Question

To define a specialization that is used for every Vector of pointers and only for
Vectors of pointers, we need a partial specialization:

 template <class T> class  Vector <T *> : private Vector<void *> {
 public:
      typedef Vector<void*> Base;
      Vector(): Base() {}
      explicit Vector(int i) : Base(i ) {}
      T *& elem(int i ) { return static_cast <T *&> (Base::elem(i)); }
      T *& opeator[](int  i) { return static_cast <T *&>(Base::operator[](i )); }
      //...
 };

This definition has me in a tizzy. This is related to partial specialization but i don’t understand the syntax. private Vector<void *> definition part looks like a parent class to me.

1. Why not specify Vector <void *> in template <class T> class Vector <void *>.
2. It would be great if anybody can breakdown the definition part. (sorry if its too much to ask)

Answer 1

Forget about the inheritance, which has nothing to do with the problem at hand.

Partial specialization means that you make a new template from an existing one which is more specialized, but still generic, by matching a more restrictive pattern. The general pattern of your example is like this:

template <typename T> class Foo;      // primary template

template <typename U> class Foo<U*>;  // partial specialization

template <> class Foo<fool>;          // full specialization

The first line is the primary template and matches everything that is not matched by a more specialized form. The third line defines an actual type (not a template!) Foo<fool> (for some given type fool). The middle line, on the other hand, is still a template, but it only matches a type of the form T = U *, i.e. a pointer:

 Foo<char> x;   // uses primary template with T = char
 Foo<fool> y;   // uses full specialization (nothing to be matched)
 Foo<int*> z;   // uses partial specialization, matching U = int

About the Vector<void*>: It just turns out that the author chooses to define the partially-specialized Vector<U*> as deriving from a fixed class Vector<void*> (which would have to be fully specialized elsewhere).