赋值运算符和用户定义的构造函数之间的关系 [英] Relationship between assignment operator and user-defined constructor

问题描述

#include <iostream>

class A{
};

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

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

此输出

constructor
constructor

至少使用C ++ 14标准和GCC.如何定义赋值运算符可以导致调用构造函数而不是operator=?赋值运算符的此属性有名称吗?

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?

推荐答案

由于您满足生成移动分配运算符的所有条件.编译器为您合成的 move-assignment运算符的形式为:

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;

回想一下，临时对象可以绑定到 const左值引用和右值引用.通过隐式转换序列，您的临时A{}被转换为临时B用于进行移动分配.您可以使用explicit构造函数禁用此功能.

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.

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