Calling Funcs with Pointer Params vs Reference Params

Question

I was hoping someone could explain to me why a function with a reference parameter would be called rather than one with a pointer parameter, given something like the following:

int Func1 ( int & a );
int Func1 ( int * a );

int main()
{
   int y = 1;
   int & x = y;
   int * z = & y;

   Func1(y); // Calls reference version
   Func1(&y); // Calls pointer version
   Func1(z); // Calls pointer version

   return 0;
}

int Func1 ( int & a )
{
    cout << "Reference param Func1 called" << endl;
    a = a + 1;

    return a + 1;
}

int Func1 ( int * a )
{
    cout << "Pointer param Func1 called" << endl;
    *a = *a + 1;

    return *a + 1;
}

I'm confused as to how the decision is made to call the pointer parameter version of Func1 for the Func1(&y) call rather than the reference parameter version of Func1. Also, why is the reference parameter version of Func1 not chosen for the Func1(z) call? If z holds an address, I don't see why the address can't be passed into the reference parameter for Func1( &a ).

Answer 1

int Func1 ( int & a ); matches calls where the argument expression has type int (and is eligible to have a non-const reference taken to it). int Func1 ( int * a ); matches calls where the argument expression has type int*.

If z holds an address, I don't see why the address can't be passed into the reference parameter for Func1( &a ).

Well, you could achieve that by calling Func1(*z).