Relationship between assignment operator and user-defined constructor

问题

#include <iostream>

class A{
};

class B: public A{
public:
    B(A&& inA){
        std::cout<<"constructor"<<std::endl;
    }
};

int main(){
    B whatever{A{}};
    whatever=A{};
}

This outputs

constructor
constructor

at least with C++14 standard and GCC. How is it defined that assignment operator can result in call to constructor instead of operator=? Is there a name for this property of assignment operator?

回答1:

Since you meet all the conditions for generating a move-assignment operator. The move-assignment operator the compiler synthesizes for you is in the form of:

B& operator=(B&&) = default;

Recall that temporaries can be bound to const lvalue references and rvalue references. By Implicit Conversion Sequences, your temporary A{} is converted to a temporary B which is used to make the move assignment. You may disable this with explicit constructors.

来源：https://stackoverflow.com/questions/46651284/relationship-between-assignment-operator-and-user-defined-constructor