Comment by monk_the_dog | Hacker News Reader

Comment by monk_the_dog | Hacker News Reader

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| but a more technical understanding requires the knowledge that rvalue references can bind to any l- or rvalue

I don't think that's true. rvalue references can only bind to rvalues.

Here's a complete program. foo(i) won't compile bacause i is not an rvalue. bar(i) will compile, and foo(1) will compile.

#include <iostream>

using namespace std;

int foo(int&& i){ return i; }

template<class T> T bar(T&& i){ return i; }

int main() { int i=0; cout << foo(i) << endl;

   return 0;

}

edit: I am replying humbly. C++ is a complex language and I might very well be wrong.

I did not mean "any rvalue-reference" and bind to "any l- or rvalue", but that you can have an "int &&", "int& &&", "const int& &&", and even "int&& &&". I updated my comment to say "can be made from" instead "can bind to".

Your code example does not work because you call "foo" instead of "bar". But consider this one:

#include <iostream>

using namespace std;

template<class T> T bar(T&& i){ return std::forward<T>(i); }

int main() {

  int i=0;

  cout << bar<int&&>(6) << endl;

  return 0;

}

Maybe you're confusing reference collapsing with universal references?

In your example, replace bar<int&&>(6) to bar<int&&>(i).

It won't compile because i is not an r value. bar(i) will compile, because the template parameter is a universal reverence that will become an lvalue (not an rvalue) through reference collapsing.

I called "bar" with "<int&&>" to demonstrate that you can bind an rvalue-ref to an rvalue-ref, not to suggest that's how it should be called.

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