trailing return type using decltype with a variadic template function

I think the problem is that the variadic function template is only considered declared after you specified its return type so that sum in decltype can never refer to the variadic function template itself. But I'm not sure whether this is a GCC bug or C++0x simply doesn't allow this. My guess is that C++0x doesn't allow a "recursive" call in the ->decltype(expr) part.

As a workaround we can avoid this "recursive" call in ->decltype(expr) with a custom traits class:

#include <iostream>
#include <type_traits>
using namespace std;

template<class T> typename std::add_rvalue_reference<T>::type val();

template<class T> struct id{typedef T type;};

template<class T, class... P> struct sum_type;
template<class T> struct sum_type<T> : id<T> {};
template<class T, class U, class... P> struct sum_type<T,U,P...>
: sum_type< decltype( val<const T&>() + val<const U&>() ), P... > {};

This way, we can replace decltype in your program with typename sum_type<T,P...>::type and it will compile.

Edit: Since this actually returns decltype((a+b)+c) instead of decltype(a+(b+c)) which would be closer to how you use addition, you could replace the last specialization with this:

template<class T, class U, class... P> struct sum_type<T,U,P...>
: id<decltype(
      val<T>()
    + val<typename sum_type<U,P...>::type>()
)>{};

Apparently you can't use decltype in a recursive manner (at least for the moment, maybe they'll fix it)

You can use a template structure to determine the type of the sum

It looks ugly but it works

#include <iostream>
using namespace std;


template<typename... T>
struct TypeOfSum;

template<typename T>
struct TypeOfSum<T> {
    typedef T       type;
};

template<typename T, typename... P>
struct TypeOfSum<T,P...> {
    typedef decltype(T() + typename TypeOfSum<P...>::type())        type;
};



template <class T>
T sum(const T& in)
{
   return in;
}

template <class T, class... P>
typename TypeOfSum<T,P...>::type sum(const T& t, const P&... p)
{
   return t + sum(p...);
}

int main()
{
   cout << sum(5, 10.0, 22.2) << endl;
}