause a buffer overflow.

OUPUT:

This C program prompts for the number of students, collects their roll numbers and ages, and then prints the entered data. Assuming 2 students are entered, the output will look similar to this, depending on user input:

Text

How many students? 2

Enter roll for student 0 101

Enter age for student 0 15

Enter roll for student 1 102

Enter age for student 1 16

Entered details are as follows:

Roll for student 0 is 101

Age for student 0 is 15

Roll for student 1 is 102

Age for student 1 is 16

Example 9.2 A C program demonstrating a simple structure.

Answer:

A complete C program in Example 9.2 where a student structure is created with two fields in it and then two structure variables are declared. We then assigned some values to them and displayed those values using printf() statements.

Answer:

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info student1, student2;

student1.roll = 1;

student1.age = 17;

student2.roll = 2;

student2.age = 19;

printf( "Roll and age of the students are: \n ");

printf( "Student 1: Roll: %d Age: %d", student1.roll, student1.age );

printf( "\n Student 2: Roll: %d Age: %d", student2.roll, student2.age );

return 0;

}

Meaning & Output:

The program defines a structure Student_Info to store a student's roll number and age. It then creates two instances, assigns values to them, and prints the details.

Output

text

Roll and age of the students are:

Student 1: Roll: 1 Age: 17

Student 2: Roll: 2 Age: 19

Explanation

1. Structure Definition: struct Student_Info contains two integer members: roll and age.

2. Variable Declaration: struct Student_Info student1, student2; creates two instances of the structure.

3. Assignment:

1. student1 is initialized with roll = 1 and age = 17.

2. student2 is initialized with roll = 2 and age = 19.

4. Output (printf):

0. printf( "Roll and age of the students are: \n ");: Prints the header line.

1. printf( "Student 1: Roll: %d Age: %d", student1.roll, student1.age );: Prints Student 1: Roll: 1 Age: 17.

2. printf( "\n Student 2: Roll: %d Age: %d", student2.roll, student2.age );: Moves to a new line and prints Student 2: Roll: 2 Age: 19.

5. Return: The function returns 0, indicating successful execution.

Example 9.3 Another C program demonstrating a simple structure.

Answer:

We may also scan values to the structure variable fields from the keyboard. Example 9.3 shows this.

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info student1, student2;

printf( "\n Enter roll of student 1: " );

scanf("%d", &student1.roll );

printf( "\n Enter age of student 1: " );

scanf("%d", &student1.age );

printf( "\n Enter roll of student 2: " );

scanf("%d", &student2.roll );

printf( "\n Enter age of student 2: " );

scanf("%d", &student2.age );

printf( "Roll and age of the students are: \n ");

printf( "Student 1: Roll: %d Age: %d", student1.roll, student1.age );

printf( "\n Student 2: Roll: %d Age: %d", student2.roll, student2.age );

return 0;

}

Meaning & Output (AI):

This is a classic example of using Structures in C. Your code correctly defines a template (Student_Info) and creates two "instances" (student1 and student2) to store data.

When you use a structure, you are essentially telling the computer to group different variables together in one block of memory.

How this works in memory

Instead of having four separate variables floating around, the computer allocates a single block for student1 and another for student2. Inside each block, the roll and age are stored side-by-side.

A Few Tips for your Code:

1. Placement of Struct Definition: Usually, it is better to define your struct outside of main(). If you define it inside main, only the main function can use it. If you move it to the top of the file, you can create other functions (like displayStudent) that also recognize the Student_Info type.

2. Formatting: Be careful with the \n (newline) and text. In your code, there are some line breaks inside the printf quotes that might cause the compiler to throw an error depending on your IDE. It's best to keep the string on one line.

Refined Version

Here is your code with the struct moved to a global scope and the formatting cleaned up:

#include <stdio.h>

// Define the structure outside so it's accessible everywhere

struct Student_Info {

int roll;

int age;

};

int main() {

struct Student_Info student1, student2;

// Student 1 Input

printf("Enter roll of student 1: ");

scanf("%d", &student1.roll);

printf("Enter age of student 1: ");

scanf("%d", &student1.age);

// Student 2 Input

printf("\nEnter roll of student 2: ");

scanf("%d", &student2.roll);

printf("Enter age of student 2: ");

scanf("%d", &student2.age);

// Output

printf("\n--- Student Details ---\n");

printf("Student 1: Roll: %d, Age: %d\n", student1.roll, student1.age);

printf("Student 2: Roll: %d, Age: %d\n", student2.roll, student2.age);

return 0;

}

~~Would you like to see how to use an "Array of Structures" so you can handle 10 or 100 students without creating~~ ~~student1~~, ~~student2~~, ~~student3~~ ~~manually?~~

OUTPUT:

When you run this code, the program will prompt you to enter the data for each student one by one. Once you provide the numbers, it will print them back to you in a formatted list.

Assuming you enter 101 and 18 for the first student, and 102 and 19 for the second, here is exactly what the console will look like:

Expected Output

Plaintext

Enter roll of student 1: 101

Enter age of student 1: 18

Enter roll of student 2: 102

Enter age of student 2: 19

Roll and age of the students are:

Student 1: Roll: 101 Age: 18

Student 2: Roll: 102 Age: 19

How the Data is Stored

Behind the scenes, the structure acts like a container. Instead of having separate variables for everything, C groups the roll and age together for each student in memory.

A Note on the "Newlines" in your Code

In the code you provided, there are literal line breaks inside your printf strings (e.g., printf("\n\nEnter...");).

In a real compiler: This might cause a "missing terminating character" error.
The Fix: Always keep your text inside the quotes on a single line. Use \n whenever you want a new line to appear in the output.

~~Would you like me to show you how to use a~~ ~~for~~ ~~loop so you can input 5 students without writing the same code 5 times?~~

Example 9.4 C program demonstrating the use of a pointer with a structure.

Example 9.4 shows a simple C program where a pointer is used to scan and display the member variables of a structure Student_Info. We see that a pointer variable of type struct Student_Info is declared in line number 10. Then this pointer is made to point to a structure variable student (declared in line number 9) in line number 11. Once the pointer ptr points to the structure variable, we use the notations discussed above for accessing the members of that structure.

Answer:

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info student;

struct Student_Info *ptr;

ptr = &student;

printf( " \n Enter roll of the student: " );

scanf(" %d", &ptr->roll );

printf( " \n Enter age of the student: " );

scanf( "%d", &ptr->age );

printf( "\n Entered details are as follows: \n \n " );

printf( " \n Roll of the student is %d", ptr->roll );

printf( " \n Age of the student is %d", ptr->age );

return 0;

}

Meaning & Output:

This C program defines a struct Student_Info to store a student's roll (integer) and age (integer). Using a pointer (ptr) to the structure, it prompts for input, reads the data, and displays the entered details, showcasing how to access structure members with pointers.

Code Breakdown:

· Structure Definition: struct Student_Info defines a data structure with roll and age fields.

· Pointer Usage: ptr = &student; assigns the address of the student structure to the pointer ptr.

· Input (scanf): The program uses ptr->roll and ptr->age to access and store the input values in the student structure using the pointer.

· Output (printf): Finally, it displays the entered roll and age using the same pointer-member access notation.

OUTPUT:

The provided C program defines a student structure, uses a pointer to take user input for roll number and age, and then displays the entered details. Assuming input 101 for roll and 20 for age, the expected output is:

text

Enter roll of the student: 101

Enter age of the student: 20

Entered details are as follows:

Roll of the student is 101

Age of the student is 20

Key Points:

· ptr = &student;: The pointer ptr is initialized to point to the student structure.

· ptr->roll: The arrow operator (->) is used to access structure members via a pointer.

· Input: The program prompts for user input twice, storing values directly into the student structure fields via ptr.

· Output: The final lines display the stored values.

Example 9.5 C program demonstrating the use of a single pointer with multiple structure variables.

Example 9.5 shows a C program where a single pointer *ptr is used to point to two structure variables student1 and student2. The pointer first points to student1 and it scans values from the keyboard. Then, it is made to point to student2. So it scans values for the next variable.

To display the student record, we do the same thing. First the pointer points to student1 (line number 22) and displays its member variables. Then, it points to student2 (line number 25) and displays its member variables.

Answer:

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info student1, student2;

struct Student_Info *ptr;

ptr = &student1; // pointer points to student1

printf( " \n Enter roll of the student 1: " );

scanf(" %d", &ptr->roll );

printf( " \n Enter age of the student 1: " );

scanf( "%d", &ptr->age );

ptr = &student2; // pointer points to student2

printf( " \n Enter roll of the student 2: " );

scanf(" %d", &ptr->roll );

printf( " \n Enter age of the student 2: " );

scanf( "%d", &ptr->age );

printf( "\n Entered details are as follows: \n \n " );

ptr = &student1; // pointer ptr points to student1 again

printf( " \n Roll of the student 1 is %d", ptr->roll );

printf( " \n Age of the student 1 is %d", ptr->age );

ptr = &student2; pointer ptr now points to student2

printf( " \n Roll of the student 2 is %d", ptr->roll );

printf( " \n Age of the student 2 is %d", ptr->age );

return 0;

}

Meaning & Output:

The provided C code demonstrates how to use pointers to access and modify structure members. However, it contains one syntax error related to comments and could be improved for better readability.

Syntax Error Correction

In the line:
ptr = &student2; pointer ptr now points to student2

The text pointer ptr now points to student2 is not a valid comment. In C, single-line comments must start with //. This will cause a compilation error.

Corrected line:
ptr = &student2; // pointer ptr now points to student2

Corrected Code

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info student1, student2;

struct Student_Info *ptr;

// Point to student1 and take input

ptr = &student1;

printf( " \n Enter roll of the student 1: " );

scanf("%d", &ptr->roll );

printf( " \n Enter age of the student 1: " );

scanf("%d", &ptr->age );

// Point to student2 and take input

ptr = &student2;

printf( " \n Enter roll of the student 2: " );

scanf("%d", &ptr->roll );

printf( " \n Enter age of the student 2: " );

scanf("%d", &ptr->age );

printf( "\n Entered details are as follows: \n \n " );

// Display student 1 details

ptr = &student1;

printf( " \n Roll of the student 1 is %d", ptr->roll );

printf( " \n Age of the student 1 is %d", ptr->age );

// Display student 2 details

ptr = &student2;

printf( " \n Roll of the student 2 is %d", ptr->roll );

printf( " \n Age of the student 2 is %d", ptr->age );

return 0;

}

Key Concepts Demonstrated

1. Structure Pointers (struct Student_Info *ptr): Declares a pointer capable of holding the address of a Student_Info struct.

2. Pointer Assignment (ptr = &student1): Assigns the memory address of a struct variable to the pointer.

3. Member Access Operator (->): Uses ptr->roll to access structure members through the pointer. This is equivalent to (*ptr).roll.

4. scanf with pointers: scanf requires the address of the variable, which is naturally provided by &ptr->roll (or ptr->roll if ptr already holds the struct address).

Output Example

text

Enter roll of the student 1: 101

Enter age of the student 1: 20

Enter roll of the student 2: 102

Enter age of the student 2: 21

Entered details are as follows:

Roll of the student 1 is 101

Age of the student 1 is 20

Roll of the student 2 is 102

Age of the student 2 is 21

Example 9.6 A C program demonstrating an array of structure variables.

Answer:

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info student[10];

int studentCount;

printf( "\n How many students? " );

scanf( "%d", &studentCount );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Enter roll for student %d ", index );

scanf(" %d", &student[index].roll );

printf( " \n Enter age for student %d ", index );

scanf( "%d", &student[index].age );

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Roll for student %d is %d", index, student[index].roll );

printf( " \n Age for student %d is %d", index, student[index].age );

}

return 0;

}

Meaning & Output:

This C program defines a Student_Info structure, creates an array of 10 students, and uses a loop to prompt for input (studentCount, roll, age) before displaying the details. It efficiently collects data for up to 10 students and prints them back, demonstrating basic struct usage and array handling.

Key Code Features:

· Struct Definition: struct Student_Info { int roll; int age; }; organizes student data.

· Array Allocation: struct Student_Info student[10]; reserves memory for 10 students.

· Input Handling: scanf is used within a for loop to capture user input for each student's roll and age.

· Output Display: A second for loop prints the entered data for each student.

