Why does this macro expression in C behave unexpectedly?

Question

I am trying to write this code in code blocks C editor but the output is coming 36 instead of 1.

#include <stdio.h>

#define square(x) x*x

int main()
{
    int p=36/square(6);
    printf("%d",p);
}

Why is the output 36 instead of 1?

Answer 1

The problem is that the definition of your macro doesn't have any parentheses in it - and macros aren't 'called', like functions are - they just cause text to be replaced.

So, in your code, the line:

int p=36/square(6);

is replaced with:

int p=36/6*6;

which is evaluated in left-to-right order (as the * and / operators have equal precedence), giving, effectively:

int p = (36/6) * 6;

To fix this, add parentheses in your macro definition:

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

Feel free to ask for further clarification and/or explanation.

Answer 2

The C preprocessor changes the line

int p=36/square(6);

into

int p=36/6*6;

You will get the desired output if your change your preprocesor macro to:

#define square(x) (x*x)

That way, the line will instead be changed to:

int p=36/(6*6);

However, this solution still is not ideal: If you for example write

int p=36/square(6+3);

then the preprocessor will change this to

int p=36/(6+3*6+3);

Therefore, the proper solution would be to change the preprocessor macro to:

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

That way, the preprocessor will correctly change it to:

int p=36/((6+3)*(6+3));

In this respect, C preprocessor macros behaves very differently from function calls, where the parentheses are not necessary.