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2013-12-22T15:23:47.817
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2013-12-22T19:46:54.327
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2013-12-22T19:46:54.327
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First solution This question remind the 'facade' design pattern. This should be re-write as this : <pre><code>class AC : public A { public: virtual void f(){ cout << "f() from A" << endl;}; }; class BC : public B ... class C : public AC, public BC {}; </code></pre> where C is the 'facade'. So in the correct calling syntax should be something like that : <pre><code>C* c = new C(); c->AC::f(); c->BC::f(); </code></pre> 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. <pre><code>template <typename C> class AC : public A { public: void f() { static_cast<C&>(*this).f_from_A(); } }; template <typename C> class BC : public B { ... }; </code></pre> so the implementation part can be done in the same class. <pre><code>class C : public AC<C>, public BC<C> { public: void f_from_A() { cout << "f_from_A" << endl; }; void f_from_B() ... }; </code></pre> The calling part is cleaner because it doesn't show any implementation details and it is closest to the question : <pre><code>C* c = new C(); ((A*) c) -> f(); ((B*) c) -> f(); </code></pre> 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.
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1. POCan I implement different implementation to same function from different base classes?
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1. COIf you remap the calls in a CRTP class, your implementations can actually be defined in `C` with full access to the object ([example at Coliru](http://coliru.stacked-crooked.com/a/1097abe9a3292dbf)).
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