47
votes

The following paper is the first proposal I found for template parameter packs.

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2004/n1603.pdf

At page 16, it talks about introducing two new operators [] and <> for accessing parameter pack elements and parameter pack types.

The suggested syntax for such an operator involves two new operators: .[] to access values and .<> to access types. For instance:

template<int N, typename Tuple> struct tuple_element;
template<int N, ... Elements>
struct tuple_element<tuple<Elements...> >
{
    typedef Elements.<N> type;
};

template<int N, ... Elements>
Elements.<N>& get(tuple<Elements...>& t)
{ return t.[N]; }

template<int N, ... Elements>
const Elements.<N>& get(const tuple<Elements...>& t)
{ return t.[N]; }

So where are these operators? If there is none, what is their replacement?

5
Their replacement is probably to put the parameter pack into a std::tuple and use std::get and std::tuple_element, which are implemented recursively. Also see this answerdyp
@DyP: std::get and std::tuple_element don't inherently need to be recursive, they can be implemented without recursion through indices and overload resolution / derived-to-base conversions.Xeo
@Xeo: Since variadic argument lists [currently] don't support access by index, I'm not aware of any implementation which isn't either recursive or types out all types (which will then fallback to a recursive approach to cover the theoretically unlimited amount of arguments). Although I'd implement the member access through base classes, determining the correct type to access would use std::tuple_element<...> which would be the recursive function.Dietmar Kühl
@Xeo As you can see from the answer below the linked answer, I did implement an equivalent to std::tuple_element with type deduction and indices -- using your O(logN) indices generator ;) But then the indices need to be created recursively (or manually).dyp

5 Answers

32
votes

C++11 doesn't have corresponding operators which is the reason they are proposed. With C++11 you'll need to either extract the corresponding information yourself or use a class which already does the necessary operation. The easiest approach is probably to just use std::tuple<T...> which already implements the corresponding logic.

If you wonder how std::tuple<T...> currently implements these operations: it is basically an exercise in functional programming using a fairly bad functional programming notation. Once you know how to get the n-th type of the sequence, getting the n-th element using inheritance from base classes parameterized on index and type is fairly trivial. Implementing something like tuple_element<N, T...> could look something like this:

template <int N, typename... T>
struct tuple_element;

template <typename T0, typename... T>
struct tuple_element<0, T0, T...> {
    typedef T0 type;
};
template <int N, typename T0, typename... T>
struct tuple_element<N, T0, T...> {
    typedef typename tuple_element<N-1, T...>::type type;
};

The actual more challenging bit in implementing something like std::tuple<T...> is conjuring up a list of indices so you got a parallel list of type and integers which can then be expanded, e.g., for a list of base classes using something like (how the internal details look exactly will differ but the basic idea of having a parallel parameters packs for the types and their indices will be somehow there):

template <typename... T, int... I>
class tuple_base<tuple_types<T...>, tuple_indices<I...>>:
     public tuple_field<T, I>... {
};
53
votes

Others have already answered that it can be done via std::tuple. If you want to access the Nth type of a parameter pack, you may find the following metafunction handy:

template<int N, typename... Ts> using NthTypeOf =
        typename std::tuple_element<N, std::tuple<Ts...>>::type;

Usage:

using ThirdType = NthTypeOf<2, Ts...>;
15
votes

Access N-th element?

Using std::forward_as_tuple:

template <int I, class... Ts>
decltype(auto) get(Ts&&... ts) {
  return std::get<I>(std::forward_as_tuple(ts...));
}

Example usage:

template<class...Ts>
void foo(Ts&&...ts){

  auto& first = get<0>(ts...);
  auto second = get<1>(ts...);

  first = 'H';
  second = 'E';

  (std::cout << ... << ts);
}

foo('h','e','l','l','o');
// prints "Hello"

This answer is to supplement Emile Cormier's answer which gives only the n-th type.

