C++11: I can go from multiple args to tuple, but can I go from tuple to multiple args?

Question

Possible Duplicate:
How do I expand a tuple into variadic template function's arguments?
“unpacking” a tuple to call a matching function pointer

In C++11 templates, is there a way to use a tuple as the individual args of a (possibly template) function?

Example:
Let's say I have this function:

void foo(int a, int b)  
{  
}

And I have the tuple auto bar = std::make_tuple(1, 2).

Can I use that to call foo(1, 2) in a templaty way?

I don't mean simply foo(std::get<0>(bar), std::get<1>(bar)) since I want to do this in a template that doesn't know the number of args.

More complete example:

template<typename Func, typename... Args>  
void caller(Func func, Args... args)  
{  
    auto argtuple = std::make_tuple(args...);  
    do_stuff_with_tuple(argtuple);  
    func(insert_magic_here(argtuple));  // <-- this is the hard part  
}

I should note that I'd prefer to not create one template that works for one arg, another that works for two, etc…

Solution 1

Try something like this:

// implementation details, users never invoke these directly
namespace detail
{
    template <typename F, typename Tuple, bool Done, int Total, int... N>
    struct call_impl
    {
        static void call(F f, Tuple && t)
        {
            call_impl<F, Tuple, Total == 1 + sizeof...(N), Total, N..., sizeof...(N)>::call(f, std::forward<Tuple>(t));
        }
    };

    template <typename F, typename Tuple, int Total, int... N>
    struct call_impl<F, Tuple, true, Total, N...>
    {
        static void call(F f, Tuple && t)
        {
            f(std::get<N>(std::forward<Tuple>(t))...);
        }
    };
}

// user invokes this
template <typename F, typename Tuple>
void call(F f, Tuple && t)
{
    typedef typename std::decay<Tuple>::type ttype;
    detail::call_impl<F, Tuple, 0 == std::tuple_size<ttype>::value, std::tuple_size<ttype>::value>::call(f, std::forward<Tuple>(t));
}

Example:

#include <cstdio>
int main()
{
    auto t = std::make_tuple("%d, %d, %d\n", 1,2,3);
    call(std::printf, t);
}

With some extra magic and using std::result_of, you can probably also make the entire thing return the correct return value.

Solution 2

Create an "index tuple" (a tuple of compile-time integers) then forward to another function that deduces the indices as a parameter pack and uses them in a pack expansion to call std::get on the tuple:

#include <redi/index_tuple.h>

template<typename Func, typename Tuple, unsigned... I>  
  void caller_impl(Func func, Tuple&& t, redi::index_tuple<I...>)  
  {  
    func(std::get<I>(t)...);
  }

template<typename Func, typename... Args>  
  void caller(Func func, Args... args)  
  {  
    auto argtuple = std::make_tuple(args...);  
    do_stuff_with_tuple(argtuple);
    typedef redi::to_index_tuple<Args...> indices;
    caller_impl(func, argtuple, indices());
  }

My implementation of index_tuple is at https://gitlab.com/redistd/redistd/blob/master/include/redi/index_tuple.h but it relies on template aliases so if your compiler doesn't support that you'd need to modify it to use C++03-style "template typedefs" and replace the last two lines of caller with

    typedef typename redi::make_index_tuple<sizeof...(Args)>::type indices;
    caller_impl(func, argtuple, indices());

A similar utility was standardised as std::index_sequence in C++14 (see index_seq.h for a standalone C++11 implementation).