Q:

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

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.

A:

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.

A:

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.