Potential Improvements:

· Boundary Checking: The code assumes studentCount will not exceed 10. Adding if (studentCount > 10) studentCount = 10; would prevent potential overflow.

· Input Validation: Validating that studentCount is a positive number.

OUTPUT:

This C program prompts for the number of students, collects their roll numbers and ages, and then prints the entered data. Assuming a user inputs 2 students with specific details, the output will look similar to the following:

text

How many students? 2

Enter roll for student 0 101

Enter age for student 0 15

Enter roll for student 1 102

Enter age for student 1 16

Entered details are as follows:

Roll for student 0 is 101

Age for student 0 is 15

Roll for student 1 is 102

Age for student 1 is 16

Key Aspects of the Program:

· Struct Definition: Defines a struct Student_Info containing roll and age.

· Array Allocation: Allocates space for 10 students (struct Student_Info student).

· Input Handling: Uses scanf to store data into the student array.

· Output Formatting: Displays the data in the same order it was entered.

Example 9.7 Another C program demonstrating an array of structure variables.

Example 9.7. In this program, we have added a new field (name) in the definition of the structure Student_Info. So we are to take input in this new field. We have taken input using a gets() function.

Answer:

#include<stdio.h>

int main()

{

struct Student_Info

{

char name[30];

int roll;

int age;

};

struct Student_Info student[10];

int studentCount;

printf( "\n How many students? " );

scanf( "%d", &studentCount );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Enter name for student %d ", index );

getchar();

gets( student[index].name );

printf( " \n Enter roll for student %d ", index );

scanf(" %d", &student[index].roll );

printf( " \n Enter age for student %d ", index );

scanf( "%d", &student[index].age );

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < studentCount; index++ )

{

printf("\n\n Student %d: ", index+1 );

printf("\n Name: ");

puts( student[index].name );

printf( " Roll: %d", student[index].roll );

printf( " \n Age: %d", student[index].age );

}

return 0;

}

Meaning & Output:

This C program defines a Student_Info structure, creates an array of 10 students, prompts for input (count, name, roll, age), and displays them. It uses getchar() to consume the newline from scanf() and gets() for string input.
Warning: Using gets() is highly discouraged in modern C as it is unsafe.

Key Aspects:

· Structure: struct Student_Info { char name[30]; int roll; int age; };

· Array: struct Student_Info student[10]; (Allows storing up to 10 students).

· Input Handling: scanf("%d", &studentCount); reads how many students. getchar(); is crucial to consume the newline character left by scanf so gets() works correctly.

· Output: The code uses puts() to print the student names.

· Limitation: If studentCount exceeds 10, the program will experience a buffer overflow.

Potential Improvements:

· Use fgets(student[index].name, 30, stdin); instead of gets().

· Add a check for studentCount <= 10

Based on the code provided, the program prompts for the number of students, then takes input (name, roll, age) for each student using gets() and scanf(). It subsequently displays the entered data formatted with "Student X", "Name", "Roll", and "Age".

Example Output (Assuming 2 students):

text

How many students? 2

Enter name for student 0 : John Doe

Enter roll for student 0 : 101

Enter age for student 0 : 20

Enter name for student 1 : Jane Smith

Enter roll for student 1 : 102

Enter age for student 1 : 19

Entered details are as follows:

Student 1:

Name: John Doe

Roll: 101

Age: 20

Student 2:

Name: Jane Smith

Roll: 102

Age: 19

Note: gets() is considered unsafe in modern C programming and can cause buffer overflows if input exceeds 30 characters.

Example 9.8 A C program for handling information about books.

Answer:

#include<stdio.h>

int main()

{

struct Book_Info

{

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalBookCount;

printf( "\n How many books in total? " );

scanf( "%d", &totalBookCount );

for ( int index = 0; index < totalBookCount; index++ )

{

printf( " \n Enter title of the book %d ", index );

getchar();

gets(book[index].title);

printf( " \n Enter Book ID %d ", index );

scanf(" %d", &book[index].bookID );

printf( " \n Enter total no of copies %d ", index );

scanf(" %d", &book[index].noOfCopies );

printf( " \n Enter author of the book %d ", index );

getchar();

gets(book[index].author);

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < totalBookCount; index++ )

{

printf("\n Book %d \n", index);

printf(" Title: ");

puts( book[index].title );

printf( " ID: %d \n Available copies: %d ", book[index].bookID,

book[index].noOfCopies );

printf(" \n Author: ");

puts( book[index].author );

}

return 0;

}

Meaning & Output:

Meaning & Output (from Gemini AI):

This code demonstrates an Array of Structures. It is much more efficient than the previous example because it uses a single array book[10] to handle multiple books using a for loop.

If you run this code and specify that you have 2 books, here is how the output will look:

Expected Output

Plaintext

How many books in total? 2

Enter title of the book 0: Hamlet

Enter Book ID 0: 101

Enter total no of copies 0: 5

Enter author of the book 0: William Shakespeare

Enter title of the book 1: 1984

Enter Book ID 1: 102

Enter total no of copies 1: 3

Enter author of the book 1: George Orwell

Entered details are as follows:

Book 0

Title: Hamlet

ID: 101

Available copies: 5

Author: William Shakespeare

Book 1

Title: 1984

ID: 102

Available copies: 3

Author: George Orwell

How it works: Array of Structures

Instead of creating book1, book2, etc., you created an array. Each "slot" in the array is a complete Book_Info structure.

, book[1] each containing title, ID, copies, and author]

Important Technical Notes:

Buffer Clearing: Your use of getchar() before gets() is a smart move. It catches the "Enter" key (newline character) left over from the scanf calls. Without it, the program would skip the title or author inputs.
Array Limits: You defined book[10]. If a user enters a "totalBookCount" greater than 10, the program will crash because it will try to access memory that doesn't belong to the array (this is called Buffer Overflow).
Modern Safety: As a reminder, gets() is considered unsafe in modern C. Most professional programmers use fgets(book[index].title, 30, stdin); because it ensures the user doesn't type more than 30 characters and break the program.

~~Would you like me to show you how to search for a specific book by its ID within this array?~~

Example 9.9 A C program for handling information about books with an issue operation.

Example 9.9. We can see that the structure is declared as a global variable along with the total number of records. We have used three userdefined functions for the following purposes.

Function name: Purpose

input(): To take input to the variables from the keyboard,

output(): To display the records,

issue(): To perform an issue operation for a book,

As we have declared the structure as global, all the functions can access it. We have called the issue() function only once. This means that issue operation is performed only once for a book.

We have called the output() function from the issue() function (line number 10 of the function definition) to display the total book record after decreasing the available quantity of the book.

After we finished our work, we wrote a return statement to come out from the function (line number 11 of the function definition). We can do so because we have already found the book record and we performed the issue operation. If we do not return, the loop will unnecessarily iterate till the last record.

Answer:

#include<stdio.h>

struct Book_Info

{

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

void input()

{

for ( int index = 0; index < totalRecord; index++ )

{

printf( " \n Enter title of the book %d ", index );

gets(book[index].title);

printf( " \n Enter Book ID %d ", index );

scanf(" %d", &book[index].bookID );

printf( " \n Enter total no of copies %d ", index );

scanf(" %d", &book[index].noOfCopies );

printf( " \n Enter author of the book %d ", index );

getchar();

gets(book[index].author);

}

void output()

{

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < totalRecord; index++ )

{

printf("\n Book %d \n\n", index);

printf(" Title: ");

puts(book[index].title);

printf( " ID: %d \n Remaining copies: %d \n", book[index].bookID,

book[index].noOfCopies );

printf(" Authors: ");

puts(book[index].author);

}

void issue()

{

int id;

printf("\n Enter the ID of the book to issue: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ )

{

if( book[index].bookID == id )

{

book[index].noOfCopies = book[index].noOfCopies - 1;

printf("\n One book issued with ID: %d", id);

output();

return;

}

int main()

{

printf( "\n How many books in total? " );

scanf( "%d", &totalRecord );

getchar();

input();

output();

issue();

return 0;

}

Meaning & Output:

This C program defines a struct Book_Info to manage library records (title, ID, copies, author) for up to 10 books using an array. It includes functions for user input (input), displaying inventory (output), and issuing a book (issue) by reducing the copy count, featuring a simple main menu flow.

Key Features and Structure:

· Struct: struct Book_Info stores Title, ID, Copies, and Author.

· Data Structure: struct Book_Info book[10]; stores up to 10 books.

· Input (input): Uses gets() for string input (unsafe, replaced in modern code) and scanf() for integers. getchar() clears the newline buffer.

· Output (output): Displays current inventory using puts().

· Issue (issue): Decrements noOfCopies for a matching bookID.

· Main Function: Manages total records and calls input, output, and issue functions sequentially.

Example Output Flow:

1. Asks for the total number of books.

2. Prompts for details for each book.

3. Displays the entered details.

4. Asks for a book ID to issue and reduces its count.

Note: The use of gets() is insecure and has been removed from the C11 standard. fgets() is recommended for safer string input.

Based on the code, the program prompts for the total number of books, accepts details (title, ID, copies, author) for each, prints them, and then reduces the copy count based on a user-inputted ID.

Sample Output:

text

How many books in total? 2

Enter title of the book 0 C Programming

Enter Book ID 0 101

Enter total no of copies 0 5

Enter author of the book 0 Dennis Ritchie

Enter title of the book 1 Data Structures

Enter Book ID 1 102

Enter total no of copies 1 3

Enter author of the book 1 E. Balagurusamy

Entered details are as follows:

Book 0

Title: C Programming

ID: 101

Remaining copies: 5

Authors: Dennis Ritchie

Book 1

Title: Data Structures

ID: 102

Remaining copies: 3

Authors: E. Balagurusamy

Enter the ID of the book to issue: 101

One book issued with ID: 101

Entered details are as follows:

Book 0

Title: C Programming

ID: 101

Remaining copies: 4

Authors: Dennis Ritchie

Book 1

Title: Data Structures

ID: 102

Remaining copies: 3

Authors: E. Balagurusamy

Key Behaviors:

· gets() Usage: gets() reads the entire line, including spaces, for titles and authors.

· Buffer Clearing: getchar() is used after scanf to consume the newline character, ensuring gets() works properly.

· Issue Functionality: It searches by bookID, decrements noOfCopies by 1, and reprints all records using output().

· Limitations: gets() is insecure and should be replaced by fgets() in modern C, though it works in this example.

Example 9.10 A C program for handling information about books with an issue operation.

Example 9.10. We see that all the portions of the program are the same except the returnBook() function. Accordingly, instead of calling issue() function, we called returnBook() function (line number 6 in main() function).

Answer:

#include<stdio.h>

struct Book_Info

{

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

void input()

{

for ( int index = 0; index < totalRecord; index++ )

{

printf( " \n Enter title of the book %d ", index );

gets(book[index].title);

printf( " \n Enter Book ID %d ", index );

scanf(" %d", &book[index].bookID );

printf( " \n Enter total no of copies %d ", index );

scanf(" %d", &book[index].noOfCopies );

printf( " \n Enter author of the book %d ", index );

getchar();

gets(book[index].author);

}

void output()

{

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < totalRecord; index++ )

{

printf("\n Book %d \n\n", index);

printf(" Title: ");

puts(book[index].title);

printf( " ID: %d \n Remaining copies: %d \n", book[index].bookID,

book[index].noOfCopies );

printf(" Authors: ");

puts(book[index].author);

}

void returnBook()

{

int id;

printf("\n Enter the ID of the book to return: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ )

{

if( book[index].bookID = = id )

{

book[index].noOfCopies = book[index].noOfCopies + 1;

printf("\n One book is returned with ID: %d", id);

output();

return;

}

int main()

{

printf( "\n How many books in total? " );

scanf( "%d", &totalRecord );

getchar();

input();

output();

returnBook();

return 0;

}

Meaning & Output:

The provided C code defines a simple library management system to input, output, and return book records using structures. It features a struct Book_Info, input/output functions, and a returnBook() function that increments the copy count. However, it contains unsafe code (gets()) and potential issues with input buffering.

Key Issues and Improvements:

· Safety Hazard: gets() is dangerous and deprecated. Use fgets() instead to prevent buffer overflows.

· Input Handling: scanf leaves newline characters in the buffer, which can cause fgets or gets to skip inputs.

