没有用于调用可变参数包函数的匹配函数

Question

Consider:

#include<tuple>

template<int N,typename... Vs,typename... Ts>
void fog( const std::tuple<Vs...>& vs , const std::tuple<Ts...> & ts )
{
}

template<typename...Vs,typename...Ts >
int gof( const std::tuple<Vs...>& vs , const std::tuple<Ts...> & ts )
{
  fog<0,Vs...,Ts...>(vs,ts);
}

int main()
{
  std::tuple<int,double> t;
  gof(t,t);
}

Why does the compiler (g++-4.6) not find the fog function and how to make it find it?

error: no matching function for call to ‘fog(const std::tuple<int, double>&, const std::tuple<int, double>&)’
note: candidate is:
note: template<int N, class ... Vs, class ... Ts> void fog(const std::tuple<Vs ...>&, const std::tuple<_Tail ...>&)

Yes, I need the integral template parameter N . (This is a boiled down example.)

Answer 1

Don't expand the parameter packs:

fog<0>(vs,ts);

Otherwise the compiler doesn't know which template parameters belong to which tuple. This way, the tuples' template parameters get deducted as usual.

Answer 2

Write:

fog<0>(vs,ts);  

instead of

fog<0,Vs...,Ts...>(vs,ts);

And let the compiler deduce the types.

As for why the second form doesn't work, because the variadic parameter can be only the last parameter. There cannot be two variadic template parameters of that form.