std::array template function

Question

I have a simple function to shuffle elements in an std::array

template <typename T, uint size>
void Shuffle(std::array<T, size> &arr) {
    for (uint i = 0; i < arr.size() - 1; i++)
    {
        std::swap(arr[i], arr[Random(i, arr.size())]);
    }
}

make/g++ doesn't like the way I'm declaring this, giving the error "variable or field Shuffle declared void." From what I've found, this is probably an irrelevant error message, but I can't figure out what's actually wrong here.

Is uint a defined type? Did you try replacing it with std::size_t? — Galik
– Galik, Commented Mar 20, 2021 at 8:37
The code will also have a problem when an empty array is passed. Since i is unsigned, arr.size() - 1 will underflow and will lead to a very long loop and invalid array access. Consider writing i + 1 < arr.size() instead. — stiar
– stiar, Commented Mar 20, 2021 at 11:31

songyuanyao · Accepted Answer · 2021-03-20 08:38:44Z

3

uint doesn't match the type of the 2nd template parameter of std::array, which is std::size_t. This causes template argument deduction failing when calling Shuffle, unless you specify template arguments explicitly.

You should declare the 2nd template parameter with type std::size_t. e.g.

template <typename T, std::size_t size>
void Shuffle(std::array<T, size> &arr)

answered Mar 20, 2021 at 8:38

songyuanyao

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