Diamond Problem

Solution 1:

Its not the same problem.

In the original problem, the overriden method can be called from A. In your problem this can't be the case because it does not exist.

In the diamond problem, the clash happens if class A calls the method Foo. Normally this is no problem. But in class D you can never know which instance of Foo needs to be called:

         +--------+
         |   A    |
         | Foo    |
         | Bar    |
         +--------+
            /  \
           /    \
          /      \
+--------+        +--------+
|   B    |        |   C    |
| Foo    |        | Foo    |
+--------+        +--------+
          \      /
           \    /
            \  /
         +--------+
         |   D    |
         |        |
         +--------+

In your problem, there is no common ancestor that can call the method. On class D there are two flavors of Foo you can chose from, but at least you know that there are two. And you can make a choice between the two.

+--------+        +--------+
|   B    |        |   C    |
| Foo    |        | Foo    |
+--------+        +--------+
          \      /
           \    /
            \  /
         +--------+
         |   D    |
         |        |
         +--------+

But, as always, you do not need multiple inheritance. You can use aggegration and interfaces to solve all these problems.

Solution 2:

In the diamond problem, class D implicitly inherits the virtual method from class A. To call it, class D would call:

A::foo()

If both classes B and C override this method, then the problem comes of which actually gets called.

In your second example however, this isn't the case as class D would need to explicitly state which was being called:

B::foo()
C::foo()

So the problems are not actually the same. In the diamond problem you aren't referencing the derived classes, but their base class, hence the ambiguity.

That's how I understand it, anyway.

Note that I'm coming from a C++ background.