typename prefix in C++ template

Below code, extended from the code in ``C++ template the complete guide'' Chp 9.3.2, shows why we need the typename prefix in C++ template

#include <iostream>


template <typename T>
class Foo 
{
public:

    enum {
        x
    };
};


template <typename T>
class Bar 
{
public:
    int y;
    Bar()
        :y(10) 
        {}
    
    void poof() 
        {
            Foo<T>::x*y;        // declaration or multiplication?
            std::cout << "poof" << y << std::endl;
            // the executabel compiled by g++ will print the poof10
            // so g++ think Foo<T>::x*y is multiplication
            // the executable compiled by CC will print the poof0
            // so CC think Foo<T>::x*y; is declaration of a pointer

            // after you add the typename prefix before Foo<T>::x*y;
            // then both g++ and CC know it is a associated type
        }
};

// evil specialization
template <>
class Foo<void> 
{
public:
    typedef int x;
};


int main() 
{
    
    Bar<void> bomb ;
    bomb.poof();
}

The compiler will be confused with

Foo<T>::x*y;

It may be multiplication if T is not void, or a pointer declaration if T is void. g++ and CC has different default assumptions, g++ think it is multiplication, but CC thinks it is declaration of a pointer.

When do we need add the typename prefix? - from ``C++ template the complete guide'' Chp 9.3.2,