c++ - `std::tuple` 有什么用？

Question

此处 ( https://stackoverflow.com/a/37550660/34509 ) 用户 @Barry可以在评论区备注 std::tuple<int[2]>并且显然不禁止实例化这种类型。我还没有听说过这个野兽，我想知道它有什么用，而不是存储 int var[2]直接或使用 std::array<int, 2> .

据报道，std::tuple<int[2]>不可复制，不可移动，也不可从 int var[2] 构造.它还有什么其他用途？

Answer 1

我很确定这是未定义的行为。请参阅Requires 和Returns 子句:

tuple.creation-10 and 12说:

Requires: For all i, U_i shall be the type cv_i tuple<<em>Args<sub>i</sub>...</em>>, where cv_i is the (possibly empty) ⁱth cv-qualifier-seq and Args_i is the parameter pack representing the element types in U_i. Let A_ik be the k_i^th type in Args_i. For all A_ik the following requirements shall be satisfied: If T_i is deduced as an lvalue reference type, then is_constructible<<em>A<sub>ik</sub></em>, <em>cv<sub>i</sub></em> <em>A<sub>ik</sub></em>&>::value == true, otherwise is_constructible<<em>A<sub>ik</sub></em>, <em>cv<sub>i</sub></em> <em>A<sub>ik</sub></em>&&>::value == true.

Returns: A tuple object constructed by initializing the k_i^th type element e_ik in e_i... with get<<em>k<sub>i</sub></em>>(std::forward<<em>T<sub>i</sub></em>>(<em>tp<sub>i</sub></em>)) for each valid k_i and each group e_i in order.

正如 Barry 所说，没有什么能阻止 std::tuple<int[2]> t;但是试图用它做任何事情可能会导致编译器出现硬错误。示例:

std::tuple<int[2]> t; // fine
std::tuple<int[2]> t{}; // fine

同时:

std::tuple<int[2]> a() {
  int a[2] = { 1, 2};
  return std::tuple<int[2]>(a);
}

int main() {
   auto x = a();
   // ...
}

给出如下错误:

error: array initializer must be an initializer list
        : _M_head_impl(std::forward<_UHead>(__h)) { }