map C++ overloaded function over heterogenous tuple?

Question

In C++, is it possible to map an overloaded function over a heterogenous tuple? For example:

double f(dobule);
size_t f(std::string);

auto t = std::make_tuple(3.14, "a string");
// should be the same as std::make_tuple(f(std::get<0>(t)), f(std::get<1>(t)));
map(f, make_tuple(3.14, "a string")); // type std::tuple<double, size_t>

I can write a map function which maps the same overloaded instance of f to each tuple element (code below), but I don't see how to defer the overloading resolution of f from the call to map to the invocation of f inside map. Has anyone figured out how to do this?

Here's my code to map a single instance of an overloaded function over a tuple (where seq and gens are taken from this answer https://stackoverflow.com/a/7858971/431282):

// basically C++14's integer sequence
template<int ...>
struct seq { };

template<int N, int ...S>
struct gens : gens<N-1, N-1, S...> { };

template<int ...S>
struct gens<0, S...> {
  typedef seq<S...> type;
};

template<typename R, typename A, typename ...B, int ...S> auto
map_1(R (*f)(A), std::tuple<B...> &&t, seq<S...>)
  -> decltype(std::make_tuple(f(std::get<S>(t))...))
{
  return std::make_tuple(f(std::get<S>(t))...);
}

template<typename R, typename A, typename ...B> auto
map(R (*f)(A), std::tuple<B...> &&t)
  -> decltype(map_1(f, std::forward<std::tuple<B...>>(t), typename gens<sizeof...(B)>::type()))
{
  return map_1(f, std::forward<std::tuple<B...>>(t), typename gens<sizeof...(B)>::type());
}

Answer 1

The issue is that it can't be done with function pointers alone, since you want to resolve the function overload when it's bound to the argument. You can't get a function pointer to function without performing overload resolution. The key is to provide a function object that performs the overload on call, as opposed to trying to get a function pointer at the start.

For this, I would declare the primary function as

template<typename Func, typename ...B> auto
map(Func&& f, std::tuple<B...> &&t)
  -> decltype(map_1(std::forward<Func>(f), std::forward<std::tuple<B...>>(t), typename gens<sizeof...(B)>::type()))
{
  return map_1(std::forward<Func>(f), std::forward<std::tuple<B...>>(t), typename gens<sizeof...(B)>::type());
}

And then defined map_1 in a similar fashion.

Then you can make a function object wrapper

struct Wrapper
{
   template<typename T>
   auto operator()(T&& t) const -> decltype( f(std::forward<T>(t)) )
   {
       return f(std::forward<T>(t));
   }
};

and call it with map(Wrapper(), make_tuple(3.14, "a string"))

Edit: if you have C++14, you can do the following (thanks @MooingDuck for inspiration)

#define BINDOVERLOADED(X) [](auto&& t) { return X(std::forward<decltype(t)>(t)); }
auto x = map(BINDOVERLOADED(f), make_tuple(3.14, "a string"));

See http://coliru.stacked-crooked.com/a/439f958827de8cf2