看 SO 上有一个比较奇怪的问题，
问题代码：
ClassX c = std::move(object_c); // Invokes move constructor ClassX&& cr = std::move(object_c); ClassX c = cr; // Invokes copy constructor
题主的疑问：cr 是右值了，为什么 `ClassX c = cr` 会触发拷贝构造，而不是移动构造
对此的解释：
If it has a name, it's an lvalue. cr is a name, it's an lvalue.
the code should do ClassX c = std::move(cr);. Keep in mind that std::move does not itself move anything, it only allows a move to occur as-if the variable were an unnamed temporary (such as a return value or result of an expression). It also allows a variable to be bound to an rvalue reference (usually something that is useful for parameter passing). If you find this to be confusing, you are not alone; it is confusing.
我测试的 demo:
#include <iostream>
class A {
public:
A() { std::cout << "Constructors" << std::endl; }
A(const A& a) { std::cout << "Copy Constructors" << std::endl; }
A& operator=(const A& a) {
std::cout << "Assignment Operators" << std::endl;
return *this;
}
A& operator=(const A&& a) {
std::cout << "Move Assignment Operators" << std::endl;
return *this;
}
};
int main() {
A a; // Constructors
A a_5; // Constructors
A a_6; // Constructors
A&& a_1 = std::move(a); // 没有提示
A& a_2 = a; // 没有提示
A a_7 = a_1; // Copy Constructors
A a_3 = a; // Copy Constructors
A a_4(a); // Copy Constructors
A a_8 = std::move(a); // Copy Constructors
a_6 = a_1; // Assignment Operators
a_5 = a; // Assignment Operators
a_1 = a_8; // Assignment Operators
a_6 = std::move(a_1); // Move Assignment Operators
return 0;
}
