Missing method in template specialization

Question

1. How does one make sure, that the specialization of a (class) template implements all functions? (Right now only when I use mul do I get an error message.)
2. What is the difference between the specialization to int for traits1/traits2. I thought they were both template specializations, but traits2 does not accept static and gives a linker error instead of a compiler error.

.

#include <iostream>

template<typename T>
struct traits1{
  static T add(T a, T b) { return a+b; } /* default */
  static T mul(T a, T b);                /* no default */
};

template<>
struct traits1<int> {
  static int add(int a, int b) { return a*b; }
  /* static int mul(int a, int b) missing, please warn */
};

template<typename T>
struct traits2{
  static T add(T a, T b);
  static T mul(T a, T b);
};

template<>
int traits2<int>::add(int a, int b) { return a*b; }

/* traits2<int>::mul(int a, int b) missing, please warn */

int main()
{
  std::cout << traits1<int>::add(40, 2) << "\n";
  // error: mul is not a member of traits1<int>
  //std::cout << traits1<int>::mul(40, 2) << "\n";

  std::cout << traits2<int>::add(40, 2) << "\n";
  // error: undefined reference to traits2<int>::mul(int, int)
  //std::cout << traits2<int>::mul(40, 2) << "\n";
  return 0;
}

Answer 1

1) Don't specialize the whole class if all you want is different behavior for one particular function. Specialize that function alone:

template<>
int traits1<int>::add(int a, int b) { return a*b; }

You can't make sure a specialization implements the all template methods, because they're unrelated.

2) You didn't provide a definition for traits2::mul , so of course you get the linker error - the declarations is there.