How to pass overloaded function to an operator?

I need to pass a function to an operator. Any unary function having correct arg type. Return type can be anything. Because this is library code, I can not wrap it or cast f to specific overload (outside of operator*). Function takes operator* 1st arg as it own argument. Artificial example below compiles and returns correct results. But it has hardcoded int return type—to make this example compile.

#include <tuple>
#include <iostream>
using namespace std;

template<typename T>
int operator* (T x,  int& (*f)(T&) ) {
    return (*f)(x);
};

int main() {
    tuple<int,int>  tpl(42,43);
    cout << tpl * get<0>;
}

Is it possible to make operator* to accept f with arbitrary return type?

UPDATE - GCC bug? Code:

#include <tuple>

template<typename T, typename U> 
U operator* (T x,  U& (*f)(T&) ) {  
    return (*f)(x);
}; 

int main() {
    std::tuple<int,int>  tpl(42,43);
    return   tpl * std::get<0,int,int>;
}  

Compiles and runs correctly with gcc462 and 453 but is reject with gcc471 and 480. So it is possible GCC regression bug. I've submitted bug report: http://gcc.gnu.org/bugzilla/show_bug.cgi?id=54111

EDIT I've changed example to use tuple as arg - it was possible trivially deduce return type in previous example.

EDIT2 Many people could not understand what is needed, so I've changed call function to operator* to make example more real.