Question

I want to define a macro at compile time based on the value of another macro. However this code is not executing as expected:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIXTEEN 16
#define TWO (SIXTEEN % 8 == 0)? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1)

int main();

int main() {
    printf("max = %d\n", TWO);
    int i;
    for (i = 0; i < TWO; i++) {
        printf("%d\n", i);
    }
    return 0;
}

This prints:

max = 2
0
1
2
...

and continues until terminated, When it should be printing simply:

max = 2
0
1

and exiting.

If I do this instead, it works:

#define TWO 2

I thought that this was an issue with the macro’s definition… however, if I do the following with the original #define, it seems to work:

...
int count = TWO;
for (i = 0; i < count; i++) {
...

Can anyone explain what’s going on here?

Answer 1

The problem is that the token TWO is replaced by the tokens with which you defined the macro, so this:

i < TWO

becomes this:

i < (SIXTEEN % 8 == 0)? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1) 

Because of operator precedence, this is read as:

(i < (SIXTEEN % 8 == 0))
    ? (SIXTEEN / 8) 
    : ((SIXTEEN / 8) + 1) 

You need extra parentheses so that when TWO is replaced by its replacement list, you get the result you want:

#define TWO ((SIXTEEN % 8 == 0)? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1))
            ^                                                       ^

When using macros, it’s best to use parentheses wherever you can to ensure the result is what you expect.