How to give properties to c++ classes (interfaces)
- by caas
Hello,
I have built several classes (A, B, C...) which perform operations on the same BaseClass. Example:
struct BaseClass {
int method1();
int method2();
int method3();
}
struct A { int methodA(BaseClass& bc) { return bc.method1(); } }
struct B { int methodB(BaseClass& bc) { return bc.method2()+bc.method1(); } }
struct C { int methodC(BaseClass& bc) { return bc.method3()+bc.method2(); } }
But as you can see, each class A, B, C... only uses a subset of the available methods of the BaseClass and I'd like to split the BaseClass into several chunks such that it is clear what it used and what is not. For example a solution could be to use multiple inheritance:
// A uses only method1()
struct InterfaceA { virtual int method1() = 0; }
struct A { int methodA(InterfaceA&); }
// B uses method1() and method2()
struct InterfaceB { virtual int method1() = 0; virtual int method2() = 0; }
struct B { int methodB(InterfaceB&); }
// C uses method2() and method3()
struct InterfaceC { virtual int method2() = 0; virtual int method3() = 0; }
struct C { int methodC(InterfaceC&); }
The problem is that each time I add a new type of operation, I need to change the implementation of BaseClass. For example:
// D uses method1() and method3()
struct InterfaceD { virtual int method1() = 0; virtual int method3() = 0; }
struct D { int methodD(InterfaceD&); }
struct BaseClass : public A, B, C // here I need to add class D
{ ... }
Do you know a clean way I can do this?
Thanks for your help
edit:
I forgot to mention that it can also be done with templates. But I don't like this solution either because the required interface does not appear explicitly in the code. You have to try to compile the code to verify that all required methods are implemented correctly. Plus, it would require to instantiate different versions of the classes (one for each BaseClass type template parameter) and this is not always possible nor desired.