· returnBook() Fix: The current if condition book[index].bookID = = id is incorrect (uses an assignment operator if written as =). It must be == to compare.

Corrected Code:

#include<stdio.h>

#include<string.h> // Required for strcspn

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

// Function to safely remove newline from fgets

void stripNewline(char *str) {

str[strcspn(str, "\r\n")] = 0;

}

void input() {

for ( int index = 0; index < totalRecord; index++ ) {

printf( "\n Enter title of the book %d: ", index );

fgets(book[index].title, sizeof(book[index].title), stdin);

stripNewline(book[index].title);

printf( " Enter Book ID %d: ", index );

scanf("%d", &book[index].bookID );

printf( " Enter total no of copies %d: ", index );

scanf("%d", &book[index].noOfCopies );

getchar(); // Clear newline from scanf

printf( " Enter author of the book %d: ", index );

fgets(book[index].author, sizeof(book[index].author), stdin);

stripNewline(book[index].author);

}

void output() {

printf( "\n --- Entered details --- \n" );

for ( int index = 0; index < totalRecord; index++ ) {

printf("\n Book %d", index);

printf("\n Title: %s", book[index].title);

printf("\n ID: %d \n Remaining copies: %d", book[index].bookID, book[index].noOfCopies );

printf("\n Authors: %s\n", book[index].author);

}

void returnBook() {

int id;

printf("\n Enter the ID of the book to return: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ ) {

// Corrected comparison operator: ==

if( book[index].bookID == id ) {

book[index].noOfCopies++;

printf(" One book is returned with ID: %d\n", id);

output();

return;

}

printf("Book ID not found.\n");

}

int main() {

printf( "How many books in total? " );

scanf( "%d", &totalRecord );

getchar(); // Clear buffer

input();

output();

returnBook();

return 0;

}

OUTPUT:

Based on the code provided, the program takes input for the number of books, their details (title, ID, copies, author), prints them, asks to return a book by ID, increases the copy count, and prints the updated inventory. Assuming a total of 2 books are entered, the output would be:

text

How many books in total? 2

Enter title of the book 0: The C Programming Language

Enter Book ID 0: 101

Enter total no of copies 0: 5

Enter author of the book 0: Ritchie

Enter title of the book 1: Data Structures

Enter Book ID 1: 102

Enter total no of copies 1: 3

Enter author of the book 1: Sahni

Entered details are as follows:

Book 0

Title: The C Programming Language

ID: 101

Remaining copies: 5

Authors: Ritchie

Book 1

Title: Data Structures

ID: 102

Remaining copies: 3

Authors: Sahni

Enter the ID of the book to return: 101

One book is returned with ID: 101

Entered details are as follows:

Book 0

Title: The C Programming Language

ID: 101

Remaining copies: 6

Authors: Ritchie

Book 1

Title: Data Structures

ID: 102

Remaining copies: 3

Authors: Sahni

Note regarding potential bugs: The code contains a space between = and = in if( book[index].bookID = = id ). To run successfully, this must be == (equality operator) instead of = =.

Figure 9.2: A running instance of the required application.

1: Display all book details

2: Display particular book details

3: Issue a book

4: Return a book

5: Quit the application

Please enter your choice: __

Now, when we want some statements to run continuously, we can use a while (1) loop as

follows.

while (1)

{

. . .

}

We need to place the options (as shown in Figure 9.2) inside the while loop so that they appear everytime. Once the user enters an option, we need to perform the required task associated with that option. For instance, the user enters 1, then we need to display all the records.

We can use a switch-case construct for this. In each case of the switch-case, we can call the appropriate function to do the required task. The following code segment shows this clearly.

We call the function displayAll() in case 1 and this will be executed when the user enters option 1. Similarly, when the user enters 4, it belongs to case 4 and we are calling the function returnBook().

while (1)

{

printf(" \n 1: Display all book details" );

printf(" \n 2: Display particular book details" );

printf(" \n 3: Issue a book" );

printf(" \n 4: Return a book" );

printf(" \n 5: Quit the application" );

printf(" \n Please enter your choice: " );

scanf( "%d", &option );

switch (option)

{

case 1:

displayAll();

break;

case 2:

displayOne();

break;

case 3:

issueBook();

break;

case 4:

returnBook();

break;

case 5:

return 0;

default:

printf(" \n Invalid choice! ");

}

Example 9.11 A C program for a mini library application.

Example 9.11. All other portions of the program are similar to the previous ones.

Answer:

#include<stdio.h>

struct Book_Info

{

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

void input()

{

for ( int index = 0; index < totalRecord; index++ )

{

printf( " \n Enter title of the book %d ", index );

gets(book[index].title);

printf( " \n Enter Book ID %d ", index );

scanf(" %d", &book[index].bookID );

printf( " \n Enter total no of copies %d ", index );

scanf(" %d", &book[index].noOfCopies );

printf( " \n Enter author of the book %d ", index );

getchar();

gets(book[index].author);

}

void displayAll()

{

printf( "\n The library has the following books: \n \n " );

for ( int index = 0; index < totalRecord; index++ )

{

printf("\n Book %d \n\n", index);

printf(" Title: ");

puts(book[index].title);

printf( " ID: %d \n Remaining copies: %d \n", book[index].bookID,

book[index].noOfCopies );

printf(" Authors: ");

puts(book[index].author);

}

void displayOne()

{

int id;

printf("\n Enter the ID of the book to issue: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ )

{

if( book[index].bookID == id )

{

printf(" Title: ");

puts(book[index].title);

printf( " ID: %d \n Remaining copies: %d \n", book[index].bookID,

book[index].noOfCopies );

printf(" Authors: ");

puts(book[index].author);

}

void issueBook()

{

int id;

printf("\n Enter the ID of the book to issue: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ )

{

if( book[index].bookID = = id )

{

book[index].noOfCopies = book[index].noOfCopies - 1;

printf("\n One book is returned with ID: %d", id);

return;

}

void returnBook()

{

int id;

printf("\n Enter the ID of the book to return: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ )

{

if( book[index].bookID = = id )

{

book[index].noOfCopies = book[index].noOfCopies + 1;

printf("\n One book is returned with ID: %d", id);

return;

}

int main()

{

printf( "\n How many books in total? " );

scanf( "%d", &totalRecord );

getchar();

input();

int option;

while (1)

{

printf(" \n 1: Display all book details" );

printf(" \n 2: Display particular book details" );

printf(" \n 3: Issue a book" );

printf(" \n 4: Return a book" );

printf(" \n 5: Quit the application" );

printf(" \n Please enter your choice: " );

scanf( "%d", &option );

switch (option)

{

case 1:

displayAll();

break;

case 2:

displayOne();

break;

case 3:

issueBook();

break;

case 4:

returnBook();

break;

case 5:

return 0;

default:

printf(" \n Invalid choice! ");

}

return 0;

}

This code ties everything together into a functional Library Management System. You have implemented a menu-driven program using a while(1) loop and a switch statement, which is the standard way to build interactive console applications in C.

However, there are a few syntax errors and logical bugs that will prevent this specific version from compiling or running correctly.

1. Critical Syntax Errors

The Assignment Bug: In issueBook() and returnBook(), you wrote if( book[index].bookID = = id ). In C, the equality operator must not have a space. It should be ==.
The String Formatting: In displayAll(), there is a line break inside your printf quotes. This will cause a compilation error.
Output Message: In issueBook(), your success message says "One book is returned" instead of "issued."

2. Logic & Safety Fixes

The Stock Issue: In issueBook(), you should check if noOfCopies > 0 before subtracting. Otherwise, your library could end up with -1 books.
The gets() Issue: As we discussed, gets() is risky. Also, mixing scanf and gets can cause the program to skip inputs because of the "newline" character left in the buffer.

Corrected & Optimized Version

Here is the fixed version of your code. I've added a few safety checks (like checking if a book exists before issuing) to make it more robust.

#include<stdio.h>

#include<string.h>

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

void input() {

for (int index = 0; index < totalRecord; index++) {

printf("\n--- Book %d ---", index + 1);

printf("\nEnter title: ");

// Using %[^\n] reads the full title including spaces

scanf(" %29[^\n]", book[index].title);

printf("Enter Book ID: ");

scanf("%d", &book[index].bookID);

printf("Enter total copies: ");

scanf("%d", &book[index].noOfCopies);

printf("Enter author: ");

scanf(" %29[^\n]", book[index].author);

}

void displayAll() {

printf("\n--- Library Records ---");

for (int index = 0; index < totalRecord; index++) {

printf("\nID: %d | Title: %s | Copies: %d | Author: %s",

book[index].bookID, book[index].title,

book[index].noOfCopies, book[index].author);

}

printf("\n");

}

void issueBook() {

int id, found = 0;

printf("\nEnter Book ID to issue: ");

scanf("%d", &id);

for (int index = 0; index < totalRecord; index++) {

if (book[index].bookID == id) {

found = 1;

if (book[index].noOfCopies > 0) {

book[index].noOfCopies--;

printf("Book issued! Remaining: %d\n", book[index].noOfCopies);

} else {

printf("Error: Out of stock!\n");

}

break;

}

if (!found) printf("Error: Book ID not found.\n");

}

void returnBook() {

int id, found = 0;

printf("\nEnter Book ID to return: ");

scanf("%d", &id);

for (int index = 0; index < totalRecord; index++) {

if (book[index].bookID == id) {

book[index].noOfCopies++;

printf("Book returned! Total copies: %d\n", book[index].noOfCopies);

found = 1;

break;

}

if (!found) printf("Error: Book ID not found.\n");

}

int main() {

printf("How many books to register (max 10)? ");

scanf("%d", &totalRecord);

input();

int option;

while (1) {

printf("\n1: All | 2: Issue | 3: Return | 4: Exit\nChoice: ");

scanf("%d", &option);

if (option == 1) displayAll();

else if (option == 2) issueBook();

else if (option == 3) returnBook();

else if (option == 4) break;

else printf("Invalid choice!\n");

}

return 0;

}

Visualizing the Program Flow

When you run this, the main() function acts as the "Air Traffic Controller," sending the program to different functions based on your input.

OUTPUT:

To show you the output, I'll simulate a session where we register 2 books and then issue one of them.

Because your code has a few syntax issues (like the space in = =), the output below assumes those are fixed so the program can actually run.

Simulation Trace

Initial Setup

Plaintext

How many books in total? 2

Enter title of the book 0: The Great Gatsby

Enter Book ID 0: 101

Enter total no of copies 0: 5

Enter author of the book 0: F. Scott Fitzgerald

Enter title of the book 1: 1984

Enter Book ID 1: 102

Enter total no of copies 1: 3

Enter author of the book 1: George Orwell

Using the Menu (Issuing a Book)

Plaintext

1: Display all book details

2: Display particular book details

3: Issue a book

4: Return a book

5: Quit the application

Please enter your choice: 3

Enter the ID of the book to issue: 102

One book is issued with ID: 102

Verifying the Change (Displaying All)

Plaintext

1: Display all book details

2: Display particular book details

3: Issue a book

4: Return a book

5: Quit the application

Please enter your choice: 1

The library has the following books:

Book 0

Title: The Great Gatsby

ID: 101

Remaining copies: 5

Authors: F. Scott Fitzgerald

Book 1

Title: 1984

ID: 102

Remaining copies: 2

Authors: George Orwell

How the Data is Managed

When the program runs, it creates an array of structures in the Stack memory. Each element in the array book[index] holds all the fields (title, ID, etc.) for that specific entry.

and book[1]]

Important Correction for your code:

In your issueBook and returnBook functions, look at this line: if( book[index].bookID = = id )

What happens: The compiler will give you an error because of the space.
The Fix: Change it to ==.
Logic Note: In issueBook, your code currently prints "One book is returned." You'll likely want to change that text to "issued" so the user doesn't get confused!

~~Would you like me to explain how the~~ ~~getchar()~~ ~~and~~ ~~gets()~~ ~~combination works to handle those book titles with spaces?~~

Quick activity: Extend the previous program of to handle some unusual conditions such as a user wants to issue a book but (i) it is not there in the library or (ii) the available quantity of the book is 0.

Answer:

Actually, your existing code already handles the second condition (available quantity is 0) quite well! However, we can refine the logic to make the feedback to the user more specific and "unusual-condition-proof."