7
votes

To get the Nth element from a pack you can write:

Option 1

Using tuple_element for getting the return type for the Nth element:

template<size_t index, typename T, typename... Ts>
inline constexpr typename enable_if<index==0, T>::type
get(T&& t, Ts&&... ts) {
    return t;
}

template<size_t index, typename T, typename... Ts>
inline constexpr typename enable_if<(index > 0) && index <= sizeof...(Ts),
          typename tuple_element<index, tuple<T, Ts...>>::type>::type
get(T&& t, Ts&&... ts) {
    return get<index-1>(std::forward<Ts>(ts)...);
}

// below is optional - just for getting a more readable compilation error
// in case calling get with a bad index

inline template<long long index, typename... Ts>
constexpr bool index_ok() {
    return index >= 0 && index < sizeof...(Ts);
}

template<long long index, typename T, typename... Ts>
inline constexpr
typename enable_if<!index_ok<index, T, Ts...>(), T>::type
get(T&& t, Ts&&... ts) {
    static_assert(index_ok<index, T, Ts...>(),
        "bad index in call to get, smaller than zero or above pack size");
    return t;
}

Option 2

Without using tuple, relying on auto return type and specifically on C++14 decltype(auto) and on using enable_if as a template parameter and not as a return type:

template<size_t index, typename T, typename... Ts,
    typename enable_if<index==0>::type* = nullptr>
inline constexpr decltype(auto) get(T&& t, Ts&&... ts) {
    return std::forward<T>(t); 
}

template<size_t index, typename T, typename... Ts,
    typename enable_if<(index > 0 && index <= sizeof...(Ts))>::type* = nullptr>
inline constexpr decltype(auto) get(T&& t, Ts&&... ts) {
    return get<index-1>(std::forward<Ts>(ts)...);
}

template<long long index, typename... Ts>
inline constexpr bool index_ok() {
    return index >= 0 && index < (long long)sizeof...(Ts);
}

// block (compilation error) the call to get with bad index,
// providing a readable compilation error
template<long long index, typename T, typename... Ts,
    typename enable_if<(!index_ok<index, T, Ts...>())>::type* = nullptr>
inline constexpr decltype(auto) get(T&& t, Ts&&... ts) {
    static_assert(index_ok<index, T, Ts...>(),
        "bad index in call to get, smaller than zero or above pack size");
    return std::forward<T>(t); // need to return something...
                               // we hope to fail on the static_assert above
}

Usage example:

template<size_t index, typename... Ts>
void resetElementN(Ts&&... ts) {
    get<index>(std::forward<Ts>(ts)...) = {}; // assuming element N has an empty ctor
}

int main() {
    int i = 0;
    string s = "hello";
    get<0>(i,2,"hello","hello"s, 'a') += get<0>(2);
    get<1>(1,i,"hello",4) += get<1>(1, 2);
    get<3>(1,2,"hello",i) += get<2>(0, 1, 2);    
    get<2>(1,2,s,4) = get<2>(0, 1, "hi");
    cout << i << ' ' << s << endl;    
    resetElementN<1>(0, i, 2);
    resetElementN<0>(s, 1, 2);
    cout << i << ' ' << s << endl;    

    // not ok - and do not compile
    // get<0>(1,i,"hello","hello"s) = 5;
    // get<1>(1,i*2,"hello") = 5;
    // get<2>(1,i*2,"hello")[4] = '!';
    // resetElementN<1>(s, 1, 2);

    // ok
    const int j = 2;
    cout << get<0>(j,i,3,4) << endl;

    // not ok - and do not compile
    // get<0>(j,i,3,4) = 5;    

    // not ok - and do not compile
    // with a readable compilation error
    // cout << get<-1>("one", 2, '3') << endl;
    // cout << get<3>("one", 2, '3') << endl;
}

Code
Option 1: http://coliru.stacked-crooked.com/a/60ad3d860aa94453
Option 2: http://coliru.stacked-crooked.com/a/09f6e8e155612f8b

2
votes

We can implement a simple function to get nth parameter directly without any recursive calls but many pure type operations in compile-time. Let's look at the key code firstly:

template<class...Ts>
struct GetImp {
  template<class T, class...Us>
  static decltype(auto) impl(Ts&&..., T&& obj, Us&&...) {
    return std::forward<T>(obj);
  }
};

template<size_t n, class...Ts>
decltype(auto) get(Ts&&...args) {
  static_assert(n<sizeof...(args), "index over range");
  return Transform<GetImp, Before_s<n, Seq<Ts...>> >
    ::impl(std::forward<Ts>(args)...);
}

What does Transform means?

