explanation of typename keyword?

In "The C++ Standard Library" Josuttis gives this example(on p.11):

Code:

template <class T>
class MyClass{
	typename T::SubType * ptr;
	...
};

He says that without the typename keyword, SubType would be considered a static member. Ok, that seems to make some sense because a static member can be referenced like this:

Code:

#include <iostream> //cout
using namespace std;

class Box  //T = Box
{
public:
	static int SubType;
};

//intialize static data member:
int Box::SubType = 4;

int main()
{
	cout<<Box::SubType<<endl;

	return 0;
}

He continues his explanation by saying:

Thus,

T::SubType * ptr;

would be a multiplication of value SubType of type T with ptr.

I don't understand why he says that SubType is a value of type T? It seems to me that when you write

T::SubType

that only says that SubType is a static member of T, and it doesn't speak to the type of SubType at all. In my example above, the statement:

Code:

cout<<Box::SubType<<endl;

does not say that SubType is of type Box. It just says to go look in the Box class for a static member variable called SubType. The type of SubType is actually int.

You need the typename keyword when you use a type that's declared in a class that hasn't been defined yet. A type, not a variable.

>> I don't understand why he says that SubType is a value of type T?

I think he means that Subtype is a value within the class T.
here's another example of how it's supposed to work:

Code:

struct A 
{
 typedef int number;
};

struct B : A 
{
 static int number;
};

int B::number = 0;

template <class T>
struct C
{ 
 C()
 {  
  typename T::number n = 2;
  T::number = n;
 }
};

int
main(void)
{
 C<B> c; 
};

oddly enough, that code chokes the Borland compiler - but it should work for you if you're using a standard-compliant one.