Here is how we can extend the issueBook() function to handle those two specific cases with clear, distinct error messages.

Updated `issueBook` Function

I have added a check to distinguish between a book that does not exist in the database versus a book that is out of stock.

void issueBook() {

    int id, found = 0;

    printf("\nEnter Book ID to issue: ");

    scanf("%d", &id);

    for (int index = 0; index < totalRecord; index++) {

        if (book[index].bookID == id) {

            found = 1; // Condition (i) solved: The book IS in the library

            // Condition (ii): Check if quantity is 0

            if (book[index].noOfCopies > 0) {

                book[index].noOfCopies--;

                printf("Success: Book '%s' issued! Remaining: %d\n",

                        book[index].title, book[index].noOfCopies);

            } else {

                // Unusual condition: Book exists, but quantity is 0

                printf("Error: The book '%s' is currently out of stock (Quantity: 0).\n",

                        book[index].title);

            break;

    // Condition (i): If the loop finishes and 'found' is still 0

    if (!found) {

        printf("Error: Book ID %d does not exist in our library records.\n", id);

🔍 Logic Flow Comparison

The table below explains how the code differentiates between your two requested scenarios:

Scenario	found variable	noOfCopies	Resulting Message
Book not in Library	Remains `0`	N/A	"Book ID does not exist"
Book exists but 0 copies	Becomes `1`	Is `0`	"Out of stock"
Normal Issue	Becomes `1`	Greater than `0`	"Book issued!"

💡 Pro-Tip for "Unusual Conditions"

In a real-world scenario, you might also want to handle input validation. For example, what happens if the user types a letter (like 'A') instead of a Book ID number?

To handle that "unusual condition," you can check the return value of scanf:

if (scanf("%d", &id) != 1) {

    printf("Invalid input! Please enter a numeric ID.\n");

    while(getchar() != '\n'); // Clear the input buffer

    return;

~~Would you like me to help you integrate a Search by Author feature, or perhaps show you how to delete a record entirely if a book is permanently removed from the library?~~

Example 9.12 Program in C to demonstrate dynamic memory allocation of structure variables.

(DYNAMIC MEMORY ALLOCATION FOR STRUCTURES)

Answer:

#include<stdio.h>

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info *ptr;

int studentCount;

printf( "\n How many students? " );

scanf( "%d", &studentCount );

ptr = (struct Student_Info*) malloc ( studentCount * sizeof(struct Student_Info) );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Enter roll for student %d ", index );

scanf(" %d", &(ptr+index)->roll );

printf( " \n Enter age for student %d ", index );

scanf( "%d", &(ptr+index)->age );

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Roll for student %d is %d", index, (ptr+index)->roll );

printf( " \n Age for student %d is %d", index, (ptr+index)->age );

}

return 0;

}

Meaning & Output by AI:

This C program dynamically allocates memory for an array of struct Student_Info structures based on user input for studentCount, then populates and prints the roll and age for each student using pointer arithmetic ((ptr+index)->member). It demonstrates proper use of malloc, structures, and pointers, though it lacks a free(ptr) call.

#include<stdio.h>

#include<stdlib.h> // Included for malloc

int main()

{

struct Student_Info

{

int roll;

int age;

};

struct Student_Info *ptr;

int studentCount;

printf( "\n How many students? " );

scanf( "%d", &studentCount );

// Dynamically allocate memory based on studentCount

ptr = (struct Student_Info*) malloc ( studentCount * sizeof(struct Student_Info) );

// Check if memory allocation was successful

if (ptr == NULL) {

printf("Memory allocation failed!");

return 1;

}

// Input data for each student

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Enter roll for student %d ", index );

scanf(" %d", &(ptr+index)->roll );

printf( " \n Enter age for student %d ", index );

scanf( "%d", &(ptr+index)->age );

}

// Output data for each student

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Roll for student %d is %d", index, (ptr+index)->roll );

printf( " \n Age for student %d is %d", index, (ptr+index)->age );

}

// Free allocated memory

free(ptr);

return 0;

}

Key Improvements and Considerations:

· #include<stdlib.h>: Added to use malloc and free.

· free(ptr): Added at the end to prevent memory leaks.

· Memory Check: Added if (ptr == NULL) to ensure malloc succeeded.

Output:

The provided C program dynamically allocates memory for a user-defined number of students, prompts for their roll numbers and ages, and then prints the entered details. Assuming the user inputs 2 students with details (Roll 101, Age 20) and (Roll 102, Age 21), the output will be:

text

How many students? 2

Enter roll for student 0 101

Enter age for student 0 20

Enter roll for student 1 102

Enter age for student 1 21

Entered details are as follows:

Roll for student 0 is 101

Age for student 0 is 20

Roll for student 1 is 102

Age for student 1 is 21

Note: The program is missing #include <stdlib.h> for malloc and does not call free(ptr) to release the allocated memory.

Quick activities

1. Extend the above program to include student names in the structure.

Answer:

To include student names, you need to add a character array (string) to your structure. I have also cleaned up the syntax slightly to make it more readable by using array notation (ptr[index]) instead of pointer arithmetic.

Here is the extended version of your program:

#include <stdio.h>

#include <stdlib.h> // Included for malloc and free

int main() {

struct Student_Info {

char name[50]; // Added field for student name

int roll;

int age;

};

struct Student_Info *ptr;

int studentCount;

printf("\n How many students? ");

scanf("%d", &studentCount);

// Dynamically allocate memory

ptr = (struct Student_Info*) malloc(studentCount * sizeof(struct Student_Info));

// Check if memory allocation was successful

if (ptr == NULL) {

printf("Memory allocation failed!\n");

return 1;

}

// Input data for each student

for (int index = 0; index < studentCount; index++) {

printf("\n--- Data for Student %d ---", index + 1);

printf("\n Enter Name: ");

scanf("%s", ptr[index].name); // Read string (no & needed for arrays)

printf(" Enter roll: ");

scanf("%d", &ptr[index].roll);

printf(" Enter age: ");

scanf("%d", &ptr[index].age);

}

// Output data for each student

printf("\n\n Entered details are as follows: \n");

printf("-----------------------------------------");

for (int index = 0; index < studentCount; index++) {

printf("\n Student %d: Name: %s, Roll: %d, Age: %d",

index + 1, ptr[index].name, ptr[index].roll, ptr[index].age);

}

printf("\n-----------------------------------------\n");

// Free allocated memory

free(ptr);

ptr = NULL;

return 0;

}

Key Changes Made:

char name[50];: Added a character array to the structure. This allocates 50 bytes per student for their name.
ptr[index].name: In the scanf for the name, we don't use the & symbol because the name of a character array already acts as a pointer to its first element.
Array Notation: I swapped (ptr+index)->roll for ptr[index].roll. Both work exactly the same way in C, but the latter is generally easier to read and less prone to typing errors.
Memory Layout: By adding the name array, each "block" in memory for a student is now larger. The sizeof(struct Student_Info) automatically calculates this new size for malloc.

~~Would you like to see how to handle names with spaces (like "John Doe") using~~ ~~fgets~~ ~~instead of~~ ~~scanf~~?

OUTPUT:

Based on the code provided, here is an example of what the execution would look like assuming you enter data for 2 students.

Sample Input/Output

text

How many students? 2

--- Data for Student 1 ---

Enter Name: Alice

Enter roll: 101

Enter age: 20

--- Data for Student 2 ---

Enter Name: Bob

Enter roll: 102

Enter age: 21

Entered details are as follows:

-----------------------------------------

Student 1: Name: Alice, Roll: 101, Age: 20

Student 2: Name: Bob, Roll: 102, Age: 21

-----------------------------------------

Explanation of Execution

1. "How many students?": You enter 2.

2. Memory Allocation: malloc allocates space for 2 struct Student_Info structures on the heap.

3. Loop 1 (Input):

1. Iterates 2 times.

2. scanf("%s", ...) reads the name (Note: %s only reads until the first space).

3. scanf("%d", ...) reads the integer roll and age.

4. Loop 2 (Output):

0. Iterates 2 times to print the data stored in ptr[0] and ptr[1].

5. free(ptr): The dynamically allocated memory is released.

2. Rewrite all the programs from Example 9.7 to 9.11 using dynamic memory allocation strategy.

Answer:

Example 9.7: A C program demonstrating an array of structure variables.

#include<stdio.h>

int main()

{

struct Student_Info

{

char name[30];

int roll;

int age;

};

struct Student_Info student[10];

int studentCount;

printf( "\n How many students? " );

scanf( "%d", &studentCount );

for ( int index = 0; index < studentCount; index++ )

{

printf( " \n Enter name for student %d ", index );

getchar();

gets( student[index].name );

printf( " \n Enter roll for student %d ", index );

scanf(" %d", &student[index].roll );

printf( " \n Enter age for student %d ", index );

scanf( "%d", &student[index].age );

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < studentCount; index++ )

{

printf("\n\n Student %d: ", index+1 );

printf("\n Name: ");

puts( student[index].name );

printf( " Roll: %d", student[index].roll );

printf( " \n Age: %d", student[index].age );

}

return 0;

}

Note: This is a solid foundational C program for managing student records using an array of structures. It covers the basics: defining a template, capturing user input, and iterating through the data to display it.

Rewrite the program Example 9.7 using dynamic memory allocation strategy. :

Switching to dynamic memory allocation is a great move. It makes your program much more flexible because you no longer have to hard-code a limit like student[10]. Instead, you can request exactly the amount of memory you need from the Heap at runtime.

Key Changes

Pointers: We replace the array student[10] with a pointer struct Student_Info *student.
malloc(): We use this function to allocate memory based on the user's input.
free(): Crucial step! Since the system doesn't automatically reclaim heap memory, we must manually release it before the program ends.
Arrow Operator (->): While you can still use student[index].name, if you were passing a single pointer to a function, you'd use the -> operator to access members.

The Optimized Code

#include <stdio.h>

#include <stdlib.h> // Required for malloc and free

struct Student_Info {

char name[30];

int roll;

int age;

};

int main() {

struct Student_Info *student; // Pointer to hold the base address

int studentCount;

printf("\nHow many students would you like to enter? ");

if (scanf("%d", &studentCount) != 1 || studentCount <= 0) {

printf("Please enter a valid number of students.\n");

return 1;

}

// Allocate memory dynamically

student = (struct Student_Info *)malloc(studentCount * sizeof(struct Student_Info));

// Check if memory allocation was successful

if (student == NULL) {

printf("Memory allocation failed! Exiting...\n");

return 1;

}

for (int index = 0; index < studentCount; index++) {

printf("\n--- Entering Data for Student %d ---", index + 1);

printf("\nName: ");

getchar(); // Clear newline buffer

fgets(student[index].name, sizeof(student[index].name), stdin);

printf("Roll: ");

scanf("%d", &student[index].roll);

printf("Age: ");

scanf("%d", &student[index].age);

}

printf("\n\n--- Displaying Records ---");

for (int index = 0; index < studentCount; index++) {

printf("\nStudent %d: %sRoll: %d | Age: %d\n",

index + 1,

student[index].name,

student[index].roll,

student[index].age);

}

// VERY IMPORTANT: Free the allocated memory

free(student);

printf("\nMemory successfully released.\n");

return 0;

}

Why use sizeof(struct Student_Info)?

When using malloc, you have to tell the computer exactly how many bytes you want. Since the size of a structure can vary depending on the operating system or compiler padding, we use sizeof to calculate it accurately.

Quick Tips for Dynamic Memory:

Always check for NULL: If you try to allocate more memory than the RAM has available, malloc returns NULL. Trying to write to a NULL pointer will cause a segmentation fault (crash).
The "Leak" Rule: For every malloc(), there should be a free(). If you forget to free(), your program will consume more and more memory over time—this is a "memory leak."

~~Would you like to see how to use~~ ~~realloc()~~ ~~to add more students to this list without restarting the program?~~

Output of the program:

Since I can't run the code on your physical machine, I’ve simulated a typical terminal session based on the dynamic memory version of the program.

In this scenario, we are entering 2 students. Notice how the program handles the name with spaces and the memory cleanup at the end.

Terminal Execution Simulation

Plaintext

How many students would you like to enter? 2

--- Entering Data for Student 1 ---

Name: Alan Turing

Roll: 101

Age: 41

--- Entering Data for Student 2 ---

Name: Ada Lovelace

Roll: 102

Age: 36

--- Displaying Records ---

Student 1: Alan Turing

Roll: 101 | Age: 41

Student 2: Ada Lovelace

Roll: 102 | Age: 36

Memory successfully released.

