为什么赋值运算符应该返回对对象的引用? [英] Why should the assignment operator return a reference to the object?

问题描述

我正在对我的 C++ 进行一些修订，我正在处理运算符重载，特别是="(赋值)运算符.我在网上查找并遇到了多个讨论它的主题.在我自己的笔记中，我把我所有的例子都记为类似

I'm doing some revision of my C++, and I'm dealing with operator overloading at the minute, specifically the "="(assignment) operator. I was looking online and came across multiple topics discussing it. In my own notes, I have all my examples taken down as something like

class Foo
{
    public:  
        int x;  
        int y;  
        void operator=(const Foo&);  
};  
void Foo::operator=(const Foo &rhs)
{
    x = rhs.x;  
    y = rhs.y;  
}

在我在网上找到的所有引用中，我注意到操作符返回对源对象的引用.为什么正确的方法是返回对对象的引用而不是完全没有?

In all the references I found online, I noticed that the operator returns a reference to the source object. Why is the correct way to return a reference to the object as opposed to the nothing at all?

推荐答案

通常的形式返回对目标对象的引用以允许赋值链接.否则，这是不可能的:

The usual form returns a reference to the target object to allow assignment chaining. Otherwise, it wouldn't be possible to do:

Foo a, b, c;
// ...
a = b = c;

仍然要记住，正确使用赋值运算符 比它更难可能看起来.

Still, keep in mind that getting right the assigment operator is tougher than it might seem.

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