Array size not passing to function

Question

When passing an array to a function and using sizeof() in the fuction, it just returns the size of a single element (4). Is there a way to fix this? And I'm limited to only using arrays, not vectors/templates.

void arraysize(int arr[]){
    cout << sizeof(arr) << endl;
}

int main(){
    int a[15];
    arraysize(a);                  //4
    cout << sizeof(a) << endl;     //60
}

Output:

4
60

The issue is that the function signature is adjusted to `void arraysize(int* arr)`, so `arr` is really a pointer to `int`. There are plenty of questions on this topic on SO. — juanchopanza, Feb 23 '15 at 08:32
@immibis Note that this is C++. You can't pass by value, but you can pass references/pointers to arrays. — juanchopanza, Feb 23 '15 at 08:35
@juachopanza Yes, but passing a reference/pointer to an array isn't passing the array. — user253751, Feb 23 '15 at 08:47
@immibis Why not? One could argue passing by value isn't passing the array either. — juanchopanza, Feb 23 '15 at 09:14

score 3 · Accepted Answer · answered Feb 23 '15 at 08:33

3

You have to pass the array by reference to keep the size info:

void arraysize(int (&arr)[15]);

with template to auto deduce the size:

template <std::size_t N>
void arraysize(int (&arr)[N]);

So without template/STL, you have to pass the the size info in some way:

void arraysize(int *arr, std::size_t size);

or

void arraysize(int *arr, int* end);

answered Feb 23 '15 at 08:33

Jarod42

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I like this. But I guess he do not understand templates(yet). – Martin Schlott Feb 23 '15 at 08:35

score 1 · Answer 2 · edited May 23 '17 at 12:28

1

You need to read up on array decay.

edited May 23 '17 at 12:28

Community

1
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answered Feb 23 '15 at 08:33

Bulletmagnet

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Also adjustment. Array decay isn't the only thing going on in OP's code. There has to be a context that requires the array decay in the first place. – juanchopanza Feb 23 '15 at 08:37
The context that requires array decay is the function call. See the first sentence of http://www.lysator.liu.se/c/c-faq/c-2.html#2-4 – Bulletmagnet Feb 23 '15 at 08:49
Only because the function takes a pointer, due to *adjustment*. If the function took something that didn't require a pointer to element of the array, there'd be no decay. – juanchopanza Feb 23 '15 at 09:01

Array size not passing to function

2 Answers2