What's happening behind the scenes?

When you run this, the computer follows a specific workflow in its memory. Here is a breakdown of the "lifecycle" of your data during that output:

Allocation: When you type 2, the program calculates $2 \times \text{sizeof(Student\_Info)}$ and carves out that exact block in the heap.
Buffer Clearing: The getchar() absorbs the "Enter" key you pressed after typing 2, ensuring fgets doesn't immediately grab an empty newline.
Data Storage: The strings and integers are packed into the allocated heap block.
Cleanup: Once the for loop prints the data, free(student) tells the OS, "I'm done with this memory; others can use it now."

Common Output Issues to Watch For:

Trailing Newlines: You might notice a small gap after the names in the output. This is because fgets() captures the \n character when you hit Enter.
Garbage Values: If you entered "3" but only filled out "2" records, the third record would display "Garbage Values" (random symbols or numbers) because that memory wasn't initialized.

~~Would you like me to show you a quick trick to remove that annoying newline character from the names using~~ ~~strtok~~ ~~or a simple~~ if ~~statement?~~

Example 9.8 A C program for handling information about books.

#include<stdio.h>

int main()

{

struct Book_Info

{

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalBookCount;

printf( "\n How many books in total? " );

scanf( "%d", &totalBookCount );

for ( int index = 0; index < totalBookCount; index++ )

{

printf( " \n Enter title of the book %d ", index );

getchar();

gets(book[index].title);

printf( " \n Enter Book ID %d ", index );

scanf(" %d", &book[index].bookID );

printf( " \n Enter total no of copies %d ", index );

scanf(" %d", &book[index].noOfCopies );

printf( " \n Enter author of the book %d ", index );

getchar();

gets(book[index].author);

}

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < totalBookCount; index++ )

{

printf("\n Book %d \n", index);

printf(" Title: ");

puts( book[index].title );

printf( " ID: %d \n Available copies: %d ", book[index].bookID,

book[index].noOfCopies );

printf(" \n Author: ");

puts( book[index].author );

}

return 0;

}

Note by AI: It looks like you've put together a solid foundation for a book management system! You are using an array of structures, which is exactly the right approach for handling a collection of related data in C.

Rewrite the program of Example 9.8 using dynamic memory allocation strategy.

Switching to dynamic memory allocation is a smart move. In your previous code, you were limited to exactly 10 books. By using malloc(), you can allocate exactly the amount of memory you need at runtime based on the user's input.

In C, we use a pointer to our structure and then allocate a block of memory equivalent to $n \times \text{sizeof(struct)}$.

Memory Management Strategy

When you use malloc, you are grabbing space from the Heap rather than the Stack. This allows your program to handle much larger datasets than a standard local array might allow.

Dynamic Memory Implementation

#include <stdio.h>

#include <stdlib.h> // Required for malloc and free

#include <string.h>

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

int main() {

struct Book_Info *books; // Pointer to structure

int totalBookCount;

printf("\nHow many books would you like to enter? ");

if (scanf("%d", &totalBookCount) != 1 || totalBookCount <= 0) {

printf("Invalid input.\n");

return 1;

}

// 1. Allocate memory dynamically

books = (struct Book_Info *)malloc(totalBookCount * sizeof(struct Book_Info));

// Check if memory allocation was successful

if (books == NULL) {

printf("Memory allocation failed! Exiting...\n");

return 1;

}

while (getchar() != '\n'); // Clear buffer

// 2. Input data using pointer arithmetic or array notation

for (int i = 0; i < totalBookCount; i++) {

printf("\n--- Book %d ---", i + 1);

printf("\nEnter title: ");

fgets(books[i].title, 30, stdin);

books[i].title[strcspn(books[i].title, "\n")] = 0;

printf("Enter Book ID: ");

scanf("%d", &books[i].bookID);

printf("Enter total no of copies: ");

scanf("%d", &books[i].noOfCopies);

while (getchar() != '\n');

printf("Enter author: ");

fgets(books[i].author, 30, stdin);

books[i].author[strcspn(books[i].author, "\n")] = 0;

}

// 3. Display the details

printf("\n--- Library Records --- \n");

for (int i = 0; i < totalBookCount; i++) {

printf("\n[%d] %s by %s (ID: %d | Copies: %d)",

i + 1, books[i].title, books[i].author,

books[i].bookID, books[i].noOfCopies);

}

// 4. Crucial step: Free the allocated memory

free(books);

printf("\n\nMemory successfully released.\n");

return 0;

}

What changed?

The Pointer: Instead of book[10], we use struct Book_Info *books. This pointer stores the starting address of our allocated memory block.
The Allocation: malloc(totalBookCount * sizeof(struct Book_Info)) calculates the exact bytes needed.
The Null Check: It's a best practice to check if malloc returned NULL. If the system is out of memory, your program should fail gracefully rather than crashing.
The Cleanup: Unlike an array, heap memory isn't automatically cleaned up. We use free(books) to return the memory to the system once the program is done with it.

~~Would you like to see how to use~~ ~~realloc~~ ~~to add more books to this list later without restarting the program?~~

OUTPUT by AI:

This output assumes that chose to enter 2 books. Notice how the program handles the strings and integers smoothly without skipping any inputs.

Simulated Console Output

Plaintext

How many books would you like to enter? 2

--- Book 1 ---

Enter title: The Great Gatsby

Enter Book ID: 101

Enter total no of copies: 5

Enter author: F. Scott Fitzgerald

--- Book 2 ---

Enter title: 1984

Enter Book ID: 102

Enter total no of copies: 3

Enter author: George Orwell

--- Library Records ---

[1] The Great Gatsby by F. Scott Fitzgerald (ID: 101 | Copies: 5)

[2] 1984 by George Orwell (ID: 102 | Copies: 3)

Memory successfully released.

How the Memory Looks Under the Hood

When you run this dynamic version, the computer sets up the memory differently than a standard array. Instead of a fixed "box" of 10 slots, it creates exactly what you asked for on the Heap.

Why this output is "cleaner"

No Buffer Slop: By using while (getchar() != '\n');, we ensure that the "Enter" key pressed after typing 101 doesn't accidentally get swallowed by the "Author" field.
Formatted Strings: The fgets and strcspn combo ensures that the author's name doesn't have a random line break in the middle of your sentence.

~~Would you like me to add a feature to save this output to a~~ ~~.txt~~ ~~file so the data persists after you close the program?~~

Example 9.9 A C program for handling information about books with an issue operation.

#include<stdio.h>

struct Book_Info

{

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

void input()

{

for ( int index = 0; index < totalRecord; index++ )

{

printf( " \n Enter title of the book %d ", index );

gets(book[index].title);

printf( " \n Enter Book ID %d ", index );

scanf(" %d", &book[index].bookID );

printf( " \n Enter total no of copies %d ", index );

scanf(" %d", &book[index].noOfCopies );

printf( " \n Enter author of the book %d ", index );

getchar();

gets(book[index].author);

}

void output()

{

printf( "\n Entered details are as follows: \n \n " );

for ( int index = 0; index < totalRecord; index++ )

{

printf("\n Book %d \n\n", index);

printf(" Title: ");

puts(book[index].title);

printf( " ID: %d \n Remaining copies: %d \n", book[index].bookID,

book[index].noOfCopies );

printf(" Authors: ");

puts(book[index].author);

}

void issue()

{

int id;

printf("\n Enter the ID of the book to issue: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ )

{

if( book[index].bookID == id )

{

book[index].noOfCopies = book[index].noOfCopies - 1;

printf("\n One book issued with ID: %d", id);

output();

return;

}

int main()

{

printf( "\n How many books in total? " );

scanf( "%d", &totalRecord );

getchar();

input();

output();

issue();

return 0;

}

Note by AI on the above program:

This is a solid start for a library management program! You’ve correctly used an array of structures to store your data and implemented basic CRUD (Create, Read, Update) logic.

Rewrite the programs of 9.9 using dynamic memory allocation strategy.:

Switching to Dynamic Memory Allocation (DMA) is a smart move. It allows your program to be flexible: instead of being locked into a maximum of 10 books, you can request exactly the amount of memory you need at runtime.

🧠 The Logic Change

Instead of struct Book_Info book[10], we will use a pointer (struct Book_Info *book). We'll then use malloc() to reserve a block of memory on the Heap based on the user's input.

💻 Updated Code with DMA

#include <stdio.h>

#include <stdlib.h> // Required for malloc and free

#include <string.h>

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

// Function prototypes

void clearBuffer();

void input(struct Book_Info *book, int totalRecord);

void output(struct Book_Info *book, int totalRecord);

void issue(struct Book_Info *book, int totalRecord);

int main() {

int totalRecord;

struct Book_Info *book; // Pointer to hold the base address

printf("How many books would you like to register? ");

if (scanf("%d", &totalRecord) != 1) return 1;

clearBuffer();

// Allocate memory dynamically

book = (struct Book_Info *)malloc(totalRecord * sizeof(struct Book_Info));

// Check if memory allocation was successful

if (book == NULL) {

printf("Memory allocation failed! Exiting...\n");

return 1;

}

input(book, totalRecord);

output(book, totalRecord);

issue(book, totalRecord);

// CRITICAL: Always free dynamically allocated memory

free(book);

printf("\nMemory freed. Goodbye!\n");

return 0;

}

void clearBuffer() {

int c;

while ((c = getchar()) != '\n' && c != EOF);

}

void input(struct Book_Info *book, int totalRecord) {

for (int i = 0; i < totalRecord; i++) {

printf("\n--- Book %d ---\n", i + 1);

printf("Title: ");

fgets(book[i].title, 30, stdin);

book[i].title[strcspn(book[i].title, "\n")] = 0;

printf("ID: ");

scanf("%d", &book[i].bookID);

printf("Copies: ");

scanf("%d", &book[i].noOfCopies);

clearBuffer();

printf("Author: ");

fgets(book[i].author, 30, stdin);

book[i].author[strcspn(book[i].author, "\n")] = 0;

}

void output(struct Book_Info *book, int totalRecord) {

printf("\n%-10s %-20s %-20s %-10s\n", "ID", "Title", "Author", "Copies");

printf("------------------------------------------------------------\n");

for (int i = 0; i < totalRecord; i++) {

printf("%-10d %-20s %-20s %-10d\n",

book[i].bookID, book[i].title, book[i].author, book[i].noOfCopies);

}

void issue(struct Book_Info *book, int totalRecord) {

int id, found = 0;

printf("\nEnter ID to issue: ");

scanf("%d", &id);

for (int i = 0; i < totalRecord; i++) {

if (book[i].bookID == id) {

found = 1;

if (book[i].noOfCopies > 0) {

book[i].noOfCopies--;

printf("Success! Issued: %s\n", book[i].title);

} else {

printf("Error: No copies available.\n");

}

break;

}

if (!found) printf("ID not found.\n");

}

🔑 Key Concepts Used

malloc(): This allocates a block of memory of a specific size. We calculate the size using $totalRecord \times \text{sizeof(struct Book\_Info)}$.
The Null Check: Always check if (book == NULL). If the computer is out of RAM, malloc returns NULL, and trying to write to it will crash your system.
free(): This is the most important part of DMA. Unlike local variables that disappear when a function ends, heap memory stays until you manually release it. Forgetting this leads to memory leaks.

~~Would you like to see how to use~~ ~~realloc()~~ ~~so you can add more books later without restarting the program?~~

OUTPUT by AI:

Since I can't "run" the code in a live terminal for you, I’ve simulated exactly how the interaction would look on your screen based on the Dynamic Memory Allocation version of the program.

🖥️ Simulated Terminal Output

Plaintext

How many books would you like to register? 2

--- Book 1 ---

Title: The Great C Basics

ID: 101

Copies: 5

Author: Ritchie Dennis

--- Book 2 ---

Title: Data Structures 101

ID: 102

Copies: 2

Author: Alan Turing

ID Title Author Copies

------------------------------------------------------------

101 The Great C Basics Ritchie Dennis 5

102 Data Structures 101 Alan Turing 2

Enter ID to issue: 102

Success! Issued: Data Structures 101

Memory freed. Goodbye!

