Question

Consider the code below. What is going on there?

#include <iostream>
#include <stdio.h>

using namespace std;

template<class T>
void test(T &t)  // but why  int&& is valid here,when  instantiated with test<int&>??
{
  puts("T&");
}

int tref(int&& param)    //compile  error
{
  puts("tref&");
}

int main()
{
  int i;
  test<int&>(i);
}

Answer 1

I am not sure I follow, if you mean int&&in the first one because T is int& you are wrong. When T is int&, the T& in the signature is still int&. Also, int&& has nothing to do with composing int& with an extra &. && has a specific meaning: rvalue-reference.

§8.3.2[dcl.ref]/5

There shall be no references to references, no arrays of references, and no pointers to references. […]

§8.3.2[dcl.ref]/6

If a typedef (7.1.3), a type template-parameter (14.3.1), or a decltype-specifier (7.1.6.2) denotes a type TR that is a reference to a type T, an attempt to create the type “lvalue reference to cv TR” creates the type “lvalue reference to T”, while an attempt to create the type “rvalue reference to cv TR” creates the type TR.