c++ - 您是否应该能够从 T 具有非平凡构造函数的 std::optional 移动？

Question

我正在尝试用 clang 编译 WebKit，我正在点击 compile errors由于本质上是以下模式:

#include <iostream>
#include <optional>

struct X {
    X() = default;
    X(const X& other) { }
};

struct Y {
    std::optional<X> x;;
};

int main() {
    Y foo;
    Y bar(std::move(foo));
}

所以，他们使用 std::optional<T>其中 T (在他们的情况下， WTF::Variant )具有非平凡的复制/移动构造函数，然后使用 std::optional移动构造函数。使用 GCC 8.1.1 可以很好地编译，但不能使用 clang 6.0.1(使用 GCC 8.1.1 的 libstdc++):

In file included from test.cpp:2:
/bin/../lib64/gcc/x86_64-pc-linux-gnu/8.1.1/../../../../include/c++/8.1.1/optional:276:9: error: call to implicitly-deleted copy constructor of 'std::_Optional_payload<X, true, true, true>'
      : _Optional_payload(__engaged
        ^                 ~~~~~~~~~
/bin/../lib64/gcc/x86_64-pc-linux-gnu/8.1.1/../../../../include/c++/8.1.1/optional:739:4: note: in instantiation of member function 'std::_Optional_payload<X, true, true, true>::_Optional_payload' requested here
        : _M_payload(__other._M_payload._M_engaged,
          ^
/bin/../lib64/gcc/x86_64-pc-linux-gnu/8.1.1/../../../../include/c++/8.1.1/optional:985:11: note: in instantiation of member function 'std::_Optional_base<X, false, false>::_Optional_base' requested here
    class optional
          ^
test.cpp:9:8: note: in implicit move constructor for 'std::optional<X>' first required here
struct Y {
       ^
test.cpp:15:7: note: in implicit move constructor for 'Y' first required here
    Y bar(std::move(foo));
      ^
/bin/../lib64/gcc/x86_64-pc-linux-gnu/8.1.1/../../../../include/c++/8.1.1/optional:288:24: note: copy constructor of '_Optional_payload<X, true, true, true>' is implicitly deleted because variant field '_M_payload' has a
      non-trivial copy constructor
          _Stored_type _M_payload;

这是有效的 C++，还是 WebKit 损坏并且 clang 正确地拒绝了这个代码？

Answer 1

考虑这个类:

struct X
{
    X(int);
    X(X&&) = delete;

    // does this need to invoke the move constructor??
    X() : X(X(0)) { }
};

根据 gcc，答案是否定的:这直接委托(delegate)给 X(int)。根据 clang，答案是肯定的，并且编译失败:

<source>:55:15: error: call to deleted constructor of 'X'    
        X() : X(X(0)) { }   
              ^ ~~~~    
<source>:52:9: note: 'X' has been explicitly marked deleted here    
        X(X&&) = delete;   
        ^

这似乎是一个潜在的核心语言问题，因为一方面 [class.base.init]/6说:

The target constructor is selected by overload resolution. Once the target constructor returns, the body of the delegating constructor is executed.

也就是说，我们专门讨论了选择一个构造函数并调用它 - 在这种情况下肯定是 X(X&&)。但另一方面，the very next paragraph说这是初始化:

The expression-list or braced-init-list in a mem-initializer is used to initialize the designated subobject (or, in the case of a delegating constructor, the complete class object) according to the initialization rules of [dcl.init] for direct-initialization.

这看起来很像 [dcl.init]/17.6 中的保证复制省略示例:

If the initializer expression is a prvalue and the cv-unqualified version of the source type is the same class as the class of the destination, the initializer expression is used to initialize the destination object. [ Example: T x = T(T(T())); calls the T default constructor to initialize x. — end example ]

我不确定哪种解释是正确的，但 clang 拒绝对我来说似乎并没有明显的错误。

---

为什么这个例子是相关的？ optional 在 libstdc++ 中的移动构造函数，对于可简单破坏和可简单复制/移动可赋值的类型(如您的 X)通过:this constructor :

  constexpr
  _Optional_payload(bool __engaged, _Optional_payload&& __other)
  : _Optional_payload(__engaged
          ? _Optional_payload(__ctor_tag<bool>{},
                      std::move(__other._M_payload))
          : _Optional_payload(__ctor_tag<void>{}))
  { }

此类型已隐式删除了复制和移动构造函数，因为它与不可简单复制构造的成员具有 union 。因此，如果此委托(delegate)构造函数必须调用隐式复制构造函数(如 clang 认为的那样)，则这是不正确的。如果不需要，只需调用一个或另一个委托(delegate)构造函数，那么此调用就可以了。