对于隐式声明的移动赋值运算符，用户声明是什么意思？

Question

声明

Implicitly-declared move assignment operator

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;

with user-declared copy assigment operator does it means only

class_name & class_name :: operator= ( class_name && )

或任何 operator=()定义？

示例：

class Bar
{
public:
   Bar() = default;
   SomeClass some;
};

class Foo
{
public:
   Foo() = default;

   Foo& operator=(Bar&& bar) : some(std::move(bar.some))
   {
   }

   SomeClass some;
};

这是否符合隐含声明的移动赋值运算符的条件？

隐式声明的移动构造函数也是如此。

Answer 1

请注意它说 "user-declared copy assignment operators"（强调我的）。并非每个赋值运算符都是复制赋值运算符。

对于class X，一个copy赋值运算符被定义为operator=，它接受一个X类型的参数、X&、const X&、volatile X& 或 const volatile X&。因此，您的 Foo::operator=(Bar&&) 不是 copy 赋值运算符，因此不会影响移动（或复制）赋值运算符的隐式声明。