6

Are there macros or builtins that can return the length of arrays at compile time in GCC?

For example:

int array[10];

For which:

sizeof(array) == 40
???(array) == 10

Update0

I might just point out that doing this in C++ is trivial. One can build a template that returns the number inside []. I was certain that I'd once found a lengthof and dimof macro/builtin in the Visual C++ compiler but cannot find it anymore.

Craig McQueen
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Matt Joiner
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  • See also SO question: [Is there a standard function in C that would return the length of an array?](http://stackoverflow.com/questions/1598773/is-there-a-standard-function-in-c-that-would-return-the-length-of-an-array) – Craig McQueen Mar 21 '11 at 00:14

4 Answers4

14
(sizeof(array)/sizeof(array[0]))

Or as a macro

#define ARRAY_SIZE(foo) (sizeof(foo)/sizeof(foo[0]))

    int array[10];
    printf("%d %d\n", sizeof(array), ARRAY_SIZE(array));

40 10

Caution: You can apply this ARRAY_SIZE() macro to a pointer to an array and get a garbage value without any compiler warnings or errors.

ndim
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2

I wouldn't rely on sizeof since aligment stuff could mess up the thing.

#define COUNT 10
int array[COUNT];

And then you could use COUNT as you want.

tibur
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    isn't alignment just important for placing the array itself? Inside a c-array, the sizeof-division shall work. – nob Aug 02 '10 at 14:37
  • I think you're right: both sizeof will take alignment into consideration. So the sizeof(blah)/sizeof(type) should be safe. – tibur Aug 02 '10 at 14:46
1
    sizeof(array) / sizeof(int)
mob
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1

im not aware of a builtin that does this, but i recently used:

sizeof(array)/sizeof(array[0])

to do just that

trh178
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