How does the preprocessor work in C?

Question

Why is the answer for the below code 16? Can anybody explain the working of this program?

#define SQUARE(n) n*n
void main()
{
    int j;      
    j =16/SQUARE(2);

    printf("\n j=%d",j);
    getch();
}

If we write the same code like below, then the answer is 4:

//the ans is 4 why?
#include<stdio.h>
#include<conio.h>

#define SQUARE(n) n*n

void main()
{
    int j;      
    j =16/(SQUARE(2));

    printf("\n j=%d",j);
    getch();
}

Answer 1

The preprocessor just replaces the text, exactly as written.

So, the macro call SQUARE(2) becomes literally 2*2.

In your case, that means the whole expression becomes 16/2*2, which because of C's precedence rules evaluates to (16/2)*2, i.e. 16.

Macros should always be enclosed in parenthesis, and have each argument enclosed as well.

If we do that, we get:

#define SQUARE(n)  ((n) * (n))

which replaces to 16/((2) * (2)), which evaluates as 16/4, i.e. 4.

The parens around each argument makes things like SQUARE(1+1) work as expected, without them a call such as 16/SQUARE(1+1) would become 16/(1+1*1+1) which is 16/3, i.e. not at all what you'd want.

Answer 2

Order of operations. Your expression is evaluating to:

 j = 16 / 2 * 2

which equals 16. Make it:

#define SQUARE(n) (n*n) 

which will force the square to be evaluated first.