为什么不调用移动构造函数？ （默认情况下只有构造函数，没有别的）

Question

Could you explain me why, in the following, code,

#include <iostream>
#include <variant>
#include <string>



class MySecondType {
public:
    MySecondType() { std::cout << "Constructeur par défaut de MySecondType\n"; }
    MySecondType(const MySecondType&) // Constructeur par copie
    {std::cout<< "Constructeur par copie de MySecondType\n";}
    MySecondType(MySecondType&&) noexcept//Constructeur par déplacement 
    {
        std::cout << "COnstructeur par déplacement de MySecondType\n";
    }


    MySecondType& operator=(MySecondType&&) noexcept
    {
        std::cout << "Opérateur d'affection par déplacement\n";
        return *this;
    }

    MySecondType& operator=(MySecondType&) 
    {
        std::cout << "Opérateur d'affection par copie\n";
        return *this;
    }



    ~MySecondType() {
        std::cout << "Destructeur de MySecondType\n";
    }

};


int main() {

    MySecondType e;
    e= MySecondType() ;
return 0;
}

I have the result I wait, with :

MySecondType e;
e= MySecondType() ;

BUT I don't have it , if I do :

MySecondType e = MySecondType() ;

I expected that the line :

MySecondType e = MySecondType() ;

would call the move constructor(after the default constructor), but it does not call it. It only creates an object with the default constructor , and that's nothing else.

Can you explain me why ?

Thank you

Answer 1

I'm asuming you actually do what @Jacob hinted at, ie MySecondType e = MySecondType(); .

If so, then the compiler is eliding the copy/move constructor. It sees that you are creating a temporary object and right after assigning it to a variable. So instead of doing that, which would be inefficient, it skips the copy/move step and directly constructs the object at its final destination.

AFAIK, this is an optional optimization pre-C++17. Starting in C++17 it's mandatory.