为什么赋值操作符返回对象的引用？ [英] Why should the assignment operator return a reference to the object?

问题描述

我正在做一些修改我的C + +，我处理的运算符重载在分钟，特别是=（赋值）运算符。我在网上看过，讨论了多个主题。在我自己的笔记中，我把所有的例子都删掉了

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;  
}

$ b $ p在我在网上找到的所有引用中，我注意到操作符返回一个引用到源对象。
为什么是正确的方法返回一个对象的引用，而不是没有什么？

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分配链接。否则，将无法执行：

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;

不过，请记住，正确分配操作符比看起来更难。

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

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