c macro, array definition as parameter

Question

I have this method

foo_l(int *array, size_t l)
{
    /*
    code
    */
}

and I wrote this macro

#define foo(X) foo_l(X,sizeof(X)/sizeof(int))

So I can use them as follows

int main()
{
    int a[]={1,2,3};
    foo(a);
    return 0;
}

and avoid writing the length of the array every time.

My question is, can I extend my macro so it can handle something like

foo_l((int[]){1,2,3}, 3);

with an array declared in the function parameter field?

Because foo((int[]){1,2,3}) doesn't work! I think that the problem is that the macro see (int[]){1,2,3} as a list of parameters and not as a unique parameter. Any Idea?

P.S. I'm pretty new to the macro world and I usually use c99.

starrify · Accepted Answer · 2014-09-14 21:18:28Z

When being passed to the preprocessor, the macro foo((int[]){1,2,3}) fails because the preprocessor believes it provided 3 parameters instead of 1:

foo((int[]){1,2,3});
// is believed to be:
// Start of macro: foo(
// Parameter 1:    (int[]){1, 
// Parameter 2:    2, 
// Parameter 3:    3}
// End of macro:   );

So it doesn't compile and gives something like:

a.c: In function ‘main’:
a.c:15:23: error: macro "foo" passed 3 arguments, but takes just 1
     foo((int[]){1,2,3});

Adding another pair of parenthesis solves the problem:

// This shall work
foo(((int[]){1,2,3}));

EDITED:

Yes I guess this may not be a good design, since people like average programmers may be very likely to pass a pointer instead of an array type to your macro foo, and it would fail as @DwayneTowell points out.
Please be careful about this. :)

Dwayne Towell · Accepted Answer · 2014-09-14 20:51:22Z

0

What you have suggested is not a good idea, because it will fail in cases like the following:

int a[] = {1,2,3,4};
int *b = a;
foo(b);        // becomes foo_l(b,sizeof(b)/(sizeof(int));
               // probably becomes foo_l(b,1)

In general the size parameter will be wrong if the first parameter is a pointer to an array instead of the array itself. Usually the expression will evaluate to 1, when sizeof(int *)==sizeof(int), which is very common but required by the standard.

answered Sep 14, 2014 at 20:51

Dwayne Towell

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8 Comments

Zilicon Over a year ago

I Agree. Also, It will not work for arrays that their size isn't known in compile time (allocated arrays). It is ok, though, to use this only for compile-time arrays.

starrify Over a year ago

The sizeof would fail since you're passing a pointer instead of an array type.

Dwayne Towell Over a year ago

sizeof(int *) does not fail, it returns the size of a pointer, usually 4 or 8 bytes these days.

starrify Over a year ago

@DwayneTowell I see, but the OP wants to calculate the array length by sizeof(X)/sizeof(int), doesn't he? That's what I mean by "fail".

Dwayne Towell Over a year ago

@starrify, fair enough, I agree my "answer" could be considered off topic from a stict OP question point-of-view, but it was such a bad idea in general, I felt someone needed to answer it in a "there be dragons ahead" sort of way instead of blindly answering the exact question posed.

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