C get array length function

Question

Making a C array size function. This is typically the standard convention for a C array size, and it works within the function in which the array was created, but it doesn't seem to work when put into another function. Why is this?

int size(char a[]) {
    return (int) (size_t) (sizeof(a)/ sizeof(char));
}

void test() {
    char a[] = {'a','b','c','d'};
    printf("%d\n", size(a));
    printf("%d\n", (int) (size_t) (sizeof(a)/ sizeof(char)));
}

int main() {
    test();
    return 0;
}

RESULT:

8       -> INCORRECT
4       -> CORRECT

`a` inside the `size` function is not an actual array anymore and thus its `sizeof` operator behaves differently. What you want can be achieved with a macro instead. — Daniel Kamil Kozar, Jun 15 '19 at 14:19
Remember, `sizeof` happens at compile-time. It can't possibly work as you intended. — ikegami, Jun 15 '19 at 14:29
It is not possible. You need to pass the size to the function as well — 0___________, Jun 15 '19 at 14:30
@klutt, Why are they casting at all? The `size_t` cast is superfluous, and they should have used `%zu` and `size_t` instead of `%d` and `int`. — ikegami, Jun 15 '19 at 14:30

klutt · Accepted Answer · 2019-06-15T14:26:52.747

It's because arrays decays to pointers in many situations, including passing them to functions. It's impossible to retrieve the size of a passed array from inside the function. If you want the function to know the size, then you need to pass it separately somehow. Many standard library functions uses this approach.

If you're only looking for a quick way to get the size of an array, use this macro:

#define SIZE(x) (sizeof(x)/sizeof(x[0]))

Sidenote: sizeof(char) is ALWAYS - by definition - equal to 1, so it is quite pointless to write. Especially if you're going to multiply or divide with it.

C get array length function

1 Answers1