C++ 中的重载赋值运算符 [英] Overloading assignment operator in C++

问题描述

据我所知，当重载 operator= 时，返回值应该是非常量引用.

As I've understand, when overloading operator=, the return value should should be a non-const reference.

A& A::operator=( const A& )
{
    // check for self-assignment, do assignment

    return *this;
}

在以下情况下允许调用非常量成员函数是非常量的:

It is non-const to allow non-const member functions to be called in cases like:

( a = b ).f();

但是为什么要返回一个引用呢?如果返回值没有声明为引用，在什么情况下会出现问题，比如说按值返回?

But why should it return a reference? In what instance will it give a problem if the return value is not declared a reference, let's say return by value?

假设复制构造函数已正确实现.

It's assumed that copy constructor is implemented correctly.

推荐答案

不返回引用是一种资源浪费，并且会产生奇怪的设计.为什么您要为您的运营商的所有用户制作一份副本，即使几乎所有用户都会丢弃该值?

Not returning a reference is a waste of resources and a yields a weird design. Why do you want to do a copy for all users of your operator even if almost all of them will discard that value?

a = b; // huh, why does this create an unnecessary copy?

此外，您的类的用户会感到惊讶，因为内置的赋值运算符不会同样复制

In addition, it would be surprising to users of your class, since the built-in assignment operator doesn't copy likewise

int &a = (some_int = 0); // works

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