How to overload and call a method in a derived class which was previously defined in its base class?

Your AddData from the base is hidden by the AddData from derived class. Either explicitly qualify the call DataLogger>::AddData(...) or bring it to scope with using DataLogger>::AddData.

DataLogger>::AddData(...)" this one worked. But I didn't understand the mechanism behind it. AddData() in the base class is not static.

How can we call it in this format? – hkBattousai Jan 6 '11 at 16:19 1 You tell the compiler where the method is from. The shorter version is to say ::AddData(..) -- use the previous parent class where AddData was declared.

– Raphael B. Jan 6 '11 at 16:24 @Rapheal B. : When I use the short version "::AddData()", compiler gives this error: "Error C2039: 'AddData' : is not a member of '`global namespace''".

– hkBattousai Jan 6 '11 at 16:27 Did you tried ::AddData(x, y)? – Raphael B. Jan 6 '11 at 19:18 @Rapheal B.

: I said I tried it. – hkBattousai Jan 6 '117 at 9:47.

Your AddData in derived class hides the function AddData in the base class, so all you need to do is, unhide the latter using using directive: class Plotter : public DataLogger> { public: //unhiding : bringing base class function into scope! Using DataLogger>::AddData; }; Read Item 33: Avoid hiding inherited names from Effective C++ by Scott Meyers.

In order to call the super class method write ::AddData(x, y);. The new Plotter::AddData method makes DataLogger::AddData invisible.

Supper class is making me hungry ;) – Moo-Juice Jan 6 '11 at 15:27.

The problem is not "that I can not access to the method in base class from derived class, even though it is defined public". The problem is that Plotter::AddData is trying to call itself (with a pair) instead of the AddData in the base class. You can make the call explicit by writing void Plotter::AddData(long double x, long double y) { DataLogger>::AddData(std::make_pair(x, y)); }.

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