模板类问题的部分专业化！

Question

The following code is confusing me

//partial specialization of vector 
template<class t>
class vector
{....
};
template<class t>
//teacher said that this partial specialization will handle all type of pointers
class vector<t*>
...
};

This is confusing me a lot, Suppose t is a char* , as the compiler will first look for the complete specialization as there is no complete specialization here , the partial specialization would be considered. Now it will become char** in partial specialization as t is char*, then it means we are not achieving the functionality that we want. Please Explain in more understandable manner.

Answer 1

The compiler will always looks for the match that is most specialized (and will fail if there's ambiguity).

Since char** matches both patterns:

char** = T*  (with T = char*)
char** = T   (with T = char**)

and the former is more specialized, that will be chosen. (And since char** is really a pointer to something, I think "we are not achieving the functionality that we want" is false.)

---

Edit.

Once the specialization is chosen, all other candidates will be thrown out (recursive substitution will not happen in the matching phase). For instance,

template<class T> struct A {
  typedef T type;
  static const int value = 0;
};
template<class T> struct A<T*> {
  typedef T type;
  static const int value = 12;
};
template<class T> struct A<T********> {
  typedef T type;
  static const int value = 1024;
};
...
A<char**>::type x;                          // <-- Line X
std::cout << A<char**>::value << std::endl; // <-- Line Y

At Line X and Y, the template A is to be instantiated. You give the first template parameter as char** . A expects either a pattern of T******** or T* or T . The 2nd one is the best match, so template<class T> struct A<T*> will be chosen. The other two will be stroke out.

# FAIL. Not specialized enough. Remove this from consideration.
//  template<class T> struct A {
//    typedef T type;
//    static const int value = 0;
//  };

# WIN. Use with T = char*.
    template<class T> struct A<T*> {
      typedef T type;
      static const int value = 12;
    };

# FAIL. Does not match char**.
// template<class T> struct A<T********> {
//   typedef T type;
//   static const int value = 1024;
// };

After performing substitution it becomes

struct A<char**> {
  typedef char* type;
  static const int value = 12;
};

and Line X and Y shall become

/*A<char**>::type*/ char* x;                       // <-- Line X
std::cout << /*A<char**>::value*/ 12 << std::endl; // <-- Line Y

---

(Note:

template<class T> class Vector

is not a partial specialization,

template<class T> class Vector<T*>

this one is.)

Answer 2

When the compiler instantiates vector<char*> , it matches the following template:

template<class T>
class vector<T*> {
  ...
};

For this template to produce a class vector<char*> it needs to be instantiated with T=char , and this is exactly what the compiler does.

When the compiler sees the type vector<char*> it will not use T=char* to instantiate the template, because this would result in the wrong type vector<char**> . It will use T=char instead, which results in vector<char*> . The compiler uses template parameters that will give the correct resulting type.

Answer 3

template<class t>
//teacher said that this partial specialization will handle all type of pointers
class vector<t*>
{
    ...
};

vector<char*> v;

Your confusion comes from assuming that the t in <class t> names the full argument that you give when instantiating. Instead, the argument is matched against <t*> and so t = char .

For example, imagine the following specialization:

template <class T> struct Foo {};
template <class T> struct Foo<vector<T> > {}; //specialization for your vector

Foo<vector<int> > f;

Again it chooses the specialization, where vector<int> is matched against vector<T> and T = int .