Perfect forwarding references

Perfect Forwarding References with class functions

Given that “beauty is in the eye of the beholder”, and having just learned about std::forward, let’s proceed to make our beautiful event class perfectly forward the operator() arguments to its _func pointer.

template<typename... TArguments>
void Event<TArguments...>::operator()(TArguments&&... args) const
{
	if(_func)
		_func(std::forward<TArguments>(args)..., _userdata);
}

Yes, very good. But then what is this error?

cannot convert argument 1 from ‘int’ to ‘int &&’

After too much time isolating the issue, it can be minimally reproduced with

template<typename T>
struct PerfectForwardingProblem {
	void ClassFunc(T&& arg) {}
};

template<typename T>
void Func(T&& arg) {}

void Demo()
{
	PerfectForwardingProblem<int> pfp;
	int x = 1;

	pfp.ClassFunc(1); // OK
	Func(1); // OK

	pfp.ClassFunc(x); //cannot convert argument 1 from 'int' to 'T &&'
	Func(x); // OK

	pfp.ClassFunc(std::move(x)); // OK
	Func(std::move(x)); // OK
}

Why? Because, to quote cppreference.com:

std::forward
forwarding references
a function argument that is declared as an rvalue reference to a cv-unqualified function template parameter

So does not work on member function using types from the class template.
It seems that many have tried, as seen from Perfect forwarding with class template argument deduction