Hiding member functions in a template class

Is it possible to hide some member functions in a template class? Let's imagine we have something like:

template <class T>
class Increment
{
public:
    void init(T initValue)
    {
         mValue = initValue;
    }  

    T increment()
    {
        ++mValue;
    }

    T increment(T delta)
    {
        mValue += delta;
    }
private:
    T mValue;
};

The objective is to use this class in a way that, in certain cases we only see the increment() function and in some other cases we only see the increment(T) member function. To do that, I can think about something with SFINAE:

class MultipleIncrement
{
    typedef int MultipleIncrement_t;
};

class SingleIncrement
{
    typedef int SingleIncrement_t;
};

template <class T, class Q>
class Increment
{
public:
    void init(T initValue)
    {
        mValue = initValue;
    }

    T increment(typename Q::SingleIncrement_t = 0)
    {
        ++mValue;
    }

    T increment(T delta, typename Q::MultipleIncrement_t = 0)
    {
        mValue += delta;
    }
private:
    T mValue;
}

And then use my template like, for example:

Increment<long, MultipleIncrement>

However, the compiler is not letting me do this. Is there any other way in which this is feasible? Would it also work if the member function is actually the constructor?

---

In this case, I would prefer using template specialization. Would something like this help you?

struct SingleIncrement;
struct MultipleIncrement;

template <
    class T, 
    class Policy = SingleIncrement // default template param
>
class Increment
{
    T mValue;
public:
    Increment(T initValue)
    :   mValue(initValue)
    {}

    T increment()
    {
        ++mValue;
    }
};

// template specialization for MultipleIncrement
template <class T>
class Increment<T,MultipleIncrement>
{
    T mValue;
public:
    Increment(T initValue)
    :   mValue(initValue)
    {}

    T increment(T delta)
    {
        mValue += delta;
    }
};

---

Template specialization is good. Inheritance sounds better. Have you considered templating on the inherited base class? (Or is this now considered a faux pax?)

#define SHOW(X)  cout << # X " = " << (X) << endl

template <class T>
class A
{
public:
  void foo(T t) {SHOW(t); }
};

template <class T, class BASE>
class B : public BASE
{
public:
  void bar(T t) {SHOW(t); }
};

int
main()
{
  B<int,A<int> > b;
  b.foo(1);
  b.bar(2);
}

---

Here is a MWE of how this could be achieved:

#include <iostream>

using namespace std;

struct multi;
struct single;

template<class T, class Q>
struct dummy {
  dummy(T value) : m_value(value) { }

  // using enable_if_t in template argument
  template<class _Q = Q, enable_if_t<is_same<_Q, single>::value && is_same<_Q, Q>::value, int> = 1>
  T increment() { return ++m_value; }

  // using enable_if_t in method return type
  template<class _Q = Q>
  enable_if_t<is_same<_Q, multi>::value && is_same<_Q, Q>::value, T>
  //enable_if_t<is_same<_Q, multi>::value, T> // (*)
  increment(T delta) { return (m_value += delta); }

  T m_value;
};

int main() {
  dummy<double, multi> m(47.10);
  //cout << m.increment() << endl; // error as expected
  cout << m.increment(.01) << endl;

  dummy<int, single> s(41);
  cout << s.increment() << endl;
  cout << s.increment<single>() << endl;
  //cout << s.increment(1) << endl; // error as expected
  //cout << s.increment<multi>(1) << endl; // not an error when using (*)
}

Output

Using c++ (Debian 6.2.1-5) 6.2.1 20161124 this yields:

47.11
42
43

Elaboration

We need to template the methods to make SFINAE work at all. We cannot use something like

std::enable_if_t<std::is_same<_Q, multiple_increment>::value, T> increment() ...

because that fails when instantiating the template dummy<T, single_increment> instead of failing when substituting method template parameters.

Further, we want the user to be able to use the methods without actually providing a method template parameter. So we make the method template parameter _Q default to Q .

Finally, to really force a compiler error when using the unwanted method even when providing a method template parameter, we only enable_if_t the method if method template parameter _Q is actually the same type as the respective class template parameter Q .

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