Could not deduce type for T

This simple code snippet fails to work as expected for some reason, instead requiring me to explicitly specify type for T (for the function call foo):

Code:

template<typename T>
void foo(typename Stuff::mpl::ArgumentHelper<T>::type Arg)
{
	cout << Stuff::mpl::type_name<Stuff::mpl::add_const<typename Stuff::mpl::ArgumentHelper<T>::const_type>::type>::name << endl;
	Arg.x = 5;
}

class A { public: int x; };

namespace mpl = Stuff::mpl;

int main()
{
	A a;
	a.x = 0;
	foo(a);
	cout << a.x << endl;
}

Does anyone know why?
ArgumentHelper is defined as follows:

Code:

		namespace detail
		{
			template<typename T, bool IsFundamental> struct ArgumentHelper { typedef T type; typedef T const_type; };
			template<typename T> struct ArgumentHelper<T, false> 
			{
				typedef typename boost::add_reference<T>::type type;
				typedef typename Stuff::mpl::add_const<type>::type const_type;
			};
		}

		template<typename T> struct ArgumentHelper
		{
			typedef typename boost::remove_reference<typename boost::remove_const<T>::type>::type raw_type;
			typedef typename detail::ArgumentHelper<T, boost::is_fundamental<raw_type>::value>::type type;
			typedef typename detail::ArgumentHelper<T, boost::is_fundamental<raw_type>::value>::const_type const_type;
		};

Explanations are greatly appreciated O_O
EDIT: And how to work around it in a good way, if possible. Otherwise the whole thing is doomed.

as far as I know, there is know way to specify a typename as a parameter. I asked a similar question in the past (I can't seem to find), but basically it's because it would force the compiler to instantiate a template in order to do the lookup...I'm sure someone around here can give a more lucid explanation of course.