Can I implement different implementation to same function from different base classes? - c++

Question

Assume the following:
class A{ virtual void f() = 0; };
class B{ virtual void f() = 0; };
Can I do the following somehow?
class C : public A, public B
{
virtual void A::f(){ printf("f() from A"); }
virtual void B::f(){ printf("f() from B"); }
};
So now I can do???
A* pa = new C();
pa->f(); // prints f() from A;
B* pb = (B*)pa;
pb->f(); // prints f() from B;
Thanks!!!

Answer 1

First solution
This question remind the 'facade' design pattern.
This should be re-write as this :
class AC : public A
{ public: virtual void f(){ cout << "f() from A" << endl;}; };
class BC : public B ...
class C : public AC, public BC {};
where C is the 'facade'.
So in the correct calling syntax should be something like that :
C* c = new C();
c->AC::f();
c->BC::f();
If you don't have any share constraint between AC & BC this should do the job as it is'nt offuscated.
Second solution
Another solution, thank to Casey (see first comment), is to use a forward declaration of the class C in a template to allow calls to methods define latter.
template <typename C>
class AC : public A {
public:
void f() { static_cast<C&>(*this).f_from_A(); }
};
template <typename C>
class BC : public B { ... };
so the implementation part can be done in the same class.
class C : public AC<C>, public BC<C> {
public:
void f_from_A() { cout << "f_from_A" << endl; };
void f_from_B() ...
};
The calling part is cleaner because it doesn't show any implementation details and it is closest to the question :
C* c = new C();
((A*) c) -> f();
((B*) c) -> f();
There is no more 'default' f() on C and it is possible to break the expected behavior of inheritance, and it is harder to read.

Answer 2

For reference, the following links seem related:
Overloading virtual functions of the same name from different base classes. Is it possible?
C++ virtual override functions with same name
Derived class defines function via base class
The OP in the second link asked if he could do something like you asked:
class Impl : public A , public B
{
public:
void A::Function () { cout << "A::Function" << endl; }
void B::Function () { cout << "B::Function" << endl; }
};
Johannes answer is:
You cannot use qualified names there. I you write void Function() {
... } you are overriding both functions. Herb Sutter shows how it can
be solved.
Another option is to rename those functions, because apparently they
do something different (otherwise i don't see the problem of
overriding both with identical behavior).
So one possible solution (by James Kanze from the first link) looks like this:
class RemapA : public A
{
virtual void fnInA() = 0;
public:
virtual void fn()
{
fnInA();
}
};
class RemapB : public B
{
virtual int fnInB() = 0;
public:
virtual int fn()
{
return fnInB();
}
};
class C : public RemapA, public RemapB
{
virtual void fnInA() { /* ... */ }
virtual void fnInB() { /* ... */ }
// ...
};
Herb Sutter's link essentially says the same thing.

Answer 3

For overriding both function you need intermediate classes. A using declaration in C selects the function of A:
#include <iostream>
struct A{ virtual void f() = 0; };
struct B{ virtual void f() = 0; };
struct IntermediateA : public A {
virtual void f() override { std::cout << "A\n"; }
};
struct IntermediateB : public B {
virtual void f() override { std::cout << "B\n"; }
};
struct C : public IntermediateA, public IntermediateB
{
using IntermediateA::f;
};
int main() {
C c;
B* b = &c;
c.f();
b->f();
return 0;
}

Answer 4

Yes. Each base class has its own virtual table.
So by casting, the compiler will check the virtual table of the type you're casting to, and get the pointer to the matching function there.
class A {public: virtual void f() = 0; };
class B {public: virtual void f() = 0; };
class C : public A, public B
{
public:
virtual void A::f(){ printf("f() from A"); }
virtual void B::f(){ printf("f() from B"); }
};
GCC doesn't support this. Only Visual Studio does.
But you will have to always use a cast or a scoped function call.
This will return an ambiguous call error:
C* c = new C();
c->f(); //ambiguous call compilation error
This will work:
((B*)c)->f();
The compiler shows the structure like this: (Visual studio flag: /d1reportSingleClassLayoutC)
class C size(8):
+---
| +--- (base class A)
0 | | {vfptr}
| +---
| +--- (base class B)
4 | | {vfptr}
| +---
+---
C::$vftable#A#:
| &C_meta
| 0
0 | &C::f
C::$vftable#B#:
| -4
0 | &CCCC::f
C::f this adjustor: 4
C::f this adjustor: 0
Basically you can see the 2 separate pointers for the matching virtual tables.
So you could do:
C* c= new C();
B* b = (B*)c;
b->f(); // calls C::B::f()
A* a = (A*)c;
a->f(); // calls C::A::f()
or even:
a = (A*)(C*)b;
a->f(); // calls C::A::f()
Note that you need to case b back to the derived class to access the virual table of the A type base class.