Unfortunately, it isn't possible Not even with a helper method. The helper method has the same problems as your second attempt: this is of type B even in the base class and will call the implementation of M in B.
Unfortunately, it isn't possible. Not even with a helper method. The helper method has the same problems as your second attempt: this is of type B, even in the base class and will call the implementation of M in B: interface I { int M(); } class A : I { int I.M() { return 1; } protected int CallM() { return (this as I).M(); } } class B : A, I { int I.M() { return CallM(); } } The only workaround would be a helper method in A that is used in A's implementation of M: interface I { int M(); } class A : I { int I.M() { return CallM(); } protected int CallM() { return 1; } } class B : A, I { int I.M() { return CallM(); } } But you would need to provide a method like this also for B if there will be a class C : B, I...
Yes that workaround is what I have added to the updated question. But it's not what I wanted to know. – M4N May 12 at 10:06 1 @M4N: That is just to make the answer a bit bigger ;-) The real answer is the first sentence: It is impossible without such workarounds.
– Daniel lgarth May 12 at 10:07 3 To add to why it's impossible: Explicitly implemented interface methods are made private - making them impossible to call from a derived class. You should probably prefer a different method like the one Daniel listed anyway, to avoid the call super code smell. – Mark H May 12 at 10:18 Thanks for explaining why it is impossible, I understand now.
– Amedio May 12 at 10:43 @M4N: Please don't forget to accept my answer if it helped you. – Daniel lgarth May 127 at 10:26.
Interface ISample {} class A : ISample {} class B : A {} ... base.fun(); ... msdn.microsoft.com/en-us/library/hfw7t1c...).aspx I have no idea its not possible call base method when it comes from implementation of interface.
No! B has a different implementation of the interface than A (it adds something to A's implementation). – M4N May 12 at 9:56 @M4N - Then you can only extend A override the method and make a base call to the method and continue implementing the method.
– Amedio May 12 at 10:00 See updated question. (explicit implementation is not necessary, but I was just wondering whether it's possible) – M4N May 12 at 10:03.
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