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?
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?
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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