loop parameter pack lambda - c++

Question

I have a small bit of code.
I want to know if i could cut out my ProcessArg function
I would like to loop the parameter pack from inside my call function.
Im i best to use a lambda function in initializer_list or something else, if so how do i do it.
Thanks
template <class R, class Arg>
R ProcessArg(Arg&& arg) {
std::cout << typeid(arg).name() << (std::is_reference<Arg>::value ? "&" : "") << std::endl;
//std::cout << std::boolalpha << std::is_reference<Arg>::value << std::endl; // not always true
return R();
}
template <typename R, typename... Args>
R CallFunction(Args&&... args) {
std::size_t size = sizeof...(Args);
std::initializer_list<R>{ProcessArg<R>(std::forward<Args>(args)) ...};
return R();
}
template<typename Fn> class FunctionBase;
template<typename R, typename... Args>
class FunctionBase <R(*)(Args...)> {
public:
FunctionBase() {}
R operator()(Args&&... args) { // Args&& is a universal reference
return CallFunction<R>(std::forward<Args>(args)...);
}
};
int foo(int a, int& b) {
std::cout << std::boolalpha << std::is_reference<decltype(a)>::value << std::endl; // falae
std::cout << std::boolalpha << std::is_reference<decltype(b)>::value << std::endl; // true
return a + b;
}
int main() {
int in = 10;
foo(1, in);
FunctionBase<decltype(&foo)> func;
func(1, in);
}

Answer 1

With c++17 fold-expressions, you can replace:
std::initializer_list<R>{ProcessArg<R>(std::forward<Args>(args)) ...};
with:
((std::cout << typeid(args).name()
<< (std::is_reference<Args>::value ? "&" : "")
<< std::endl
), ...);
Or with a lambda expression to improve readability:
auto process = [](auto&& arg) {
std::cout << typeid(arg).name()
<< (std::is_lvalue_reference<decltype(arg)>::value ? "&" : "")
<< std::endl;
};
(process(std::forward<Args>(args)), ...);
In c++20:
auto process = [] <typename Arg> (Arg&& arg) {
std::cout << typeid(arg).name()
<< (std::is_reference<Arg>::value ? "&" : "")
<< std::endl;
};
(process(std::forward<Args>(args)), ...);