c++ - 具有已删除特殊成员函数的自定义类型 std::variant 的赋值运算符？

Question

考虑:

#include <variant>

struct A { 
  A() = default;
  A(A&&) = delete;
};

struct B { 
  B() = delete;
  B(A&&) {};
};

int main() {
  std::variant<A, B> v{};
  v = A{};
}

MSVC 接受了， while GCC and Clang rejected it with the same error message :

opt/compiler-explorer/gcc-snapshot/lib/gcc/x86_64-linux-gnu/12.0.0/../../../../include/c++/12.0.0/variant:1465:3: error: call to deleted member function 'operator='
                operator=(variant(std::forward<_Tp>(__rhs)));
                ^~~~~~~~~
<source>:15:5: note: in instantiation of function template specialization 'std::variant<A, B>::operator=<A>' requested here
  v = A{};
    ^

我应该信任哪个编译器？

Answer 1

编辑
最初，没有 language-lawyer标签，这就是我使用 cppreference 进行分析的原因。但是，查看最新草案( relevant section )，我没有看到任何会使它无效的内容。

我相信 MSVC 是不正确的。根据documentation :

3. Converting assignment.

• Determines the alternative type T_j that would be selected by overload resolution for the expression F(std::forward<T>(t)) if there was an overload of imaginary function F(T_i) for every T_i from Types... in scope at the same time, except that:

  • An overload F(T_i) is only considered if the declaration T_i x[] = { std::forward<T>(t) }; is valid for some invented variable x;

一些虚函数具有转发引用参数并用 A{} 调用它参数， A x[] = { std::forward<T>(t) }; B x[] = { std::forward<T>(t) }; 时无效是。因此， T_j应解析为 B .现场演示: https://godbolt.org/z/fM67e7oGj .
然后:

• If *this already holds a T_j...

这不适用，因为 v不持有 B .
下一个:

• Otherwise, if std::is_nothrow_constructible_v<T_j, T> || !std::is_nothrow_move_constructible_v<T_j> is true...

这也不适用，因为这个表达式是 false ;现场演示: https://godbolt.org/z/x674rnbcj (并且，MSVC 同意: https://godbolt.org/z/5Techn8jG )。
最后:

• Otherwise, equivalent to this->operator=(variant(std::forward<T>(t))).

但是，此调用无法使用 MSVC 编译: https://godbolt.org/z/cWr4f6EhK .