隐式声明的移动赋值运算符

Question

我无法理解 cpp 参考中的这个简单规则；

If no user-defined move assignment operators are provided for a class type (struct, class, or union), and all of the following is true:

there are no user-declared copy constructors;

there are no user-declared move constructors;

there are no user-declared copy assignment operators;

there are no user-declared destructors;

the implicitly-declared move assignment operator would not be defined as deleted, (until C++14)

THEN the compiler WILL declare a move assignment operator as an inline public member of its class with the signature T& T::operator=(T&&).

考虑到这一点，考虑

struct bar
{

   bar(int i) : _i(i) {};
   bar (const bar& other) { };
   bar (bar& other) { };

   int _i;
};

后面跟着这个，说;

   bar b2(2);
   bar b3(3);
   cout << "b3._i " << b3._i << endl << endl;
   b3 = std::move(b2);
   cout << "b3._i " << b3._i << endl;

我们得到的输出是；

b3._i 3

b3._i 2

所以我们有一个动作发生了；

b3 = std::move(b2);

我没有定义那个移动赋值运算符，所以它是由编译器隐式为我定义的。但是我已经打破了规定的条件，我有一个用户定义的复制构造器....但是编译器生成的移动仍然发生了。我明显误解了文字，谁能开导我一下？

谢谢，祝你有愉快的一天。

G

Answer 1

So here we have a move happening;
b3 = std::move(b2);

不，这是副本。 std::move(b2) 只是将 b2 转换为 bar&&。 bar&& 然后在 implicitly-generated copy-assignment 运算符中绑定到 const bar&。

live example on wandbox

隐式声明的移动赋值运算符

Implicitly-declared move assignment operator

move

c++11