Develop a C program to evaluate enlisted expression

(1) b*c/d-2+b/c (where b=4,c=2,d=2)

#include <stdio.h>
int main()
{
    int b=4, c=2, d=2;
    printf("%d\n",b*c/d-2+b/c);
    return 0;
}

Output

(2) 20.0-(5.5/(0.5*10.0))+15.0

#include <stdio.h>
int main()
{
    printf("%f\n",20.0-(5.5/(0.5*10.0))+15.0);
    return 0;
}

Output

(3) (++a) - (--b) * (++a) / (b++) where a=5 and b=5

#include <stdio.h>
int main()
{
    int a=5, b=5, result;
    result = (++a) - (--b) * (++a) / (b++);
    printf("%d\n", result);
    return 0;
}

Output

Try to understand order of evaluation

PracticalServer