Get the number of elements of an array passed by parameter in C?

Question

Can someone help me figure this out?

I think this is not possible... Unless i pass another parameter with the size? I'm looking to avoid it.

How could I do this?

int size(int *array){
    return (sizeof(array) / sizeof(int));
};

int main(int argc, const char * argv[])
{
    int array[] = {1,2,3,4};

    printf("%d", size(array) );

    return 0;
}

Thank you!

score 7 · Accepted Answer · answered Oct 16 '13 at 22:27

7

You can not do that as a function. You can do it as a macro though.

#define SIZEOF_ARRAY(array) (sizeof(array) / sizeof(*array))

answered Oct 16 '13 at 22:27

Bill Lynch

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2

Every C programmer writes that macro. Naming choices are interesting, haven't seen your version before. Please ignore the moron that voted you down, good answer. – Hans Passant Oct 16 '13 at 22:30
2

You should add that this only works on arrays *before* they decay to a pointer to the first element; that's not obvious to C newbies. (Although neither is array decay.) – detly Oct 16 '13 at 22:35
1

Minor: Consider `printf("%u", SIZEOF_ARRAY(array) )` or `printf("%zu", SIZEOF_ARRAY(array) )`. – chux - Reinstate Monica Oct 16 '13 at 22:37
I'm curious, why can't SIZEOF_ARRARY be implemented as a function? – tlehman Oct 16 '13 at 22:49
1

Because an array name decays into a pointer, when passed as an argument to a function. – wildplasser Oct 16 '13 at 22:59
1

Ah, so sizeof(ptr) would be the size of the pointer, which is not right. – tlehman Oct 17 '13 at 00:11

Get the number of elements of an array passed by parameter in C?

1 Answers1