template class member function only specialization

Question

I am reading the Complete Guide on Templates and it says the following:

Where it is talking about class template specialization.

Although it is possible to specialize a single member function of a class template, once you have done so, you can no longer specialize the whole class template instance that the specialized member belongs to.

I'm actually wondering how this is true, cause you can specialize without any member functions at all. Is it saying that you cannot have a specialization with only one member function and then another with all member functions?

Can someone please clarify?

Solution 1

I think it is referring to the following case:

template <typename T>
struct base {
   void foo() { std::cout << "generic" << std::endl; }
   void bar() { std::cout << "bar" << std::endl; }
};
template <>
void base<int>::foo() // specialize only one member
{ 
   std::cout << "int" << std::endl; 
}
int main() {
   base<int> i;
   i.foo();         // int
   i.bar();         // bar
}

Once that is done, you cannot specialize the full template to be any other thing, so

template <>
struct base<int> {};  // error

Solution 2

I think what is meant is that you can either:

• specialize the whole class and all members (data and functions, static or not, virtual or not) have to be declared and defined even if they are the same as for the non specialized version,

• specialize some function members, but then you can't specialize the whole class (i.e. all members are declared in the same way as for the non specialized case, you just provide the implementation for some function members).

Comments

[rlc]

You should perhaps note that if you don't just specialize the member and you use the full template specialization you have above, the specialization is still valid - the calling code that depends on foo and bar being there no longer is. I.e., the same code you posted above, modified to not specialize the member and not call foo and bar will work fine - see ideone.com/3kHa3

[Tony The Lion]

so I guess this one member specialization creates an instance and therefore you can no longer have another specialization?

[Bo Persson]

@Tony - Right, but you can of course have another specialization for a different type, like base<float>.