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

Thomas picture Thomas · May 26, 2012 · Viewed 24.6k times · Source

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…

Answer

Kerrek SB picture Kerrek SB · May 26, 2012

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.