🔍 What happened behind the scenes?

Memory Allocation: The moment you typed 2, the program calculated that it needed space for two Book_Info structures and reserved that exact spot on the Heap.
Input Phase: It gathered your strings and integers. Notice that the titles and authors can have spaces (like "The Great C Basics") because we used fgets instead of scanf.
The "Issue" Update: When you issued ID 102, the program looked through the allocated memory, found the match, and decremented the noOfCopies for that specific memory address from 2 to 1.
Cleanup: Before the program closed, the free(book) command handed that memory back to your operating system.

⚠️ Common Errors to Watch For

If you try to run this and it behaves strangely, check these two things:

ID Mismatch: If you enter a character (like 'A') when it asks for an ID (integer), the scanf will fail and the program might enter an infinite loop.
Buffer Overflow: If you enter a title longer than 30 characters, fgets will safely cut it off, protecting your memory from corruption.

~~Would you like me to add a "Save to File" feature so your library data doesn't disappear every time you close the program?~~

Example 9.10 A C program for handling information about books with an issue operation.

Corrected Code:

#include<stdio.h>

#include<string.h> // Required for strcspn

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

// Function to safely remove newline from fgets

void stripNewline(char *str) {

str[strcspn(str, "\r\n")] = 0;

}

void input() {

for ( int index = 0; index < totalRecord; index++ ) {

printf( "\n Enter title of the book %d: ", index );

fgets(book[index].title, sizeof(book[index].title), stdin);

stripNewline(book[index].title);

printf( " Enter Book ID %d: ", index );

scanf("%d", &book[index].bookID );

printf( " Enter total no of copies %d: ", index );

scanf("%d", &book[index].noOfCopies );

getchar(); // Clear newline from scanf

printf( " Enter author of the book %d: ", index );

fgets(book[index].author, sizeof(book[index].author), stdin);

stripNewline(book[index].author);

}

void output() {

printf( "\n --- Entered details --- \n" );

for ( int index = 0; index < totalRecord; index++ ) {

printf("\n Book %d", index);

printf("\n Title: %s", book[index].title);

printf("\n ID: %d \n Remaining copies: %d", book[index].bookID, book[index].noOfCopies );

printf("\n Authors: %s\n", book[index].author);

}

void returnBook() {

int id;

printf("\n Enter the ID of the book to return: ");

scanf("%d", &id);

for ( int index = 0; index < totalRecord; index++ ) {

// Corrected comparison operator: ==

if( book[index].bookID == id ) {

book[index].noOfCopies++;

printf(" One book is returned with ID: %d\n", id);

output();

return;

}

printf("Book ID not found.\n");

}

int main() {

printf( "How many books in total? " );

scanf( "%d", &totalRecord );

getchar(); // Clear buffer

input();

output();

returnBook();

return 0;

}

Note by AI for the program:

This is a solid C program for a basic Library Management System! You've handled the classic "newline buffer" issue quite well by using getchar() after scanf and creating a custom stripNewline function.

Rewrite the program of 9.10 using dynamic memory allocation strategy.

Switching to Dynamic Memory Allocation is a great move. It makes your program much more flexible because you aren't locked into a fixed array size of 10. Instead, you can request exactly the amount of memory you need at runtime.

In C, this is primarily handled using malloc() (Memory Allocation) and free() to release that memory back to the system when you're done.

Key Changes

1. Pointer usage: We change struct Book_Info book[10] to a pointer: struct Book_Info *book.

2. Heap Allocation: We use malloc to reserve space based on user input.

3. Arrow Operator: When using a pointer to a struct, we use -> instead of . to access members (though in an array-indexed pointer like book[index].title, the dot operator still works).

The Dynamic Program

#include <stdio.h>

#include <stdlib.h> // Required for malloc and free

#include <string.h>

struct Book_Info {

    char title[30];

    int bookID;

    int noOfCopies;

    char author[30];

};

// Global pointer to hold our dynamic array

struct Book_Info *book = NULL;

int totalRecord;

void stripNewline(char *str) {

    str[strcspn(str, "\r\n")] = 0;

void input() {

    for (int index = 0; index < totalRecord; index++) {

        printf("\n--- Book %d ---\n", index + 1);

        printf("Enter title: ");

        fgets(book[index].title, sizeof(book[index].title), stdin);

        stripNewline(book[index].title);

        printf("Enter Book ID: ");

        scanf("%d", &book[index].bookID);

        printf("Enter total copies: ");

        scanf("%d", &book[index].noOfCopies);

        getchar(); // Clear newline

        printf("Enter author: ");

        fgets(book[index].author, sizeof(book[index].author), stdin);

        stripNewline(book[index].author);

void output() {

    printf("\n--- Library Records ---\n");

    for (int index = 0; index < totalRecord; index++) {

        printf("\n[%d] Title: %s | ID: %d | Copies: %d | Author: %s",

               index + 1, book[index].title, book[index].bookID,

               book[index].noOfCopies, book[index].author);

    printf("\n");

int main() {

    printf("How many books would you like to register? ");

    if (scanf("%d", &totalRecord) != 1 || totalRecord <= 0) {

        printf("Invalid input.\n");

        return 1;

    getchar();

    // Dynamic Allocation

    // We calculate size: (number of books) * (size of one struct)

    book = (struct Book_Info *)malloc(totalRecord * sizeof(struct Book_Info));

    // Check if memory was successfully allocated

    if (book == NULL) {

        printf("Memory allocation failed! Exiting...\n");

        return 1;

    input();

    output();

    // Clean up: Always free dynamic memory!

    free(book);

    book = NULL;

    printf("\nMemory freed. System closed.\n");

    return 0;

How Memory Changes

In your first version, the memory was allocated on the Stack (fixed and automatic). In this version, it's allocated on the Heap.

Why `malloc` is powerful here:

· Efficiency: If you only have 2 books, you only use 2 slots of memory.

· Scalability: If you have 10,000 books, you won't get a "Stack Overflow" error; you just use the available RAM.

· Calculated Size: The formula used is:

$$\text{Total Memory} = \text{totalRecord} \times \text{sizeof(struct Book\_Info)}$$

Important Note: Always remember to call free(book). If you don't, your program will have a memory leak, which means it occupies RAM even after it's finished running.

~~Would you like to see how to use~~ realloc ~~so you can add more books to the list later without restarting the program?~~

OUTPUT by AI:

Since I can't "run" the code in a live terminal for you, I have simulated the exact interaction based on the logic of the dynamic memory version.

If you were to input 2 books, here is exactly how your terminal screen would look:

Simulated Terminal Output

Plaintext

How many books would you like to register? 2

--- Book 1 ---

Enter title: The Great Gatsby

Enter Book ID: 101

Enter total copies: 5

Enter author: F. Scott Fitzgerald

--- Book 2 ---

Enter title: 1984

Enter Book ID: 102

Enter total copies: 3

Enter author: George Orwell

--- Library Records ---

[1] Title: The Great Gatsby | ID: 101 | Copies: 5 | Author: F. Scott Fitzgerald

[2] Title: 1984 | ID: 102 | Copies: 3 | Author: George Orwell

Memory freed. System closed.

What’s happening "Under the Hood"?

When you run this, the program performs specific operations on the system's memory. Because we used malloc, the operating system looks for a contiguous block of empty space in the Heap large enough to fit your structs.

Why this output is "clean":

Buffer Clearing: Notice how "Enter title" for Book 2 didn't get skipped? That's because the getchar() in your code successfully swallowed the \n (Enter key) hit after typing the number of copies for Book 1.
String Formatting: The stripNewline function ensures that when we print the records, the Author and Title stay on the same line rather than creating accidental line breaks.

~~Would you like to try adding a search function next so you can find a book's details by entering its ID?~~

Example 9.11 A C program for a mini library application.

#include<stdio.h>

#include<string.h>

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

struct Book_Info book[10];

int totalRecord;

void input() {

for (int index = 0; index < totalRecord; index++) {

printf("\n--- Book %d ---", index + 1);

printf("\nEnter title: ");

// Using %[^\n] reads the full title including spaces

scanf(" %29[^\n]", book[index].title);

printf("Enter Book ID: ");

scanf("%d", &book[index].bookID);

printf("Enter total copies: ");

scanf("%d", &book[index].noOfCopies);

printf("Enter author: ");

scanf(" %29[^\n]", book[index].author);

}

void displayAll() {

printf("\n--- Library Records ---");

for (int index = 0; index < totalRecord; index++) {

printf("\nID: %d | Title: %s | Copies: %d | Author: %s",

book[index].bookID, book[index].title,

book[index].noOfCopies, book[index].author);

}

printf("\n");

}

void issueBook() {

int id, found = 0;

printf("\nEnter Book ID to issue: ");

scanf("%d", &id);

for (int index = 0; index < totalRecord; index++) {

if (book[index].bookID == id) {

found = 1;

if (book[index].noOfCopies > 0) {

book[index].noOfCopies--;

printf("Book issued! Remaining: %d\n", book[index].noOfCopies);

} else {

printf("Error: Out of stock!\n");

}

break;

}

if (!found) printf("Error: Book ID not found.\n");

}

void returnBook() {

int id, found = 0;

printf("\nEnter Book ID to return: ");

scanf("%d", &id);

for (int index = 0; index < totalRecord; index++) {

if (book[index].bookID == id) {

book[index].noOfCopies++;

printf("Book returned! Total copies: %d\n", book[index].noOfCopies);

found = 1;

break;

}

if (!found) printf("Error: Book ID not found.\n");

}

int main() {

printf("How many books to register (max 10)? ");

scanf("%d", &totalRecord);

input();

int option;

while (1) {

printf("\n1: All | 2: Issue | 3: Return | 4: Exit\nChoice: ");

scanf("%d", &option);

if (option == 1) displayAll();

else if (option == 2) issueBook();

else if (option == 3) returnBook();

else if (option == 4) break;

else printf("Invalid choice!\n");

}

return 0;

}

Note by AI of the program:

This is a solid, functional C program for a basic Library Management System. You've handled the fundamental logic of C structures and basic file-less data management quite well.

Rewrite the program of Example 9.11 using dynamic memory allocation strategy.:

Switching to Dynamic Memory Allocation is a smart move. It transforms your program from one with a hard limit of 10 books to one that can handle as many books as your computer's RAM allows.

In C, this is primarily achieved using malloc() (Memory Allocation) and free() to give the memory back to the system when you're done.

🧠 How Dynamic Allocation Works

Instead of reserving a fixed block of 10 slots at compile time, we ask the computer for a specific block of memory at runtime based on the user's input.

💻 The Updated Program

#include <stdio.h>

#include <stdlib.h> // Required for malloc and free

#include <string.h>

struct Book_Info {

char title[30];

int bookID;

int noOfCopies;

char author[30];

};

// Global pointer instead of a fixed array

struct Book_Info *book = NULL;

int totalRecord;

void input() {

for (int index = 0; index < totalRecord; index++) {

printf("\n--- Book %d ---", index + 1);

printf("\nEnter title: ");

scanf(" %29[^\n]", book[index].title);

printf("Enter Book ID: ");

scanf("%d", &book[index].bookID);

printf("Enter total copies: ");

scanf("%d", &book[index].noOfCopies);

printf("Enter author: ");

scanf(" %29[^\n]", book[index].author);

}

void displayAll() {

printf("\n--- Library Records ---");

for (int index = 0; index < totalRecord; index++) {

printf("\nID: %d | Title: %s | Copies: %d | Author: %s",

book[index].bookID, book[index].title,

book[index].noOfCopies, book[index].author);

}

printf("\n");

}

int main() {

printf("How many books would you like to register? ");

if (scanf("%d", &totalRecord) != 1 || totalRecord <= 0) {

printf("Invalid number of books!\n");

return 1;

}

// Allocate memory dynamically

// Formula: (Number of items) * (Size of one item)

book = (struct Book_Info *)malloc(totalRecord * sizeof(struct Book_Info));

// Safety check: always ensure malloc didn't return NULL

if (book == NULL) {

printf("Memory allocation failed! Exiting...\n");

return 1;

}

input();

int option;

while (1) {

printf("\n1: All | 2: Exit\nChoice: ");

scanf("%d", &option);

if (option == 1) displayAll();

else if (option == 2) break;

else printf("Invalid choice!\n");

}

// IMPORTANT: Clean up the memory before the program ends

free(book);

printf("Memory freed. Goodbye!\n");

return 0;

}

🔑 Key Changes Explained

The Pointer: We changed struct Book_Info book[10] to struct Book_Info *book. This is now a "remote control" that can point to any size block of memory.
malloc: The line book = (struct Book_Info *)malloc(totalRecord * sizeof(struct Book_Info)); asks the OS for exactly enough space to hold the number of books the user requested.
The Safety Check: We check if (book == NULL). If your computer is out of memory, malloc fails; without this check, your program would crash.
free(book): This is the most important part of dynamic allocation. Unlike local arrays, dynamic memory isn't deleted automatically. If you don't free it, you get a memory leak.

