On Wed, May 4, 2011 at 10:02 AM, Arno Schödl <aschoedl@think-cell.com> wrote:
Hello,
is there a way to get the result type of an overloaded member function? Something like:
template< class T, class A >
typename boost::result_of< T::func(A) >::type invoke( T& t, A& a ) {
return t.func(a);
}
Not exactly. The template parameters to result_of are types. In your example, assuming func is the name of an overloaded member function, T::func is not a type. To call a pointer to member function, assuming func is a template, you could do the following. template< class T, class F, class A > typename boost::result_of< F(A) >::type invoke_ptr(T t, F f, A a) { return (t.*f)(a); } struct S0 { template<class T> T func(T x) { return x; } }; invoke_ptr(S0(), &S0::func<int>, 1); invoke_ptr(S0(), &S0::func<float>, 0.1); It would also work with non-template overloads, but then you need to specify the type of the member function pointer explicitly. struct S1 { int func(int x) { return x; } float func(float x) { return x; } }; invoke_ptr<S1, int(S1::*)(int), int>(S1(), &S1::func, 1); invoke_ptr<S1, float(S1::*)(float), float>(S1(), &S1::func, 0.1); Now, let's assume instead that func is actually a callable object type; e.g. a functor. Then you could do something like the following, which is closer to your original example. template< class T, class A > typename boost::result_of< typename T::func(A) >::type invoke(T t, A a) { return typename T::func()(a); } struct S2 { struct func { template<class T> T operator()(T x) { return x; } }; }; invoke(S2(), 1); invoke(S2(), 0.1); I compiled these examples on boost trunk with g++ 4.5 with c++0x and BOOST_RESULT_OF_USE_DECLTYPE defined. But you could use the TR1 result_of protocol to make the last example work in C++03 as well. Hope that helps! Daniel Walker