For example, if we have a type T that is std::tuple<int,double,float>, then Transform<GetImp,T> would be GetImp<int,double,float>. note that I define another empty struct "Seq" instead of std::tuple to do the same thing with less compile time.(In fact both of them could be compiled very quickly,but I guess an empty struct would be more effectively) So Before_s<n,Seq<Ts...>> generate a Seq<?> and then we transform it into GetImp, so that we can know what type of [0]~[n-1] parameters are, and then drop them off to index the nth parameter directly. For example, Before_s<3,Seq<T0,T1,T2,T3,T4...>> is Seq<T0,T1,T2>, Before_s<2,Seq<T0,T1,T2,T3,T4...>> is Seq<T0,T1> etc. We use Before_s to deal with our Seq type to reduce compile time when we use one meta function to implement another meta function for less compile time.

Implementation

#define OMIT_T(...) typename __VA_ARGS__::type

template<class...Args>
struct Seq { };

template< template<class...> class Dst >
struct TransformImp{
    template< template<class...>class Src, class...Args >
    static Dst<Args...> from(Src<Args...>&&);
};
template< template<class...> class Dst, class T>
using Transform = decltype(TransformImp<Dst>::from(std::declval<T>()));
template<class T>
using Seqfy = Transform<Seq, T>;


template<class...>struct MergeImp;
template<class...Ts, class...Others>
struct MergeImp<Seq<Ts...>, Seq<Others...>>
{
  using type = Seq<Ts..., Others...>;
};
template<class first, class second>
using Merge = OMIT_T(MergeImp<Seqfy<first>, Seqfy<second> >);
template<class T, class U>
using Merge_s = OMIT_T(MergeImp<T, U>);

template<size_t, class...>struct BeforeImp;

template<size_t n, class T, class...Ts>
struct BeforeImp<n, Seq<T, Ts...>> {
    static_assert(n <= sizeof...(Ts)+1, "index over range");
    using type = Merge_s<Seq<T>, OMIT_T(BeforeImp<n - 1, Seq<Ts...>>)>;
};

template<class T, class...Ts>
struct BeforeImp<1, Seq<T, Ts...>> {
    using type = Seq<T>;
};
template<class T, class...Ts>
struct BeforeImp<0, Seq<T, Ts...>> {
    using type = Seq<>;
};
template<size_t n>
struct BeforeImp<n, Seq<>> {
    using type = Seq<>;
};

template<size_t n, class T>
using Before = OMIT_T(BeforeImp<n, Seqfy<T>>);
template<size_t n, class T>
using Before_s = OMIT_T(BeforeImp<n, T>);

Edited: Advanced Implementation

We needn't use Before_s to calculate n-1 types before nth type,instead, we can ignore them:

struct EatParam{
    constexpr EatParam(...)noexcept{}
};

template<size_t n>
struct GenSeqImp {
  using type = Merge_s<OMIT_T(GenSeqImp<n / 2>), OMIT_T(GenSeqImp<n - n / 2>)>;
};
template<>
struct GenSeqImp<0> {
  using type = Seq<>;
};
template<>
struct GenSeqImp<1> {
  using type = Seq<EatParam>;
};

template<size_t n>
using GenSeq = OMIT_T(GenSeqImp<n>);


template<class...Ts>
struct GetImp {
  template<class T>
  static constexpr decltype(auto) impl(Ts&&..., T&& obj, ...)noexcept {
    return std::forward<T>(obj);
  }
};


template<size_t n, class...Ts>
constexpr decltype(auto) get(Ts&&...args)noexcept {
  static_assert(n<sizeof...(args), "index over range.");
  //return Transform<GetImp, Before_s<n, Seq<Ts...>> >
  return Transform<GetImp, GenSeq<n>>
    ::impl(std::forward<Ts>(args)...);
}

In addition , there is a very interesting article about implementation of getting nth type:

Thanks for their work, I didn't know we could use (...) to do the hack before.