Preprocessors in C

Question

I have a doubt in the output of a code. The code uses preprocessors of c language. The code is given below,

#include <stdio.h>
#define sqr(x) x*x

int main() {

    int x = 16/sqr(4);
    printf("%d", x);
}

The result of sqr(4) is equal to 16. So, the value in x must be 1 (16/16=1). But if I print the value of x, the output is 16. Please explain me why this is happening.

I am attaching the screenshot of output pane. Output Window

score 1 · Answer 1 · answered Dec 27 '19 at 10:52

1

After macro replaces your expression, it becomes int x = 16/4*4 and is evaluated accordingly. i.e. (16/4)*4 which is 16.

sqr(x) doesn't calculate 4*4, it is not a function.

answered Dec 27 '19 at 10:52

Breakpoint

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Alexander van Oostenrijk · Accepted Answer · 2019-12-27T10:56:16.517

1

In C, macros are filled into your code upon compilation. Let's consider what happens in your case:

Your macro is:

#define sqr(x) x*x

When it gets filled into this:

int x = 16/sqr(4);

You would get:

int x = 16/4*4;

Operator precedence being equal for / and *, this gets evaluated as (16/4)*4 = 16, that is, from left to right.

However, if your macro were this:

#define sqr(x) (x*x)

then you would get:

int x = 16/(4*4);

... and that reduces to 1.

However, when the argument for your macro is more complex than a simple number, e.g. 2+3, it would still go wrong. A better macro is:

#define sqr(x) ((x)*(x))

edited Dec 27 '19 at 10:56

answered Dec 27 '19 at 10:53

Alexander van Oostenrijk

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But would give an unexpected result in `sqr(2 + 3)`. – bipll Dec 27 '19 at 10:54
Well spotted. I've added your observation. – Alexander van Oostenrijk Dec 27 '19 at 10:56
I just missed the expansion of macro concept while solving that code. So, I just asked such a silly question. Thank you everyone for answering. – Hem Bhagat Dec 27 '19 at 11:14

score 0 · Answer 3 · answered Dec 27 '19 at 10:52

0

You are actually doing 16/4*4 which is in fact 16.

You can change your define to

#define sqr(x) (x*x)

Now you will be doing 16/(4*4).

answered Dec 27 '19 at 10:52

André Cascais

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score 0 · Answer 4 · answered Dec 27 '19 at 10:53

0

It is expanded to

int x = 16/4*4;

Something like

#define sqr(x) ((x) * (x))

would be more safe, precedence and associativity-wise, still works poorly in expressions like sqr(a++). Inline functions are simpler to use in this context.

answered Dec 27 '19 at 10:53

bipll

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Preprocessors in C

4 Answers4