C++ Why should each function have a reference qualifier when defining overloads and reference functions

Question

I can't understand ：Why if a member function has a reference qualifier, all versions with the same list must have a reference qualifier. Is it because multiple definitions are created when this function is called?

class A{
   public:
         A sorter() &&;
         A sorter();
         A operator+ (const A&);
       }

Although there is only function declaration above, it should be enough

A a , b;
(a+b).sorter();

Does the definition and calling of a function like the above make the compiler unable to recognize which member function to use？

It is the same for regular parameters: `void f(int)` versus `void f(int&&)`. — Jarod42, Apr 25 '20 at 08:14

John Zwinck · Accepted Answer · 2020-04-25T07:22:11.297

3

When one overload is reference-qualified, the others must be as well. This works:

     A sorter() &&;
     A sorter() &;

Put another way: you can choose between two sets of possible overloads:

const and non-const (unqualified).
const& (synonymous with const in 1), & (non-const), and && (rvalue/temporary).

edited Apr 25 '20 at 07:22

answered Apr 25 '20 at 07:15

John Zwinck

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That's what I'm asking about，What's the meaning of doing this? To eliminate syntax errors, if only one member function is defined with a reference, will double definitions be created? Or just the syntax of C + + – Nater Apr 25 '20 at 07:27
1

@Nater: This is because a unqualified member function corresponds to both `&` and `&&`, i.e. it would be ambiguous. For more on this: https://stackoverflow.com/questions/35389911/why-is-overloading-on-just-one-ref-qualifier-not-allowed – John Zwinck Apr 25 '20 at 08:04

C++ Why should each function have a reference qualifier when defining overloads and reference functions

1 Answers1