Depending on a class template parameter, define or not define a function in the class

Question

Suppose we have a class:

template <class Type>
class A
{
public:
    void function1(float a, Type b);
    void function1(float a, float b);
};

Now instantiate the class like this:

A<int> a;

It's fine, this class will have 2 overloaded functions with these parameters: (float a, int b); (float a, float b);

But when you instantiate the class like this:

A<float> a;

You get compile error:

member function redeclared.

So, depending on the type of Type, I wan't (or don't want) the compiler to define a function, something like this:

template <class Type>
class A
{
public:
    void function1(float a, Type b);

    #if Type != float
    void function1(float a, float b);
    #endif
};

But, of course, the syntax above doesn't work. Is it possible to perform such a task in C++? If possible, please provide an example.

Answer 1

You can use some C++11 std::enable_if :

template <class Type>
class A
{
public:
    template<typename t = Type, 
       typename std::enable_if<!std::is_same<t, float>::value, int>::type = 0>
    void function1(float a, Type b) {
    }

    void function1(float a, float b) {
    }
};