Question

Is it possible to write a C++ template owner_of<...> such that given this code:

struct X { int y; }

owner_of<&X::y>::type is X?

Answer 1

You can almost do that (or at least I could not find a better solution so far):

#include <string>
#include <type_traits>

using namespace std;

template<typename T>
struct owner_of { };

template<typename T, typename C>
struct owner_of<T (C::*)>
{
    typedef C type;
};

struct X
{
    int x;
};

int main(void)
{
    typedef owner_of<decltype(&X::x)>::type should_be_X;
    static_assert(is_same<should_be_X, X>::value, "Error" );
}

If you mind the use of decltype, maybe a macro could do:

#define OWNER_OF(p) owner_of<decltype( p )>::type

int main(void)
{
    typedef OWNER_OF(&X::x) should_be_X;
    static_assert(is_same<should_be_X, X>::value, "Error" );
}

An alternative solution based on decltype:

template<typename T, typename C>
auto owner(T (C::*p)) -> typename owner_of<decltype(p)>::type { }

int main(void)
{
    typedef decltype(owner(&X::x)) should_be_X;
    static_assert(is_same<should_be_X, X>::value, "Error" );
}