Variadic template function: argument number for each argument

Question

I'm playing around with variadic function templates in C++11 and have got the basic idea with code something like:

void helper() { std::cout << "No args" << std::endl; } template< typename T > void helper( T&& arg ) { size_t n = 0; std::cout << "arg " << n << " = " << arg << std::endl; helper(); } template< typename T, typename... Arguments > void helper( T&& arg, Arguments&& ... args ) { size_t n = sizeof...( args ); std::cout << "arg " << n << " = " << arg << std::endl; helper( args... ); }

However, what I want is for the argument number (the variable n in the code) to count up rather than down. How can I do this elegantly? I could write a wrapper function that creates a 'hidden' argument count but I feel there should be a neater way?

Thanks!

WhozCraig · Accepted Answer · 2014-06-17 16:54:01Z

Do you mean something like this? I think I understand what you're looking for, but if not I'll not hesitate to drop this answer:

#include <iostream> #include <utility> void helper() { std::cout << "no args" << std::endl; } template<size_t N, typename T> void helper(T&& arg) { std::cout << "arg " << N << " = " << arg << std::endl; } template<size_t N, typename T, typename... Args> void helper(T&& arg, Args&&... args) { helper<N>(std::forward<T>(arg)); helper<N+1, Args...>(std::forward<Args>(args)...); } template<typename T, typename... Args> void helper(T&& arg, Args&& ... args) { helper<0>(std::forward<T>(arg), std::forward<Args>(args)... ); } int main() { helper(); std::cout << '\n'; helper("single"); std::cout << '\n'; helper("one", 2U); std::cout << '\n'; helper(1,"two", 3.0, 4L); std::cout << '\n'; return 0; }

Output

no args arg 0 = single arg 0 = one arg 1 = 2 arg 0 = 1 arg 1 = two arg 2 = 3 arg 3 = 4

See it live

This is almost identical to the solution I've currently been using with a 'hidden' N so not quite what I'm getting at. Don't drop the answer though because it's imo the neatest so far :)
@user2746401 Sure. The thing I was trying to get to was calculating the parameter index via a template-argument that carries it along. It doesn't introduce any run-time variable, which I understood (perhaps incorrectly) was your goal in doing this while addressing ascend-vs-descend index direction. In the end the template arguments at compile-time manage the index; not a passed-along var. Its all a compile-time expansion (as yours is) but in the right direction. Glad it was worth some brain-food.

Jarod42 · Accepted Answer · 2014-06-17 15:53:46Z

You may do the following: live example

#if 1 // Not in C++11 // make_index_sequence #include <cstdint> template <std::size_t...> struct index_sequence {}; template <std::size_t N, std::size_t... Is> struct make_index_sequence : make_index_sequence<N - 1, N - 1, Is...> {}; template <std::size_t... Is> struct make_index_sequence<0u, Is...> : index_sequence<Is...> {}; #endif // make_index_sequence namespace detail { template<std::size_t...Is, typename... Ts> void helper(index_sequence<Is...>, Ts&&...args) { int dummy[] = {0, ((std::cout << "arg " << Is << " = " << args << std::endl), 0)...}; static_cast<void>(dummy); } } void helper() { std::cout << "No args" << std::endl; } template<typename ... Ts> void helper(Ts&&... args) { detail::helper(make_index_sequence<sizeof...(Ts)>(), std::forward<Ts>(args)...); }

Thanks but, as mentioned in my question, I was hoping to avoid writing such a wrapper that then creates a hidden argument with the argument numbers.
@Xeo: imo I don't think the answer is particularly clean and I think my current solution (similar to WhozCraig's) is cleaner and more readable.Perhaps I didn't phrase my question as best I could. I'm trying to find out if there is an inbuilt way - integral to variadic templates - that carries the original argument number along in the parameter pack.

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