CODEWITHSUNDEEP
carraysmacros
GiuMaz
GiuMaz

Reputation: 63

c macro, array definition as parameter

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.

Upvotes: 0

Views: 3384

Answers (2)

starrify
starrify

Reputation: 14751

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. :)

Upvotes: 3

Dwayne Towell
Dwayne Towell

Reputation: 8583

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.

Upvotes: 0

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