CODEWITHSUNDEEP
templatesc++11
JadziaMD
JadziaMD

Reputation: 2760

Selecting Member Function Using Enable_if

I have a class which I need to perform different action based upon the value supplied in the template. But I'm getting an error message "prototype does not match any in class..."

#include <iostream>
#include <type_traits>
using namespace std;

template<int t>
struct A
{
    template<typename EqualTwo>
    void test();
};

template<int t>
template<typename std::enable_if<t == 2>::value>
void A<t>::test()
{
    cout << "T is equal to two.";
}

template<int t>
template<typename std::enable_if<t != 2>::value>
void A<t>::test()
{
    cout << "T is not equal to two.";
}

int main() {
    A<5> five;
    A<2> two;
    cout << five.test() << endl;
    cout << two.test() << endl;
    return 0;
}

Upvotes: 1

Views: 706

Answers (1)

Dimitrios Bouzas
Dimitrios Bouzas

Reputation: 42929

Your code is messed up, probably what you want to do is the following:

template<int t, typename Enable = void>
struct A;

template<int t>
struct A<t, typename std::enable_if<t == 2>::type> {
  void test() { cout << "T is equal to two." << endl; }
};

template<int t>
struct A<t, typename std::enable_if<t != 2>::type> {
  void test() { cout << "T is not equal to two." << endl; }
};

and in main():

int main() {
  A<5> five;
  A<2> two;
  five.test();
  two.test();
}

Live Demo

Edit:

Unfortunately, you can't specialize a member function of a template class unless you specialize the class itself.

You could however do the following:

template<int t>
struct A {
  void test() {
    if(t == 2) {
      cout << "T is equal to two." << endl;
    } else {
      cout << "T is not equal to two" << endl;
    }
  }
};

Upvotes: 2

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