C Exercises: Calculate the total, percentage and division to take marks of three subjects
C Conditional Statement: Exercise-12 with Solution
Write a C program to read roll no, name and marks of three subjects and calculate the total, percentage and division.
Sample Solution:
C Code:
#include <stdio.h>
#include <string.h>
void main()
{
int rl,phy,che,ca,total;
float per;
char nm[20],div[10];
printf("Input the Roll Number of the student :");
scanf("%d",&rl);
printf("Input the Name of the Student :");
scanf("%s",nm);
printf("Input the marks of Physics, Chemistry and Computer Application : ");
scanf("%d%d%d",&phy,&che,&ca);
total = phy+che+ca;
per = total/3.0;
if (per>=60)
strcpy(div,"First");
else
if (per<60&&per>=48)
strcpy(div,"Second");
else
if (per<48&&per>=36)
strcpy(div,"Pass");
else
strcpy(div,"Fail");
printf("\nRoll No : %d\nName of Student : %s\n",rl,nm);
printf("Marks in Physics : %d\nMarks in Chemistry : %d\nMarks in Computer Application : %d\n",phy,che,ca);
printf("Total Marks = %d\nPercentage = %5.2f\nDivision = %s\n",total,per,div);
}
Sample Output:
Input the Roll Number of the student :784 Input the Name of the Student :James Input the marks of Physics, Chemistry and Computer Application : 70 80 90 Roll No : 784 Name of Student : James Marks in Physics : 70 Marks in Chemistry : 80 Marks in Computer Application : 90 Total Marks = 240 Percentage = 80.00 Division = First
Flowchart:

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C Programming: Tips of the Day
Why do C and C++ compilers allow array lengths in function signatures when they're never enforced?
It is a quirk of the syntax for passing arrays to functions.
Actually it is not possible to pass an array in C. If you write syntax that looks like it should pass the array, what actually happens is that a pointer to the first element of the array is passed instead.
Since the pointer does not include any length information, the contents of your [] in the function formal parameter list are actually ignored.
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