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Asked: 2026-05-10T15:28:32+00:00

I have defined an interface in C++, i.e. a class containing only pure virtual

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I have defined an interface in C++, i.e. a class containing only pure virtual functions.

I want to explicitly forbid users of the interface to delete the object through a pointer to the interface, so I declared a protected and non-virtual destructor for the interface, something like:

class ITest{ public:     virtual void doSomething() = 0;  protected:     ~ITest(){} };  void someFunction(ITest * test){     test->doSomething(); // ok     // deleting object is not allowed     // delete test;  }

The GNU compiler gives me a warning saying:

class ‘ITest’ has virtual functions but non-virtual destructor

Once the destructor is protected, what is the difference in having it virtual or non-virtual?

Do you think this warning can be safely ignored or silenced?

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1 Answer

  1. It’s more or less a bug in the compiler. Note that in more recent versions of the compiler this warning does not get thrown (at least in 4.3 it doesn’t). Having the destructor be protected and non-virtual is completely legitimate in your case.

    See here for an excellent article by Herb Sutter on the subject. From the article:

    Guideline #4: A base class destructor should be either public and virtual, or protected and nonvirtual.

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