[Help] A Question About Template Specialization

compiled the following code

Code:

class t {
public:
   template <typename T>
   void tt(T i) { }
   template <>
   void tt(int i) { }
};

int main()
{
}

and I got the following error message

5 invalid explicit specialization before '>' token
5 explicit specialization in non-namespace scope `class t'
6 invalid member function declaration

but if I moved function tt out of the class t, say,

Code:

template <typename T>
void tt(T i) { }
template <>
void tt(int i) { }

int main()
{
}

then it compiled well.

I wonder if C++ dosen't support such kind of specialization inside a class? Why? For any good reason?

Anyway, I do need to declare a template function, and a special version of this function, inside a normal class. How? The only alternative is to make it an overloaded function? say,

Code:

class t {
public:
   template <typename T>
   void tt(T i) { }

   void tt(int i) { }
};

int main()
{
}

then it compiled well

You can't partially specialize a template method, any more than you can a template function.

Partial specialization is for templates CLASSES only. To specialize functions you just use overloading.

but as you can see here, this is a complete specialization, not partial.
Partial specialization is OK for function too.

finally, i got how to specialize template methods

Code:

class t {
public:
   template <typename T>
   void tt(T i) { }
};

template <>
void t::tt(int i) { }

int main()
{
}

Originally Posted by Antigloss

but as you can see here, this is a complete specialization, not partial.
Partial specialization is OK for function too.

No it is not. Try compiling this:

Code:

template <typename A, typename B>
void func(A a, B b) { }

template <typename B>
void func<int, B>(int a, B b) { }

The message my compiler gives is:

t.cc:5: partial specialization `func<int, B>' of function template

As I said, you can't partially specialize a function template this way. You have to overload it.