Template specialization based on arguments of template parameter member function [duplicate] - c++

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This question already has answers here:
Where and why do I have to put the "template" and "typename" keywords?
(8 answers)
Closed 6 years ago.
I can run this code, however when I enable the 3 out-commented lines it does not compile anymore and gives the following error:
1>d:\git\testprojekt\testprojekt\testprojekt.cpp(41): warning C4346: 'first_argument<F>::type': dependent name is not a type
1> d:\git\testprojekt\testprojekt\testprojekt.cpp(41): note: prefix with 'typename' to indicate a type
1> d:\git\testprojekt\testprojekt\testprojekt.cpp(43): note: see reference to class template instantiation 'Bla<Type>' being compiled
1>d:\git\testprojekt\testprojekt\testprojekt.cpp(41): error C2923: 'DoStuff': 'first_argument<F>::type' is not a valid template type argument for parameter 'Arg'
1> d:\git\testprojekt\testprojekt\testprojekt.cpp(22): note: see declaration of 'first_argument<F>::type'
My idea why it isnt working is that the compiler wants to make sure Bla compiles for all sorts of template parameters, but first_argument can only handle template parameters that have the operator() defined.
Does anybody know how to make this example work?
I need it to select a class, here doStuff, based on whether the template parameters operator() accepts an argument or not, inside another templated class, here Bla.
#include <iostream>
template<typename F, typename Ret>
void helper(Ret(F::*)());
template<typename F, typename Ret>
void helper(Ret(F::*)() const);
template<typename F, typename Ret, typename A, typename... Rest>
char helper(Ret(F::*)(A, Rest...));
template<typename F, typename Ret, typename A, typename... Rest>
char helper(Ret(F::*)(A, Rest...) const);
template<typename F>
struct first_argument {
typedef decltype(helper(&F::operator())) type;
};
template <typename Functor, typename Arg = first_argument<Functor>::type>
struct DoStuff;
template <typename Functor>
struct DoStuff<Functor, char>
{
void print() { std::cout << "has arg" << std::endl; };
};
template <typename Functor>
struct DoStuff<Functor, void>
{
void print() { std::cout << "does not have arg" << std::endl; };
};
template <typename Type>
struct Bla
{
//DoStuff<typename Type> doStuff;
//void print() { doStuff.print(); };
};
int main()
{
struct functorNoArg {
void operator() () {};
};
struct functorArg {
void operator()(int a) { std::cout << a; };
};
auto lambdaNoArg = []() {};
auto lambdaArg = [](int a) {};
std::cout << std::is_same<first_argument<functorArg>::type,int>::value <<std::endl; // this works
DoStuff<functorArg> doStuff;
doStuff.print();
DoStuff<functorNoArg> doStuff2;
doStuff2.print();
DoStuff<decltype(lambdaArg)> doStuff3;
doStuff3.print();
DoStuff<decltype(lambdaNoArg)> doStuff4;
doStuff4.print();
Bla<functorArg> bla;
//bla.print();
return 0;
}
Thanks to all template nerds for helping :)

Answer 1

In your struct Bla you should say DoStuff<Type> doStuff; (typename is not needed or allowed here).
In (corrected version):
template <typename Functor, typename Arg = typename first_argument<Functor>::type>
struct DoStuff;
You were missing typename before first_argument<Functor>::type.