What does && mean at the end of a function signature (after the closing parenthesis)? [duplicate]

Question

In Workarounds for no 'rvalue references to *this' feature, I see the following member function (a conversion operator):

template< class T > struct A {     operator T&&() && // <-- What does the second '&&' mean?     {         // ...     } }; 

What does the second pair of && mean? I am not familiar with that syntax.

Answer 1

This is a ref-value qualifier. Here is a basic example:

// t.cpp #include <iostream>  struct test{   void f() &{ std::cout << "lvalue object\n"; }   void f() &&{ std::cout << "rvalue object\n"; } };  int main(){   test t;   t.f(); // lvalue   test().f(); // rvalue } 

Output:

$ clang++ -std=c++0x -stdlib=libc++ -Wall -pedantic t.cpp $ ./a.out lvalue object rvalue object 

Taken from here.

Answer 2

It indicates that the function can be invoked only on rvalues.

struct X {       //can be invoked on lvalue       void f() & { std::cout << "f() &" << std::endl; }        //can be invoked on rvalue       void f() && { std::cout << "f() &&" << std::endl; } };  X x;  x.f();  //invokes the first function         //because x is a named object, hence lvalue  X().f(); //invokes the second function           //because X() is an unnamed object, hence rvalue 

Live Demo output:

f() & f() && 

Hope that helps.