c++ - Enable method based on boolean template parameter

I want to implement a private function based on a boolean template parameter. Something like that:

#include 

using namespace std;

template 

class Aggregator {
public:
    void fun(int a) {
        funInternal(a);
    }

private:
    void funInternal(int a, typename std::enable_if::type* = 0) {
        std::cout << "Feature is enabled!" << std::endl;
    }


    void funInternal(int a, typename std::enable_if::type* = 0) {
        std::cout << "Feature is disabled!" << std::endl;
    }
};

int main()
{
   Aggregator a1;
   Aggregator a2;


   a1.fun(5);
   a2.fun(5);

   return 0;
}

But the program above does not compile: error: no type named 'type' in 'struct std::enable_if' void funInternal(int a, typename std::enable_if::type* = 0).

Is it possible to realize the desired behavior with enable_if?

Answer

The following is an adaptation of the solution (http://coliru.stacked-crooked.com/a/480dd15245cdbb6f) provided by @chris in the comments, which seems to meet your needs.

#include 

template
class Aggregator
{
public:

    void fun(int a)
    {
        funInternal(a);
    }

private:
    template
    void funInternal(typename std::enable_if::type a)
    {
        std::cout << "Feature is enabled!" << std::endl;

    }

    template
    void funInternal(typename std::enable_if::type a)
    {
        std::cout << "Feature is disabled!" << std::endl;
    }
};

int main()

{
    Aggregator a1;
    Aggregator a2;

    a1.fun(5);
    a2.fun(5);

    return 0;
}