Invert tuple arguments

Question

I want to write a template class InvTuple which defines type as a tuple of class arguments in inverse order. So it should work like

InvTuple<T1, T2, T3, ...>::type   --->   tuple<..., T3, T2, T1>

I defined it like so

template<class...T>
struct InvTuple;

template<class T1, class...T>
struct InvTuple < T1, T... >
{
    template<class... U>
    using doInvert = typename InvTuple<T...>::doInvert < U..., T1 > ;  
                     // <--- unrecognizable template declaration/definition, 
                     // syntax error : '<'

    using type = doInvert<>;
};

template<>
struct InvTuple <>
{
    template<class... U>
    using doInvert = tuple<U...>;

    using type = doInvert < > ;
};

But this does not compile due to the error as shown in the code. Please help me to understand what's wrong.

Answer 1

You need the template keyword:

using doInvert = typename InvTuple<T...>::template doInvert < U..., T1 > ;

and you also need to switch U... and T1 in the same line to have that work properly:

#include <iostream>
#include <tuple>
#include <typeinfo>
using namespace std; // Don't try this at home

template<class...T>
struct InvTuple;

template<class T1, class...T>
struct InvTuple < T1, T... >
{
    template<class... U>
    using doInvert = typename InvTuple<T...>::template doInvert < T1, U... >;

    using type = doInvert<>;
};

template<>
struct InvTuple <>
{
    template<class... U>
    using doInvert = tuple<U...>;

    using type = doInvert < > ;
};

int main()
{
    InvTuple<int,char,bool> obj;
    InvTuple<int,char,bool>::type obj2;
    cout << typeid(obj).name() << endl; // InvTuple<int, char, bool>
    cout << typeid(obj2).name() << endl; // std::tuple<bool, char, int>
}

Example