具有抑制移动 construct/assign 的类型如何仍然被认为是可移动的？

Question

struct copyable { // and movable
  copyable() = default;
  copyable(copyable const&) { /*...*/ };
  copyable& operator=(copyable const&) { /*...*/ return *this; }
};

由于复制构造函数和复制赋值操作函数是显式定义的，这意味着编译器不能隐式定义移动构造函数和移动赋值函数，因此不允许移动操作。

请问我上面的理解是否正确？

Answer 1

it signifies that move constructor and move assignment function cannot be implicitly defined by the compiler

是的，没错。

and hence move operation is not allowed.

不，移动操作仍然可以通过复制构造函数和复制赋值运算符执行（尽管这可能不是您预期的情况），因为右值总是可以绑定到 const&.

更准确地说，class copyable 仍然是 MoveConstructible and MoveAssignable 。

A class does not have to implement a move constructor to satisfy this type requirement: a copy constructor that takes a const T& argument can bind rvalue expressions.

If a MoveConstructible class implements a move constructor, it may also implement move semantics to take advantage of the fact that the value of rv after construction is unspecified.

和

The type does not have to implement move assignment operator in order to satisfy this type requirement: a copy assignment operator that takes its parameter by value or as a const Type&, will bind to rvalue argument.

If a MoveAssignable class implements a move assignment operator, it may also implement move semantics to take advantage of the fact that the value of rv after assignment is unspecified.

正如@Curious 指出的那样，您可以显式声明移动构造函数和移动赋值运算符 delete 以使 copyable 不可移动；然后与右值表达式一起使用 deleteed 重载将被选中并且编译将失败。

具有抑制移动 construct/assign 的类型如何仍然被认为是可移动的？

How can a type with suppressed move construct/assign still be considered moveable?

c++

move-semantics

c++11