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Home/ Questions/Q 204439
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T17:28:04+00:00 2026-05-11T17:28:04+00:00

I’ve got a large set of inherited classes (criteria) which inherit from a base

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I’ve got a large set of inherited classes (criteria) which inherit from a base class (criterion). Here’s criterion‘s code

class criterion
{
public:
    virtual unsigned __int32 getPriorityClass() const = 0;
    virtual BOOL include(fileData &file) const = 0;
    virtual void reorderTree() = 0;
    virtual unsigned int directoryCheck(const std::wstring& directory) const = 0;
    virtual std::wstring debugTree() const = 0;
};

Some examples of derived classes from this one:

class fastFilter : public criterion
{
public:
    void reorderTree() {};
    unsigned int  directoryCheck(const std::wstring& /*directory*/) const { return DIRECTORY_DONTCARE; };
    unsigned __int32 getPriorityClass() const { return PRIORITY_FAST_FILTER; };
};

class isArchive : public fastFilter
{
public:
    BOOL include(fileData &file) const
    {
        return file.getArchive();
    }
    std::wstring debugTree() const
    {
        return std::wstring(L"+ ISARCHIVE\n");
    };
};

Since I don’t have a destructor here at all, but yet this is supposed to be a base class, do I need to insert an empty virtual destructor, I.e. like this?:

virtual void ~criterion() = 0;

If that virtual destructor declaration is needed, do all intermediate classes need one as well? I.e. would fastFilter above need a virtual destructor as well?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T17:28:04+00:00Added an answer on May 11, 2026 at 5:28 pm

    Yes – the base class needs a virtual destructor, even if it’s empty. If that is not done, then when something delete‘s a derived object through a base pointer/reference, the derived object’s member objects will not get a chance to destroy themselves properly.

    Derived classes do not need to declare or define their own destructor unless they need something other than default destructor behavior.

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