passing array (not vector) to function [duplicate]

Question

I want to do this

void DoSomething (int arr[])
{
// do something
}

but instead of arr[] it's std::array

void DoSomething (std::array <int> arr) // <--- incorrect way to declare an array
{
// do something
}

Using std::vector instead of std::array works, but I expect a workaround with std::array, since arr[] works just fine.

This works too

const int arrSize;
void DoSomething (std::array <int,arrSize> arr)
{
// do something
}

but again, I expect a way to do this without any extra declarations.

Show a minimal reproducible example, how you are going to call the function and pass arguments. — 3CEZVQ
– 3CEZVQ, Commented Nov 7, 2023 at 2:17
void DoSomething (int arr[]) ends up passing a pointer and loses all size info so DoSomething does not know how many elements to operate on. — drescherjm
– drescherjm, Commented Nov 7, 2023 at 2:18
Pass a pointer to the first slot, then also pass the capacity of the array. When passing an array to a function, the capacity property is lost. — Thomas Matthews
– Thomas Matthews, Commented Nov 7, 2023 at 2:18
std::array is an array of a fixed size that must be specified. — Evg
– Evg, Commented Nov 7, 2023 at 2:20

Remy Lebeau · Accepted Answer · 2023-11-07 02:22:05Z

3

std::array is a template. If you want to accept a std::array of any size, your function will have to become a template as well, eg:

template <size_t N>
void DoSomething (std::array<int, N> arr)
{
    // do something
}

The compiler can then deduce N for you, based on whatever std::array you pass in.

answered Nov 7, 2023 at 2:22

Remy Lebeau

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1 Comment

Consider passing the array by (const) reference (const) std::array<int,N>& arr to avoid copying the array