Unexpected output in expression that uses macros

Question

My code:

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

int main()
{
    int i = 3, j, k, l;
    j = PRODUCT(i + 1);
    k = PRODUCT(i++);
    l = PRODUCT(++i);
    printf("%d %d %d %d", i, j, k, l);
    return 0;
}

I am not able to comprehend why the output is:

7 7 12 49.

Is there any error in macro or some other problem?

Using this macro with `i++` results in undefined behavior. Anything can happen. — Fiddling Bits, May 05 '20 at 18:27
Macros are evil. Try this: `j = PRODUCT("Hello");` Functions are so much better. — Thomas Matthews, May 05 '20 at 18:27
Does this answer your question? [#define in C: what is happening](https://stackoverflow.com/questions/61491522/define-in-c-what-is-happening) — user12986714, May 05 '20 at 18:27
@user12986714 your reference has no link with the current problem — bruno, May 05 '20 at 18:29

anastaciu · Answer 1 · 2020-05-05T19:00:58.143

6

Your code has undefined behavior, operations in i:

k=PRODUCT(i++);
l=PRODUCT(++i);

lack a sequence point.

As for:

j=PRODUCT(i+1);

It expands to i+1*i+1 which is i+i+1 which is 7. I assume it's not the expected result, in the future also include that in your question.

edited May 05 '20 at 19:00

answered May 05 '20 at 18:26

anastaciu

23,467
7
28
53

cigien · Accepted Answer · 2020-05-05T18:35:37.243

Your macro is incorrect. The following expression:

PRODUCT(i+1)

will expand to

(i+1 * i+1)

which is 2*i+1.

Your macro should be:

#define PRODUCT(x) ((x)*(x))

I strongly suggest you stop using macros for this sort of thing. You could easily write this as a function:

int product(int x)
{
  return x * x;
}

Note that this will only work for the example I gave. If you try

PRODUCT(i++)

you will get

( (i++) * (i++) )

which invokes undefined behaviour, as this expression lacks a sequence point between the 2 increments.

Unexpected output in expression that uses macros

2 Answers2