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 · 2014-06-17T16:54:01.723

2

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

edited Jun 17 '14 at 16:54

answered Jun 17 '14 at 16:18

WhozCraig

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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 Jun 18 '14 at 15:08
@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. – WhozCraig Jun 19 '14 at 03:19

score 0 · Answer 2 · answered Jun 17 '14 at 15:53

0

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)...);
}

answered Jun 17 '14 at 15:53

Jarod42

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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. – user2746401 Jun 17 '14 at 15:59
@user2746401: Why? It's a pretty clean way. – Xeo Jun 17 '14 at 16:28
@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. – user2746401 Jun 18 '14 at 15:13

Variadic template function: argument number for each argument

2 Answers2