Accept any kind of callable and also know argument type

Question

I\'m not sure if it\'s possible, so that\'s what I want to find out.

I\'d like to create a function which accepts any kind of functor/callable object, but I want to

Answer 1

template < typename T >
void option( function< void(T) > )
{
    cout << typeid( T ).name() << endl;
}

template < typename T >
void option( void (*func)(T) )
{
    option( function< void(T) >( func ) );
}

template< typename F, typename A >
void wrapper( F &f, void ( F::*func )( A ) const )
{
    option( function< void(A) >( bind( func, f, placeholders::_1 ) ) );
}

template< typename F, typename A >
void wrapper( F &f, void ( F::*func )( A ) )
{
    option( function< void(A) >( bind( func, f, placeholders::_1 ) ) );
}

template < typename T >
void option( T t )
{
    wrapper( t, &T::operator() );
}

void test( int )
{
}

struct Object
{
    void operator ()( float )
    {
    }
};

int main( int, char *[] )
{
    Object obj;

    option( test );
    option( [](double){} );
    option( obj );

    return 0;
}

Based on information found here c++0x: overloading on lambda arity, which I found through @dyps link

This isn't the best solution, since it requires overloads for const/non-const/volatile etc. It does get the job done in terms of the original problem I was trying to solve...