What is the output of below code and why?

Question

#include <stdio.h>
#define square(x) x*x
int main() {
printf("%d",square(4-1));
return 0;
}

Answer 1

The code in question (found in OP's profile)

https://code.sololearn.com/chDFYl81R58B/?ref=app

* Output:
-1 (minus one)

* Explanation:
Every macro calls will be replaced with the macro body definition "as-is". In that code, you define "x*x" as the macro body, but it yields unexpected output, because the compiler calculates <x> [which is an expression -> (4-1)] differently than what you think, instead of doing (4-1)*(4-1) it simply does 4-1*4-1, taking into account the operator precedence, that is 4-(1*4)-1, or 4-4-1, which yields -1.

To prevent such things from happening wrap the macro argument within parentheses, as follows:

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

This will ensure argument <x> (which was passed in as an expression rather than a value) is calculated before it is multiplied by itself. Hence it will be calculated as (4-1)*(4-1), and you'll have the right result : )

Hth, cmiiw

Answer 2

FYI - If you put the URL in the tags area, we can't click on it or copy the text.

For the benefit of others looking in on this thread:
https://code.sololearn.com/chdfyl81r58b/?ref=app

Answer 3

square(4-1) => 4-1*4-1 => 4-4-1 => -1

Macro are expanded with simple substitution

Answer 4

4-1*4-1->4-4+1+->-1