c++ - 具有抑制 move 构造函数/分配的类型如何仍被视为可 move 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 move 构造函数和 move 赋值函数不能被编译器隐式定义，因此不允许 move 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， move 操作仍然可以通过复制构造函数和复制赋值操作符执行(尽管这可能不是您所期望的)，因为右值总是可以绑定(bind)到 const&。

更准确地说，类 copyable 仍然是 MoveConstructible和 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 所指出的，您可以声明 move 构造函数和 move 赋值运算符delete明确地使 copyable 不可 move ；然后使用右值表达式，将选择 deleteed 重载并且编译将失败。