不传递参数实例的可变参数递归

出于某种原因，我似乎无法将此操作与 C++11 一起使用。

template<typename T>
int value(){ return 1;}


template<typename T1>
constexpr int SumCpp11(){
  return value<T1>();
}

template<typename T1, typename... T>
constexpr int SumCpp11(){
  return value<T1>() + SumCpp11<T...>();
}

int main(int argc, char* argv[])
{

   return  SumCpp11<int, int, double, double>();
   
}

编译器愉快地展开递归直到终止，然后抱怨不明确的声明。我已经反复修改了这个。我不能使用初始化列表技巧，因为我需要留在 C++11 中，这需要是一个 constexpr，因为这个最终值在代码的另一部分被用作模板参数。

clang++ --std=c++11 tailVar.cpp 
tailVar.cpp:13:24: error: call to 'SumCpp11' is ambiguous
  return value<T1>() + SumCpp11<T...>();
                       ^~~~~~~~~~~~~~
tailVar.cpp:13:24: note: in instantiation of function template specialization 'SumCpp11<double, double>' requested here
tailVar.cpp:13:24: note: in instantiation of function template specialization 'SumCpp11<int, double, double>' requested here
tailVar.cpp:20:12: note: in instantiation of function template specialization 'SumCpp11<int, int, double, double>' requested
      here
   return  SumCpp11<int, int, double, double>();
           ^
tailVar.cpp:7:15: note: candidate function [with T1 = double]
constexpr int SumCpp11(){
              ^
tailVar.cpp:12:15: note: candidate function [with T1 = double, T = <>]
constexpr int SumCpp11(){
              ^
1 error generated.

回答

template<typename T1, typename T2, typename... T>
constexpr int SumCpp11(){
  return value<T1>() + SumCpp11<T2, T...>();
                                //  ^^^^ when T... is empty, 
                                // the first overload is chosen
}