What will be the output? And why?

Question

#include<stdio.h>
#define square(x) (x * x)
int main()
{
int x, y=2;

x= square(y+1);
printf("%d\n",x);

return 0;

}

Answer 1

The output will be 5

The line x = square(y + 1);

Gers replaced by "square" macro:

x = y + 1 * y + 1;

=> x = y + y + 1; => x = 2 + 2 + 1; ≠> x = 5

If yoi want square(x) gives the "right" answe, then you must add ( ) like this:

#define square(x) (x)*(x)

Now the line will be replaced by:

x = (y + 1) * (y + 1);

Answer 2

Given that square (x) which considered as (y+1) and macro has (x*x)
Since we know x=y+1;
(X*X) means ( y+1*y+1 ) which is not equal to (y+1)*(y+1) ; so answer will be y+1*y+1=y+y+1;
Given:y=2;
Then 2y+1 gives [2*2+1=5]