Question

I have a templated C++ class that exposes a number of methods, e.g

template<int X, int Y>
class MyBuffer {
public:
    MyBuffer<X,Y> method1();
};

Now, I want to expose additional methods to this class if X == Y. I have done this by subclassing MyBuffer,

template<int X>
class MyRegularBuffer : public MyBuffer<X,X> {
public:
    MyRegularBuffer method2();
};

Now, the problem is that I want to be able to do e.g.

MyRegularBuffer<2> buf = ...
MyRegularBuffer<2> otherBuf = buf.method1().method2();

But I am not sure how to accomplish this. I tried to think of copy constructors, conversion operators, etc, but my C++ skills are unfortunately a bit rusty.

EDIT: I should add that creation of these objects is relatively cheap (and also, it won’t happen a lot), which means it would be OK to do something like this:

MyRegularBuffer<2> buf = ...
MyRegularBuffer<2> temp = buf.method1(); // Implicit conversion
MyRegularBuffer<2> otherBuf = temp.method2();

The question is then, how can I define the conversion like that. The conversion operator needs to be in MyBuffer, I think, but I want it to be available only if X==Y.

Answer 1

You don’t need a separate class to represent the special behaviour. Partial specialization allows you to treat some of the MyBuffer <X,Y> cases specially and give them extra methods.

Keep your original declaration of MyBuffer<X,Y> and add this:

template<int Y>
class MyBuffer<Y, Y> {
public:
    MyBuffer<Y,Y> method1();
    MyBuffer<Y,Y> method2();
};

MyBuffer<1,2> m12; m12.method2(); // compile fail, as desired, as it doesn't have such a method because 1 != 2
MyBuffer<2,2> m22; m22.method2(); // compile success

Edit: my final lines weren’t very useful after all, as pointed out by Georg in the comments, so I’ve deleted them.