c++ - 如何从 initializer_list 初始化二维数组

Question

typedef unsigned char u8;

template <u8 N>
class matrix {
    public:
        matrix(const std::initializer_list<std::initializer_list<int>>& l);
    private:
        std::array<std::array<int, N>, N> arr;
};

template <u8 N>
matrix<N>::matrix(const std::initializer_list<std::initializer_list<int>>& l) : arr(l) {}

error C2664: 'std::array,2>::array(std::array,2> &&)': cannot convert argument 1 from 'const std::initializer_list>' to 'const std::array,2> &'

如何从 2D std::initializer_list 初始化 2D std::array(或 C 数组)？

Answer 1

std::array 是一个集合。这意味着您可以从 braced-init-list 中初始化它

std::array<int, 4> foo = {1,2,3,4};

在您的构造函数中，虽然您没有braced-init-list，但您有一个std::initlizer_list，它不能用于初始化一个std::数组。您需要做的是将 std::initlizer_list 中的数据复制到数组中，例如

template <u8 N>
matrix<N>::matrix(std::initializer_list<std::initializer_list<int>> l)
{
    u8 counter = 0;
    for (auto row : l)
        std::copy(row.begin(), row.end(), arr[counter++].begin());
}

另请注意，您不必传递 by std::initlizer_list 引用。它是一个轻量级对象，可以按值传递，不会复制底层元素。

---

您可以更改您的类以将二维数组作为参数，这样您就可以直接用它初始化 arr 但因为 how you have to initialize aggregates of aggregates你会陷入花括号 hell ，因为你的代码会变成

typedef unsigned char u8;

template <u8 N>
class matrix {
    public:
        matrix(std::array<std::array<int, N>, N> l);
    private:
        std::array<std::array<int, N>, N> arr;
};

template <u8 N>
matrix<N>::matrix(std::array<std::array<int, N>, N> l) : arr(l) {}


int main()
{
    std::array<std::array<int, 2>, 2> arr{{{{1, 2}}, {{3, 4}}}}; // this is how you have to initialize a 2d std::array

    matrix<2> m = {{{{{1, 2}}, {{3, 4}}}}}; // we need one more brace here since we are initialzing the matrix, which needs its own brace
}