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

{ Double-spacing corrected. -mod }

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.

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