someone plz explain how this is resulting in 3

Question

#include <stdio.h>
#define square(x)(x*x)
int main() {
int x,y=1;
x=square(y+1);
printf("%d\n",x);
return 0;
}

Answer 1

square(y+1) becomes (y+1*y+1)
And that becomes 1+1*1+1 which is 3.

So y+1 isnt evaluated but just placed in the spots where x stands.

Answer 2

thanks

Answer 3

https://www.sololearn.com/discuss/1830715/?ref=app

Answer 4

It goes like this
1. Substituting in the function.
2. Substituting the values.
3. evaluation.
So according to the operator precedence first multiplication will be done and then addition.