Specialize function template result

Question

I'm pretty new to template metaprogramming and can't find my thinking error in this approach:

template <typename T>
    typename T::ReturnType Query(const std::string& Str);

template <>
ResultTypeRowCount Query(const std::string& Str) { return this->queryRowCount(Str); }

ResultTypeRowCount implements a public typedef with the name ReturnType

Thankyou for reading

Answer 1

Specializing your template should follow this pattern:

template<typename T>
  void foo() {
  }

template<>
  void foo<int>() {
  }

Answer 2

It should be:

template <>
ResultTypeRowCount::ReturnType Query<ResultTypeRowCount>(const std::string& Str) { return this->queryRowCount(Str); }