i) x= a*b%c/d
ii) y = a*((b%c)/d)
iii) z=a*b+a*(c-d)
Solution:
In this program, read the values of a, b, c, d as an input and compute the result of above expression.
#include<stdio.h>
int main()
{
int a,b,c,d;
float x=0,y=0,z=0;
// If we do not assign zero then in some cases compiler might be taken garbage value with result.
printf("Enter Value for a = ");
scanf("%d",&a);
printf("Enter Value for b = ");
scanf("%d",&b);
printf("Enter Value for c = ");
scanf("%d",&c);
printf("Enter Value for d = ");
scanf("%d",&d);
x=a*b%c/d; // Expression-1 for accurate result we can do type-casting here
y=a*((b%c)/d); // Expression-2
z=a*b+a*(c-d); // Expression-3
printf("\nResult of Expression-1 = %.2f\n",x);
printf("Result of Expression-2 = %.3f\n",y);
printf("Result of Expression-3 = %.2f\n",z);
return 0;
}
Output:
Note - Above output is obtained without type-casting the variables in the expression.
Above output is obtained after type-casting the variables in an expression. After type-casting the variables, expression is written as:
x = a*b%c/(float)d;
y = a*((b%c)/(float)d);
