在示波器分辨率运算符的左侧使用“＆”号？

Question

I'm trying to understand a piece of code. I know how to use scope resolution operator. The syntax is

return-type class-name::operator+(argument list)

In my code I have the following form

const Rectangle & Rectangle ::operator =(const Rectangle & rhs)

I want to know that what is the purpose of using & on the left hand side of scope resolution operator.

Answer 1

Lets take that argument , it's declared as

const Rectangle & rhs

The argument name is rhs , it's type is const Rectangle & .

Now lets go back to the return type. The return type is

const Rectangle &

Note that the ampersand is part of the return type .

It might be easier to see if it's written as

const Rectangle&

(without space between Rectangle and the & ).

That means the function returns a reference to a constant Rectangle object.

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Perhaps it's easier to illustrate using your

return-type class-name::operator =(argument list)

Placing the function and the above line next to each other (with some added spacing) we have:

return-type       class-name::operator =(argument list)
const Rectangle & Rectangle ::operator =(const Rectangle & rhs)