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Home/ Questions/Q 769153
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T18:17:27+00:00 2026-05-14T18:17:27+00:00

In C++, you can do the following: class base_class { public: virtual void do_something()

  • 0

In C++, you can do the following:

class base_class
{
public:
    virtual void do_something() = 0;
};

class derived_class : public base_class
{
private:
    virtual void do_something()
    {
        std::cout << "do_something() called";
    }
};

The derived_class overrides the method do_something() and makes it private. The effect is, that the only way to call this method is like this:

base_class *object = new derived_class();
object->do_something();

If you declare the object as of type derived_class, you can’t call the method because it’s private:

derived_class *object = new derived_class();
object->do_something(); 
// --> error C2248: '::derived_class::do_something' : cannot access private member declared in class '::derived_class'  

I think this is quite nice, because if you create an abstract class that is used as an interface, you can make sure that nobody accidentally declares a field as the concrete type, but always uses the interface class.

Since in C# / .NET in general, you aren’t allowed to narrow the access from public to private when overriding a method, is there a way to achieve a similar effect here?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T18:17:28+00:00Added an answer on May 14, 2026 at 6:17 pm

    If you explicitly implement an interface, this will at least encourage people to use the interface type in the declaration.

    interface IMyInterface
    {
        void MyMethod();
    }
    
    class MyImplementation : IMyInterface
    {
        void IMyInterface.MyMethod()
        {
        }
    }
    

    One will only see MyMethod after casting the instance to IMyInterface. If the declaration uses the interface type, there is no casting needed in subsequent uses.

    MSDN page on explicit interface implementation (thanks Luke, saves me a few seconds^^)

    IMyInterface instance = new MyImplementation();
    instance.MyMethod();
    
    MyImplementation instance2 = new MyImplementation();
    instance2.MyMethod();  // Won't compile with an explicit implementation
    ((IMyInterface)instance2).MyMethod();
    
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