c++ - What does the single ampersand after the parameter list of a member function declaration mean?

Question

From the answer here.

class wrap {
public:
   operator obj() const & { ... }   //Copy from me.
   operator obj() && { ... }  //Move from me.
private:
   obj data_;
};

I know the && means that the member will be invoked when the object is an rvalue reference. But what does the single ampersand mean? How is it different than without the ampersand?

Answer 1

It means the member will be invoked when the object is an lvalue reference.

[C++11: 9.3.1/5]: A non-static member function may be declared with a ref-qualifier (8.3.5); see 13.3.1.

[C++11: 13.3.1/4]: For non-static member functions, the type of the implicit object parameter is

• “lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier
• “rvalue reference to cv X” for functions declared with the && ref-qualifier

where X is the class of which the function is a member and cv is the cv-qualification on the member function declaration. [..]

(and some more rules that I can't find)

Without a ref-qualifier, the function can always be invoked, regardless of the value category of the expression through which you're invoking it:

struct foo
{
    void bar() {}
    void bar1() & {}
    void bar2() && {}
};

int main()
{
    foo().bar();  // (always fine)
    foo().bar1(); // doesn't compile because bar1() requires an lvalue
    foo().bar2();
    
    foo f;
    f.bar();      // (always fine)
    f.bar1();
    f.bar2();     // doesn't compile because bar2() requires an rvalue
}

Live demo ^{(thanks Praetorian)}