为什么我们在赋值运算符重载而不是在正负运算中使用引用返回? [英] Why we use reference return in assignment operator overloading and not at plus-minus ops?

问题描述

正如我在书本和网络上阅读的那样，在C ++中，我们可以使用这些原型(作为class Money的成员函数)重载"plus"或"minus"运算符:

const Money operator +(const Money& m2) const;

const Money operator -(const Money& m2) const;

，并为赋值运算符提供以下信息:

const Money& operator =(const Money& m2);

为什么在赋值运算符重载而不在加号和减号运算符中使用对Money对象的引用作为返回值?

解决方案

从分配中返回引用允许链接:

a = b = c;  // shorter than the equivalent "b = c; a = b;"

(如果操作员返回新值的副本，这在大多数情况下也可以使用，但是通常效率较低.)

我们无法从算术运算中返回引用，因为它们会产生新的值.唯一(明智的)返回新值的方法是按值返回它.

像您的示例一样，返回常量值可以防止移动语义，所以不要这样做.

As I read in books and in the web, in C++ we can overload the "plus" or "minus" operators with these prototypes (as member functions of a class Money):

const Money operator +(const Money& m2) const;

const Money operator -(const Money& m2) const;

and for the assignment operator with:

const Money& operator =(const Money& m2);

Why use a reference to a Money object as a return value in the assignment operator overloading and not in the plus and minus operators?

解决方案

Returning a reference from assignment allows chaining:

a = b = c;  // shorter than the equivalent "b = c; a = b;"

(This would also work (in most cases) if the operator returned a copy of the new value, but that's generally less efficient.)

We can't return a reference from arithmetic operations, since they produce a new value. The only (sensible) way to return a new value is to return it by value.

Returning a constant value, as your example does, prevents move semantics, so don't do that.

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