Why '(&b) -> f() ' is static binding?

I am puzzled by a face test, which source code is like below:

#include <iostream>

using namespace std ;

class base

{

public:

   virtual void f() { cout << "base::f()" << endl ; }

};

class derive : public base

{

public:

   void f() { cout << "derive::f()" << endl ; }

};

int main()

{

   base b ;

   (&b) -> f() ;//why this f() is static bound?

   (&b) -> ~base() ;

   new (&b) derive ;

   (&b) -> f() ;//why this f() is static bound?

   base *p = &b ;

   p->f() ;//it is easy to understand this line

   system("pause");

   return 0 ;

}

The result of this short program is so peculiar and like this:

base::f()

base::f()

derive::f()

In my opinion, the result of expression '(&b)' is a 'base*' type, so
it will cause a polymorphic behavior. But why '(&b) -> f();' is static
bound?

After referring some authoritative documents, I still can not find the
reason. So I am doubting that the standards did not clearly state this
situation. I am eager to know the reason and the fact.