Implementation of interface in C++

Question

I need little help with inheritance in C++. I have code with same structure like this:

class IBase {
public:
    virtual long Dimensions() = 0;
};

class IShape : public IBase {
    virtual long Area() = 0;
};

class Rectangle : public IShape  {
private:
    long x;
public:
    long Dimensions() {return x};
    long Area() {return x*x};
};

class Rhombus: public IShape  {
private:
    long x;
    long fi;
public:
    long Dimensions() {return x};
    long Area() {return x*x*sin(fi)};
};

As you can see, implementation of Dimensions() is same for both classes. Now I wan't to do something like this:

class BaseImplementation : public IBase {
protected:
    long x;
public:
    virtual long Dimensions() {return x};
};

class Rectangle : public IShape, public BaseImplementation {
public:
    long Area() {return x*x};
};

class Rhombus: public IShape, public BaseImplementation {
private:
    long fi;
public:
    long Area() {return x*x*sin(fi)};
};

Is possible to insert implementation of method Dimensions() into class Rhombus from BaseImplementation? Is this supported in some version of C++ standard? Thx.

@CharlesB It’s not a bad idea in general to separate interface inheritance and mix-ins (default implementations). It’s a pretty common technique. — Konrad Rudolph, Jan 27 '12 at 11:15
@user1173308 What do you mean exactly by "insert implementation of method into class"? Do you want to override the behaviour of `Dimensions()` for the `Rhombus` class? — CharlesB, Jan 27 '12 at 11:15
@KonradRudolph: thanks, I'm not used to this. deleted comment as it was wrong. — CharlesB, Jan 27 '12 at 11:16

score 3 · Accepted Answer · answered Jan 27 '12 at 11:14

The problem with your hierarchy is that your Rectangle now inherits IBase twice: once through IShape, and once through BaseImplementation. C++ feature called Virtual Inheritance is designed to deal with situations like that. Note that you need to make IShape's inheriting IBase virtual public as well.

score 1 · Answer 2 · answered Jan 27 '12 at 11:14

1

If Dimensions() is implemented in BaseImplementation, and is at least protected in access, it should be visible to any derived object.

So Rhombus will be able to use the Dimension() function if it is derived from BaseImplementation. If you want to have a specific implementation for Dimension() in case of a Rhombus, your Dimension() should be virtual and you should override it in Rhombus

answered Jan 27 '12 at 11:14

Tony The Lion

61,704
67
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If a try to create object Rhombus. The GCC(g++) shows this error: cannot declare variable 'xyz' to be of abstract type 'Rhombus' note: because the following virtual functions are pure within 'Rhombus': virtual long int IBase::Dimensions() – zdenek Jan 27 '12 at 11:35

score 1 · Answer 3 · answered Jan 27 '12 at 11:44

You could simply implement a layer in between that defines Dimension:

class IBase {
public:
    virtual long Dimensions() = 0;
};

class IShape : public IBase {
public:
    virtual long Area() = 0;
};

class IShapesDefinedDimension : public IShape
{
public:
   long Dimensions() { return x; }
protected:
    long x;
};

class Rectangle : public IShapesDefinedDimension  {
public:
    long Area() {return x*x;}
};

class Rhombus: public IShapesDefinedDimension  {
public:
    long Area() {return x*x*sin(fi); }
...
    };

score 0 · Answer 4 · answered Jan 27 '12 at 11:15

0

There is no need to explicitly insert the implementation. You can already call myRhombus->Dimension() because it is inherited.

answered Jan 27 '12 at 11:15

Kit Fisto

4,385
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Implementation of interface in C++

4 Answers4