protected

This question already has an answer here: Why can't I access C# protected members except like this? 7 answers It may seems rather newbie question, but can you explain why method Der.B() cannot access protected Foo via Base class variable? This looks weird to me: public class Base { protected int Foo; } public class Der : Base { private void B(Base b) { Foo = b.Foo; } // Error: Cannot access protected member private void D(Der d) { Foo = d.Foo; } // OK } Thanks! This is a frequently asked question. To figure out why this is illegal, think about what could go wrong. Suppose you had another

Here is a sample of code that annoys me: class Base { protected: virtual void foo() = 0; }; class Derived : public Base { private: Base *b; /* Initialized by constructor, not shown here Intended to store a pointer on an instance of any derived class of Base */ protected: virtual void foo() { /* Some implementation */ }; virtual void foo2() { this->b->foo(); /* Compilator sets an error: 'virtual void Base::foo() is protected' */ } }; How do you access to the protected overrided function? Thanks for your help. :o) Protected members in a base-class are only accessible by the current object. Thus,

I want to invoke a protected method of another instance from within a subclass of the class providing this protected method. See the following example: public class Nano { protected void computeSize() { } } public class NanoContainer extends Nano { protected ArrayList<Nano> children; } public class SomeOtherNode extends NanoContainer { // {Nano} Overrides protected void computeSize() { for (Nano child: children) { child.computeSize(); // << computeSize() has protected access in nanolay.Nano } } } javac tells me that computeSize() has protected access in Nano . I can't see the reason for this

I changed my C++ base class to be protected inheritance and my dynamic_cast (s) stopped working. Why should changing the inheritance to protected change the behavior of dynamic_cast ? struct Base { static Base *lookupDerived(); // Actually returns a Derived * object. }; struct Derived : protected /* Switch this to public to get it working */ Base { static void test() { Base *base = lookupDerived(); if (dynamic_cast<Derived *>(base)) { std::cout << "It worked (we must be using public inheritance)." << std::endl; } else { std::cout << "It failed (we must be using protected inheritance)." << std:

This question already has an answer here: What's the real reason for preventing protected member access through a base/sibling class? 6 answers Consider you have the following code: public abstract class MenuItem { protected string m_Title; protected int m_Level; protected MenuItem m_ParentItem; public event ChooseEventHandler m_Click; protected MenuItem(string i_Title, int i_Level, MenuItem i_ParentItem) { m_Title = i_Title; m_Level = i_Level; m_ParentItem = i_ParentItem; } } and public class ContainerItem : MenuItem { private List<MenuItem> m_SubMenuItems; public ContainerItem(string i_Title

Is it possible to define properties that are only available to the class they are defined in, and that class's subclasses? Stated another way, is there a way to define protected properties? Technically, no. Properties are really just methods, and all methods are public. The way we "protect" methods in Objective-C is by not letting other people know about them. Practically, yes. You can define the properties in a class extension, and still @synthesize them in your main implementation block. This is possible by using a class extension (not category) that you include in the implementation files