#include <stdio.h> #define sqr(x) x*x int main() { printf("%d",sqr(3+1)); return 0; }

Question

Can any one explain this C program?

Answer 1

Output: 7

Here's how:

#define sqr(x) x*x

Here value of x = (3+1) the thing we are passing on sqrt() as parameters.

x = 3+1
So it will executed like this:
x*x
3+1*3+1
3+3+1 //using operator precedence..
7

I hope u understand
Happy Coding 💕💕

Answer 2

Thank you