⚡ Pro Tip: realloc

If you wanted to add more books later without restarting the program, you could use realloc(). It grows the existing block of memory while keeping your current data intact.

~~Would you like me to show you how to add a "Add New Book" feature that uses~~ ~~realloc~~ ~~to expand the list on the fly?~~

OUTPUT by AI:

Since I can't "run" the code in a live terminal for you, I have simulated the exact interaction based on the logic of the dynamic memory version.

Here is what the terminal would look like if you registered 2 books and then viewed the records.

🖥️ Simulated Terminal Output

Plaintext

How many books would you like to register? 2

--- Book 1 ---

Enter title: The C Programming Language

Enter Book ID: 101

Enter total copies: 5

Enter author: Dennis Ritchie

--- Book 2 ---

Enter title: Clean Code

Enter Book ID: 102

Enter total copies: 3

Enter author: Robert Martin

1: All | 2: Exit

Choice: 1

--- Library Records ---

ID: 101 | Title: The C Programming Language | Copies: 5 | Author: Dennis Ritchie

ID: 102 | Title: Clean Code | Copies: 3 | Author: Robert Martin

1: All | 2: Exit

Choice: 2

Memory freed. Goodbye!

🔍 What’s Happening Behind the Scenes?

When you run this, the program follows a specific lifecycle in your computer's RAM. Understanding the "Heap" versus the "Stack" is key to mastering dynamic allocation.

Allocation Phase: When you enter 2, the program calculates:

$2 \times \text{sizeof(Book\_Info)}$ bytes. It finds a gap in the Heap memory large enough to fit that data.

Input Phase: The scanf functions write directly into that Heap space.
Access Phase: Even though book is a pointer, you can still use book[index] syntax. C treats pointers and arrays almost identically when accessing offsets.
Cleanup Phase: The free(book) command tells the OS, "I'm done with this specific block," making it available for other programs (like your browser or music player).

🛠️ Common Output Errors to Watch For

Buffer Overrun: If you try to enter a title longer than 29 characters, the %29[^\n] will stop reading to prevent a crash.
Input Mismatch: If you type "Two" instead of 2 for the book count, the program will trigger the "Invalid number of books!" error we added.

~~Would you like me to add a sorting feature so the output displays the books in alphabetical order by title?~~

Exercise

1. Define structure in the context of a programming language. How is it different from an array?

Answer:

A structure (often called a struct) in programming is a composite data type used to group a collection of related data items of potentially different data types under a single name [1, 2]. These items, known as members or fields, allow for the modeling of real-world entities that have multiple, varied properties.

A structure differs from an array in the following key ways:

Feature	Structure (Struct)	Array
Data Types	Stores elements of different (heterogeneous) data types (e.g., an integer id, a string name, a float salary) [1].	Stores elements of the same (homogeneous) data type (e.g., an array of only integers, or only strings) [3].
Accessing Elements	Elements are accessed by their unique member names using the dot (.) or arrow (->) operators [1].	Elements are accessed by their index (position) using square brackets ([]) [3].
Purpose	Used to create a complex, single unit representing an entity (e.g., a Student record) [1].	Used to store a collection of similar data items in a linear sequence (e.g., a list of all student IDs) [3].

In summary, a structure is designed for combining different types of data that logically belong together to describe a single object, while an array is designed for collecting multiple items of the exact same type in an ordered list [2, 3].

2. Is structure a built-in data type? Can we apply basic arithmetic operations such as addition, subtraction to structure variables? Show with a simple C program.

Answer:

Structure is a user-defined data type in C, not a built-in one. This means the programmer must define its layout and members [1].

Basic arithmetic operations such as addition and subtraction cannot be directly applied to structure variables as a whole. You must perform these operations on the individual members of the structure. The language does not inherently understand how to add or subtract two entire composite structures [1, 2].

Example C Program

The following program demonstrates how to work with structure members and why direct operations fail.

#include <stdio.h>

// Define a structure to represent a point in 2D space

struct Point {

int x;

int y;

};

int main() {

// Declare and initialize structure variables

struct Point p1 = {10, 20};

struct Point p2 = {5, 8};

struct Point sum;

// --- Applying arithmetic operations to individual members ---

// We can add the 'x' members and 'y' members separately

sum.x = p1.x + p2.x;

sum.y = p1.y + p2.y;

printf("Point 1: (%d, %d)\n", p1.x, p1.y);

printf("Point 2: (%d, %d)\n", p2.x, p2.y);

printf("Sum of points: (%d, %d)\n", sum.x, sum.y); // Output: Sum of points: (15, 28)

// --- Attempting direct arithmetic operation on structure variables (will cause a compilation error) ---

// If you uncomment the line below, the program will not compile because

// the '+' operator is not defined for 'struct Point' types.

// sum = p1 + p2; // Error: invalid operands to binary + (have 'struct Point' and 'struct Point')

return 0;

}

Explanation:

· The line sum.x = p1.x + p2.x; is valid because p1.x and p2.x are integers (built-in types), for which addition is defined.

· The commented-out line sum = p1 + p2; is invalid and results in a compilation error because C does not know how to add two struct Point variables automatically [2].

If you need to treat structures as single units for operations, you would typically write a custom function to handle the logic, or in C++, you could overload operators.

3. Identify the members of the structure from the below code segment.

struct Account

{

char acNo[15] ;

char ifsc[15];

char acType[7];

double balance;

double minBalance;

};

struct Account account1, account2, account3, account[10];

Answer:

The members of the struct Account are defined within the braces and represent the data components of the structure.

Members:

· char acNo[15] (Account Number)

· char ifsc[15] (IFSC Code)

· char acType[7] (Account Type)

· double balance (Account Balance)

· double minBalance (Minimum Balance)

These members represent the data type's internal components, while account1, account2, account3, and account[10] are instances (variables) of this structure.

4. Identify the structure variables from the below code segment.

struct Account

{

char acNo[15] ;

char ifsc[15];

char acType[7];

double balance;

double minBalance;

};

struct Account account1, account2, account3, account[10];

Answer:

The structure variables from the provided code segment are the identifiers declared after the structure definition, which are instances of the struct Account type.

The structure variables are:

· account1

· account2

· account3

· account (this is an array of 10 struct Account variables)

5. Consider the structure below and write statements for the following.

a. to declare a variable of the structure

b. to display the age of the teacher

struct Teacher

{

char name[30] ;

int age;

};

Answer:

Based on the struct Teacher definition, here are the requested statements in C/C++:

a. To declare a variable of the structure:

struct Teacher t1;

(This creates a variable named t1 of type struct Teacher.)

b. To display the age of the teacher:

printf("Age: %d", t1.age);

(This uses the dot operator (.) to access the age member of the t1 variable.)

Question No. 5. Consider the structure below and write statements for the following.

a. to declare a variable of the structure

b. to display the age of the teacher

struct Teacher

{

char name[30] ;

int age;

};

Answer:

Based on the struct Teacher definition, here are the requested statements in C/C++:

a. To declare a variable of the structure:

struct Teacher t1;

(This creates a variable named t1 of type struct Teacher.)

b. To display the age of the teacher:

printf("Age: %d", t1.age);

(This uses the dot operator (.) to access the age member of the t1 variable.)

6. Declare a pointer for structure Teacher (from Question No. 5.) and dynamically allocate memory for 10 records.

Answer:

To declare a pointer for the Teacher structure and dynamically allocate memory for 10 records, you can use the malloc function (in C) or the new operator (in C++).

1. Using C (with malloc)

In C, you must include <stdlib.h> and use malloc to request memory from the heap.

// Declare the pointer

struct Teacher *ptr;

// Allocate memory for 10 Teacher structures

ptr = (struct Teacher*) malloc(10 * sizeof(struct Teacher));

2. Using C++ (with new)

In C++, the process is more streamlined as the new operator automatically handles the size calculations.

C++

// Declare and allocate memory for 10 Teacher objects

Teacher *ptr = new Teacher[10];

Memory Visualization

When you allocate memory for 10 records, the pointer ptr points to the start of a contiguous block of memory large enough to hold 10 individual Teacher structures.

Key Points to Remember:

Accessing Data: Since ptr is a pointer, you access members using the arrow operator (->) for a single record (e.g., ptr->age) or array indexing (e.g., ptr[i].age).
Memory Management: Always free the memory once you are done to avoid memory leaks:

In C: free(ptr);
In C++: delete[] ptr;

~~Would you like to see a code example of how to use a loop to input names and ages into these 10 allocated records?~~

7. Consider the structure below and write statements for the following.

a. to declare a pointer for the above structure and display the salary.

b. to declare a single pointer for two different variables of the higher structure and display the details of the employee whose salary is more.

struct Employee

{

char name[30];

double salary;

};

Answer:

Here are the requested C language statements for the struct Employee.

a. To declare a pointer for the above structure and display the salary:

struct Employee emp1 = {"John Doe", 55000.50};

struct Employee* empPtr = &emp1; // Declare and initialize pointer

printf("Salary: %.2f\n", empPtr->salary); // Display salary using -> operator

b. To declare a single pointer for two different variables of the higher structure and display the details of the employee whose salary is more:

struct Employee emp1 = {"Alice", 60000.0};

struct Employee emp2 = {"Bob", 75000.0};

struct Employee* maxPtr;

// Use pointer to compare and point to the higher salary

maxPtr = (emp1.salary > emp2.salary) ? &emp1 : &emp2;

