为什么包含 <utility> 会破坏 GCC 中的结构化绑定？

Question

考虑：

struct Point { int x, y; };

int main()
{
    const auto [x, y] = Point{};
}

此代码在 C++17 模式下使用 gcc 7.1 编译良好，但是这个：

#include <utility>

struct Point { int x, y; };

int main()
{
    const auto [x, y] = Point{};
}

报错：

bug.cpp: In function 'int main()':
bug.cpp:7:16: error: 'std::tuple_size<const Point>::value' is not an integral constant expression
     const auto [x, y] = Point{};
                ^~~~~~

这是怎么回事？编译器错误，或者这就是结构化绑定的工作方式？

Answer 1

结构化绑定背后的核心思想是 std::tuple_size<T> 定义了您从解包中获得的组件数量 T，并且 T::get<N> 应该访问第 N 个元素。毫不奇怪，这个 std::tuple_size<T> 是 <utility> 中基本模板的特化。

现在在这种情况下，Point 不支持结构化绑定，但这是 C++17 规定的一种特殊情况（所有 public 非静态成员）不需要特殊的拆包支持。这是上述规则的一个例外。

编译器在这里被自己绊倒了，当它看到来自 <utility> 的非专用 std::tuple_size 时，它试图使用通用规则。

Answer 2

这是编译器错误 78939. Although it's a bit more complicated than that - there were a few issues between the core language and the library that were mutually contradictory (GB 20, LWG 2770, and LWG 2446)，它会导致 gcc/libstdc++ 在此处展示的那种行为。代码在有或没有 #include <utility> 的情况下都可以正常工作，这只是标准措辞是否正确的问题。

是的，类和所有 public 非匿名联合成员应该可以在结构化绑定声明中使用 [dcl.struct.bind]/4:

Otherwise, all of E's non-static data members shall be public direct members of E or of the same unambiguous public base class of E, E shall not have an anonymous union member, and the number of elements in the identifier-list shall be equal to the number of non-static data members of E. Designating the non-static data members of E as m0, m1, m2, ... (in declaration order), each vi is the name of an lvalue that refers to the member mi of e and whose type is cv Ti, where Ti is the declared type of that member; the referenced type is cv Ti. The lvalue is a bit-field if that member is a bit-field. [ Example:
struct S { int x1 : 2; volatile double y1; };
S f();
const auto [ x, y ] = f();

这与包含 <utility> 完全无关，此代码中的任何内容都不依赖于任何库功能 - 成员是直接获取的，而不是通过 get/tuple_size机制。

为什么包含 <utility> 会破坏 GCC 中的结构化绑定？

Why does including <utility> break structured bindings in GCC?

c++

gcc

c++17

structured-bindings

gcc7