C++ 为什么赋值运算符应该返回一个 const ref 以避免 (a=b)=c [英] C++ why the assignment operator should return a const ref in order to avoid (a=b)=c

问题描述

我正在阅读一本关于 C++ 的书，更准确地说是关于运算符重载的书.

I am reading a book about C++ and more precisely about the operator overloading.

示例如下:

const Array &Array::operator=(const Array &right)
{
// check self-assignment
// if not self- assignment do the copying
return *this; //enables x=y=z
}

书中关于返回const ref而不是ref的解释是为了避免诸如(x=y)=z之类的赋值.我不明白为什么我们应该避免这种情况.我知道 x=y 在这个例子中首先被评估，因为它返回一个 const 引用，=z 部分不能被执行.但为什么?

The explanation provided by the book about returning const ref instead of ref is to avoid assignments such as (x=y)=z. I don't understand why we should avoid this. I understand that x=y is evaluated first in this example and since it returns a const reference the =z part cannot be executed. But why?

推荐答案

(x=y) 的意思是x.operator=(y)，返回对象<代码>x.因此，(x=y)=z 的意思是 (x.operator=(y)).operator=(z).parens中的表达式将x设置为y并返回x，然后外位将x设置为z.它不会像您期望的那样将 y 设置为 z，而表达式 x = y = z 也是如此.

(x=y) means x.operator=(y), which returns the object x. Therefore, (x=y)=z means (x.operator=(y)).operator=(z). The expression in parens sets x to y and returns x, and then the outer bit sets x to z. It does not set y to z as you might expect, and as the expression x = y = z does.

这种行为是违反直觉的(它们在赋值后应该都相等，对吗?)；返回一个 const 引用使它不可能并避免了这个问题.

This behavior is counter-intuitive (they should all be equal after the assignment, right?); returning a const reference makes it impossible and avoids the problem.

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