Question

In C99, we have compound literals, and they can be passed to functions as in:

f((int[2]){ 1, 2 });

However, if f is not a function but rather a function-like macro, gcc barfs on this due to the preprocessor parsing it not as one argument but as two arguments, “(int[2]){ 1” and “2 }“.

Is this a bug in gcc or in the C standard? If it’s the latter, that pretty much rules out all transparent use of function-like macros, which seems like a huge defect…

Edit: As an example, one would expect the following to be a conforming program fragment:

fgetc((FILE *[2]){ f1, f2 }[i]);

But since fgetc could be implemented as a macro (albeit being required to protect its argument and not evaluate it more than once), this code would actually be incorrect. That seems surprising to me.

Answer 1

This “bug” has existed in the standard since C89:

#include <stdio.h>

void function(int a) {
    printf("%d\n", a);
}

#define macro(a) do { printf("%d\n", a); } while (0)

int main() {
    function(1 ? 1, 2: 3); /* comma operator */
    macro(1 ? 1, 2: 3);    /* macro argument separator - invalid code */
    return 0;
}

I haven’t actually looked through the standard to check this parse, I’ve taken gcc’s word for it, but informally the need for a matching : to each ? trumps both operator precedence and argument list syntax to make the first statement work. No such luck with the second.