// Display details of the higher paid employee

printf("Higher Salary Employee: %s, Salary: %.2f\n", maxPtr->name, maxPtr->salary);

``` [3]

Question No. 7. Consider the structure below and write statements for the following.

a. to declare a pointer for the above structure and display the salary.

b. to declare a single pointer for two different variables of the higher structure and display the details of the employee whose salary is more.

struct Employee

{

char name[30];

double salary;

};

8. Rewrite the program of Question No. 7 to facilitate dynamic memory allocation for N number of record where N in a user input.

Answer:

To handle a dynamic number of records based on user input, we use functions like malloc() (from stdlib.h). This allows us to allocate a block of memory on the heap that acts like an array, but with a size determined at runtime.

Dynamic Memory Allocation for N Records

Here is the complete program to achieve this:

#include <stdio.h>

#include <stdlib.h> // Required for malloc and free

struct Employee {

char name[30];

double salary;

};

int main() {

int n, i;

struct Employee *ptr;

printf("Enter the number of records (N): ");

scanf("%d", &n);

// Dynamic memory allocation for N records

ptr = (struct Employee*) malloc(n * sizeof(struct Employee));

// Check if memory allocation was successful

if (ptr == NULL) {

printf("Memory not allocated. Exiting...");

return 1;

}

// Input details for N employees

for (i = 0; i < n; i++) {

printf("\nEnter details for Employee %d:\n", i + 1);

printf("Name: ");

scanf("%s", (ptr + i)->name);

printf("Salary: ");

scanf("%lf", &(ptr + i)->salary);

}

// Displaying the records using the pointer

printf("\n--- Employee Records ---\n");

for (i = 0; i < n; i++) {

printf("Employee %d: %s, Salary: %.2f\n", i + 1, (ptr + i)->name, (ptr + i)->salary);

}

// Deallocate the memory

free(ptr);

return 0;

}

Key Concepts Used

malloc(n * sizeof(struct Employee)): This calculates the total bytes required for N structures and reserves that space in the memory.
Pointer Arithmetic: We use (ptr + i) to navigate to the ith record.
The -> Operator: Used to access members when dealing with a pointer to a structure.
free(ptr): This is a crucial step in dynamic memory management to prevent memory leaks by releasing the memory back to the system once the program is done with it.

~~Would you like me to show you how to rewrite this using~~ ~~calloc()~~ ~~to automatically initialize the salaries to zero?~~

OUTPUT OF THE PROGRAM:

To show the output, let’s assume the user inputs 2 for the number of records. Here is how the console interaction and output would look:

Sample Program Execution

Plaintext

Enter the number of records (N): 2

Enter details for Employee 1:

Name: Alice

Salary: 65000.50

Enter details for Employee 2:

Name: Bob

Salary: 42000.75

--- Employee Records ---

Employee 1: Alice, Salary: 65000.50

Employee 2: Bob, Salary: 42000.75

Understanding the Memory Layout

When you run this program, the computer organizes the data in the Heap section of the memory. Unlike a standard array where the size is fixed at the start, the pointer ptr points to a flexible block of memory that grows based on your input $N$.

Key Observations in the Output:

Precision: The salary is displayed with two decimal places because of the %.2f format specifier.
Indexing: Even though C uses 0-based indexing (0 and 1), the output uses i + 1 to make it more readable for the user (Employee 1 and 2).
Sequential Storage: The records are stored contiguously in memory. If ptr points to address 1000, the second record starts exactly sizeof(struct Employee) bytes later.

~~Would you like me to show you how to add a search feature to this program so you can find a specific employee by name?~~

9. Consider the below structure and design a simple banking system that supports the following operations.

a. opening of accounts (Hint: input to the variables)

b. displaying details based on account number

c. displaying all account details

d. displaying details of all accounts whose balance is more than 1000

e. depositing an amount to an account

f. withdrawing some amount from an account

struct Account

{

char acNo[15] ;

char ifsc[15];

char acType[7];

double balance;

double minBalance;

};

Answer:

Here is a C program that designs a simple banking system supporting all the requested operations using an array of the provided struct Account.

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

// Define the Account structure

struct Account {

char acNo[15];

char ifsc[15];

char acType[7];

double balance;

double minBalance;

};

// Global array to store accounts and a counter

struct Account accounts[100]; // Supports up to 100 accounts

int numAccounts = 0;

// Function prototypes

void openAccount();

void displayDetailsByAcNo();

void displayAllAccountDetails();

void displayAccountsAbove1000();

void depositAmount();

void withdrawAmount();

int findAccountIndex(char acNo[]);

int main() {

int choice;

do {

printf("\n*** Simple Banking System Menu ***\n");

printf("1. Open an account\n");

printf("2. Display details based on account number\n");

printf("3. Display all account details\n");

printf("4. Display accounts with balance more than 1000\n");

printf("5. Deposit an amount\n");

printf("6. Withdraw some amount\n");

printf("7. Exit\n");

printf("Enter your choice: ");

scanf("%d", &choice);

switch (choice) {

case 1:

openAccount();

break;

case 2:

displayDetailsByAcNo();

break;

case 3:

displayAllAccountDetails();

break;

case 4:

displayAccountsAbove1000();

break;

case 5:

depositAmount();

break;

case 6:

withdrawAmount();

break;

case 7:

printf("Exiting program. Goodbye!\n");

break;

default:

printf("Invalid choice. Please try again.\n");

}

} while (choice != 7);

return 0;

}

// Helper function to find the index of an account by its account number

int findAccountIndex(char acNo[]) {

for (int i = 0; i < numAccounts; i++) {

if (strcmp(accounts[i].acNo, acNo) == 0) {

return i; // Account found, return its index

}

return -1; // Account not found

}

// a. Opening of accounts

void openAccount() {

if (numAccounts < 100) {

printf("Enter Account Number: ");

scanf("%s", accounts[numAccounts].acNo);

printf("Enter IFSC Code: ");

scanf("%s", accounts[numAccounts].ifsc);

printf("Enter Account Type (e.g., Sav, Cur): ");

scanf("%s", accounts[numAccounts].acType);

printf("Enter Initial Balance: ");

scanf("%lf", &accounts[numAccounts].balance);

accounts[numAccounts].minBalance = 1000.0; // Set minimum balance requirement

printf("Account opened successfully!\n");

numAccounts++;

} else {

printf("Cannot open more accounts. System limit reached.\n");

}

// Helper function to print account details

void printAccountDetails(struct Account acc) {

printf("\nAccount Details:\n");

printf("Account Number: %s\n", acc.acNo);

printf("IFSC Code: %s\n", acc.ifsc);

printf("Account Type: %s\n", acc.acType);

printf("Balance: %.2lf\n", acc.balance);

printf("Minimum Balance: %.2lf\n", acc.minBalance);

}

// b. Displaying details based on account number

void displayDetailsByAcNo() {

char acNo[15];

printf("Enter account number to display details: ");

scanf("%s", acNo);

int index = findAccountIndex(acNo);

if (index != -1) {

printAccountDetails(accounts[index]);

} else {

printf("Account not found.\n");

}

// c. Displaying all account details

void displayAllAccountDetails() {

if (numAccounts > 0) {

printf("\n*** All Account Details ***\n");

for (int i = 0; i < numAccounts; i++) {

printAccountDetails(accounts[i]);

printf("---------------------------\n");

}

} else {

printf("No accounts to display.\n");

}

// d. Displaying details of all accounts whose balance is more than 1000

void displayAccountsAbove1000() {

int count = 0;

printf("\n*** Accounts with Balance > 1000 ***\n");

for (int i = 0; i < numAccounts; i++) {

if (accounts[i].balance > 1000.0) {

printAccountDetails(accounts[i]);

printf("---------------------------\n");

count++;

}

if (count == 0) {

printf("No accounts found with balance more than 1000.\n");

}

// e. Depositing an amount to an account

void depositAmount() {

char acNo[15];

double amount;

printf("Enter account number to deposit into: ");

scanf("%s", acNo);

int index = findAccountIndex(acNo);

if (index != -1) {

printf("Enter amount to deposit: ");

scanf("%lf", &amount);

if (amount > 0) {

accounts[index].balance += amount;

printf("Amount deposited successfully. New balance is %.2lf.\n", accounts[index].balance);

} else {

printf("Invalid amount. Must be positive.\n");

}

} else {

printf("Account not found.\n");

}

// f. Withdrawing some amount from an account

void withdrawAmount() {

char acNo[15];

double amount;

printf("Enter account number to withdraw from: ");

scanf("%s", acNo);

int index = findAccountIndex(acNo);

if (index != -1) {

printf("Enter amount to withdraw: ");

scanf("%lf", &amount);

// Basic validation for positive amount and sufficient balance

if (amount > 0 && accounts[index].balance - amount >= accounts[index].minBalance) {

accounts[index].balance -= amount;

printf("Amount withdrawn successfully. New balance is %.2lf.\n", accounts[index].balance);

} else if (amount <= 0) {

printf("Invalid amount. Must be positive.\n");

} else {

printf("Insufficient funds or minimum balance requirement not met (min: %.2lf).\n", accounts[index].minBalance);

}

} else {

printf("Account not found.\n");

}

The code provides a complete, menu-driven C program to perform all the specified banking operations, managing the accounts in a global array of the defined struct Account.

This system stores all data in memory and will reset upon program termination.

Would you like me to show you how to save account data to a file so it persists between runs? Shall we implement file handling next?

Meaning & OUTPUT:

This C program is a well-structured implementation of a Simple Banking System. It uses a struct to represent bank accounts and an array to manage them.

Here is a breakdown of how the code works and a few suggestions to make it even more robust.

1. Code Architecture

The program is organized into three main layers:

1. Data Structure: The struct Account stores the core data (Account Number, IFSC, Type, Balance, and Min Balance).

2. Helper Functions: findAccountIndex and printAccountDetails are used to keep the main logic clean and avoid repeating code.

3. Core Features: Functions for opening, searching, listing, depositing, and withdrawing.

2. Key Components Explained

Feature	Logic Used
Account Storage	Uses a global array `accounts[100]` and a counter `numAccounts` to track the current index.
Searching	Uses `strcmp()` from `<string.h>` to compare the user input string with the stored account numbers.
Withdrawal Rule	Enforces a minimum balance check: `balance - amount >= minBalance`.
Validation	Checks if the amount is positive for both deposits and withdrawals.

3. Potential Improvements

While the code is functionally sound, here are a few "pro-tips" for a C programmer:

· Buffer Overflow Protection: Instead of scanf("%s", ...);, use scanf("%14s", ...);. This prevents the user from entering more characters than the array can hold (which causes crashes).

· Case Sensitivity: strcmp is case-sensitive. If a user enters "sav" but the account was opened as "Sav", it might cause confusion. You could use strcasecmp (on Linux) or convert input to lowercase.

· Data Persistence: Currently, all data is lost when the program closes. To make this a "real" app, you would eventually want to use File I/O (fopen, fwrite, fread) to save the accounts to a .dat or .txt file.

4. Quick Fix for a Small Syntax Error

In the code you provided, there are a couple of spacing issues likely caused by copy-pasting (e.g., case1:, case2:, and elseif). Make sure there is a space:

// Change this:

case1:

// To this:

case 1:

// And change this:

elseif (amount <= 0)

// To this:

else if (amount <= 0)

~~Would you like me to show you how to add File I/O so that your account data stays saved even after you close the program?~~

OUTPUT (by AI):

Since I cannot run the code directly on your machine, I have simulated the program's execution based on the logic provided. This demonstrates a typical "session" where an account is opened, a deposit is made, and the details are viewed.

Simulated Program Output

Plaintext

*** Simple Banking System Menu ***

1. Open an account

2. Display details based on account number

3. Display all account details

4. Display accounts with balance more than 1000

5. Deposit an amount

6. Withdraw some amount

7. Exit

Enter your choice: 1

Enter Account Number: 101ABC

Enter IFSC Code: BANK0001

Enter Account Type (e.g., Sav, Cur): Sav

Enter Initial Balance: 1500

Account opened successfully!

*** Simple Banking System Menu ***

1. Open an account

2. ... (other options)

Enter your choice: 5

Enter account number to deposit into: 101ABC

Enter amount to deposit: 500

Amount deposited successfully. New balance is 2000.00.

*** Simple Banking System Menu ***

1. Open an account

2. ... (other options)

Enter your choice: 2

Enter account number to display details: 101ABC

Account Details:

Account Number: 101ABC

IFSC Code: BANK0001

Account Type: Sav

Balance: 2000.00

Minimum Balance: 1000.00

*** Simple Banking System Menu ***

1. Open an account

...

Enter your choice: 7

Exiting program. Goodbye!

How the Data is Managed (Memory View)

When you run this program, the computer allocates a block of memory for the accounts array. Here is a conceptual look at how the data is organized in the system's memory after the session above:

Index	acNo	ifsc	balance	numAccounts (Counter)
0	"101ABC"	"BANK0001"	2000.00	1
1	(Empty)	(Empty)	0.00
...	...	...	...

Logic Check: The Withdrawal Limit

If you were to try and withdraw 1500 from the account above, the program would stop you:

Current Balance: 2000.00
Withdrawal Request: 1500.00
Calculation: $2000 - 1500 = 500$
Result: Since 500 is less than the Minimum Balance (1000), the withdrawAmount() function will trigger the else block and deny the transaction.

~~Would you like me to modify the code to include a Transfer function so you can move money between two different accounts?~~

PODMESWAR HELP SECTOR

Pages

Chapter - 9 Structure in C, Computer Science, Class 10 /Class ten, ASSEB (SEBA), Prepared by Podmeswar through AI

Updated `issueBook` Function

🔍 Logic Flow Comparison

💡 Pro-Tip for "Unusual Conditions"

Key Changes

The Dynamic Program

How Memory Changes

Why `malloc` is powerful here:

1. Code Architecture

2. Key Components Explained

3. Potential Improvements

4. Quick Fix for a Small Syntax Error

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Pages

Chapter - 9 Structure in C, Computer Science, Class 10 /Class ten, ASSEB (SEBA), Prepared by Podmeswar through AI

Updated issueBook Function

🔍 Logic Flow Comparison

💡 Pro-Tip for "Unusual Conditions"

Key Changes

The Dynamic Program

How Memory Changes

Why malloc is powerful here:

1. Code Architecture

2. Key Components Explained

3. Potential Improvements

4. Quick Fix for a Small Syntax Error

Updated `issueBook` Function

Why `malloc` is powerful here: