trailing return type using decltype with a variadic template function

Question

I want to write a simple adder (for giggles) that adds up every argument and returns a sum with appropriate type. Currently, I've got this:

#include <iostream>
using namespace std;

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

template <class T, class... P>
auto sum(const T& t, const P&... p) -> decltype(t + sum(p...))
{
   return t + sum(p...);
}

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

On GCC 4.5.1 this seems to work just fine for 2 arguments e.g. sum(2, 5.5) returns with 7.5. However, with more arguments than this, I get errors that sum() is simply not defined yet. If I declare sum() like this however:

template <class T, class P...>
T sum(const T& t, const P&... p);

Then it works for any number of arguments, but sum(2, 5.5) would return integer 7, which is not what I would expect. With more than two arguments I assume that decltype() would have to do some sort of recursion to be able to deduce the type of t + sum(p...). Is this legal C++0x? or does decltype() only work with non-variadic declarations? If that is the case, how would you write such a function?

This is an interesting problem. Maybe you should ask in the Usenet group comp.std.c++ whether this kind of "recursive call" in ->decltype(expr) is supposed to work or not. — sellibitze
It's not supposed to work by the current wording. The point of declaration of functions/variables etc.. is after their declarator. Thus, sum in the late specified return type cannot find the sum template being defined. — Johannes Schaub - litb
@Johannes: But isn't lookup simply delayed (until the 2nd phase) due to the expression's dependence on template parameters? — sellibitze
@sellibitze that's a good point, but it will depend on the types of the template parameters, because only argument dependent lookup is done at the instantiation context. If they are int and double like here, the function template won't be found. If there is a globally declared class among the arguments, the global sum will be found. So this is rather "random" when it finds the "sum", it doesn't work in general. — Johannes Schaub - litb
@DeagMG: Point taken. But this is a little bit outside the scope of the question. — sellibitze

sellibitze sellibitze · Accepted Answer · 2010-09-19T06:29:29

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>()
)>{};

trailing return type using decltype with a variadic template function

7 Answers