解压std :: array [英] Unpacking a std::array

问题描述

我试图通过 std :: tie ：

#include <tuple>
#include <array>

int main() {
    std::array<int, 3> arr = {1, 2, 3};
    int a, b, c;
    std::tie(a, b, c) = arr;
    return 0;
}

此方法在clang中有效，但无法在g ++ 5.4中编译： operator = 没有匹配项。编译选项为 -std = c ++ 11 。

This works in clang, but fails to compile in g++ 5.4: no match for ‘operator=’. Compile options are -std=c++11.

1. 为什么这样做


2. 如何使用clang而不是g ++？


3. 如何像一个人可能要拆开一个元组那样，方便地解包一个 std :: array ？ li>
   

1. Why does this work in clang but not g++?
2. How can I portably unpack a std::array, like one might unpack a tuple?

谢谢您的帮助！

推荐答案

我将创建专用功能以将数组转换为元组。 C ++ 14代码可能如下所示：

I would create dedicated function to transform array to tuple. C++14 code could look as follows:

template <class T, std::size_t N, std::size_t... Is>
auto unpack_impl(std::array<T, N> &arr, index_sequence<Is...>) -> decltype(std::make_tuple(arr[Is]...)) {
    return std::make_tuple( arr[Is]... );
}

template <class T, std::size_t N>
auto unpack(std::array<T, N> &arr) -> decltype(unpack_impl(arr, make_index_sequence<N>{})) {
    return unpack_impl(arr, make_index_sequence<N>{});
}

然后使用它：

std::array<int, 3> arr = {{1, 2, 3}};
int a, b, c;
std::tie(a, b, c) = unpack(arr);

在c ++ 11中，您将需要实现 integer_sequence ，因为它不是标准中的开箱即用...

In c++11 you would need however to implement integer_sequence as it does not come out of the box in the standard...

此处，您可以找到完整的c ++ 11解决方案。

Here you can find complete c++11 solution.

编辑：

如果数组包含一些更复杂的对象，则可能要避免不必要的复制。为此，您可以使用const引用的元组代替 make_tuple ，或者如果constness不打扰您，则可以简单地将数组元素捆绑起来：

If an array contains some more complex objects you might want to avoid unnecessary copy. To do that instead of make_tuple you may use tuple of const references or if the constness doesn't bother you you could simple tie array elements up:

template <class T, std::size_t N, std::size_t... Is>
auto unpack_impl(std::array<T, N> &arr, index_sequence<Is...>) -> decltype(std::tie( arr[Is]... )) {
    return std::tie( arr[Is]... );
}

Edit2

此也在VS上编译

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