How to create specialization template using scope resolution operator in cpp

Question

template<class t> class Temp{
    static t x;
    public:
      Temp(){};
      t increment();
      ~Temp(){/*body of destructor is important.*/};
};

template<class t>t Temp<t>::x;

template<class t> t Temp<t>::increment(){
    return ++x;
}

/*Template specialization starts.*/
template<>class Temp<int>{
    int x;
    public:
      Temp():x(0){};
      int increment();
      ~Temp(){};
};
/*Below is the error part.*/
template<>int Temp<int>::increment(){
    return 0;
}

The problem is the last block of code. Compilation Error ->error: template-id 'increment<>' for 'int Temp::increment()' does not match any template declaration

Nitheesh George · Accepted Answer · 2020-05-05T18:40:48.467

0

You don't have to use template<> with specialized member function, because the compiler knows that you are specializing the Temp for int type. So an empty template<> giving the error.

int Temp<int>::increment() {
  return ++x;
}

template is used to tell the compiler that T is template param, that's all. But in your case your are specializing for int type, so you don't have to specify template<>. template<> is applicable only for the class and not for member functions defining out side the class.

edited May 05 '20 at 18:40

answered May 05 '20 at 18:22

Nitheesh George

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so use void or nothing? using void gave same type of error. – akash May 05 '20 at 18:28
see the code I have posted, the return type is int because you are passing t as 'int' – Nitheesh George May 05 '20 at 18:29
Sorry that was a typo – Nitheesh George May 05 '20 at 18:36
problem was solved, But I have a question to ask. Why template is used in non-specilatization case for defining member functions outside the class? – akash May 05 '20 at 18:50
That is because the return type is 'T' and we have to tell the compiler what it is. So using template we are telling the compiler that the return T given is a template param. – Nitheesh George May 05 '20 at 18:59

How to create specialization template using scope resolution operator in cpp

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