Average of Marks in C

Algorithm

1. Read the number of subjects n; reject values below 1.
2. Read n marks and add them to a running total.
3. Divide the total by n, using floating-point division.
4. Print the average with two decimal places.

C program

#include <stdio.h>

int main(void) {
    int n, i, marks;
    int total = 0;
    double average;

    printf("Enter the number of subjects:\n");
    if (scanf("%d", &n) != 1 || n < 1) {
        printf("Error: enter at least one subject\n");
        return 1;
    }

    printf("Enter the marks of each subject:\n");
    for (i = 1; i <= n; i++) {
        if (scanf("%d", &marks) != 1) {
            printf("Error: invalid marks\n");
            return 1;
        }
        total += marks;
    }

    average = (double)total / n;
    printf("Average marks: %.2f\n", average);
    return 0;
}

Sample input and output

3
60 75 90

Enter the number of subjects:
Enter the marks of each subject:
Average marks: 75.00

2
0 0

Enter the number of subjects:
Enter the marks of each subject:
Average marks: 0.00

How the program works

The total is an int, but the average is computed as (double)total / n. The cast forces floating-point division, so an average such as 76.67 is not truncated to 76.

The running total avoids storing every mark, so the program needs only one variable for the sum.

Viva questions

Common mistakes

• Dividing two ints and printing the truncated result.
• Forgetting to reject n = 0, which divides by zero.