Why do we need a pure virtual destructor in C++?


I understand the need for a virtual destructor. But why do we need a pure virtual destructor? In one of the C++ articles, the author has mentioned that we use pure virtual destructor when we want to make a class abstract.

But we can make a class abstract by making any of the member functions as pure virtual.

So my questions are

  1. When do we really make a destructor pure virtual? Can anybody give a good real time example?

  2. When we are creating abstract classes is it a good practice to make the destructor also pure virtual? If yes..then why?

8/3/2009 5:51:26 AM

Accepted Answer

  1. Probably the real reason that pure virtual destructors are allowed is that to prohibit them would mean adding another rule to the language and there's no need for this rule since no ill-effects can come from allowing a pure virtual destructor.

  2. Nope, plain old virtual is enough.

If you create an object with default implementations for its virtual methods and want to make it abstract without forcing anyone to override any specific method, you can make the destructor pure virtual. I don't see much point in it but it's possible.

Note that since the compiler will generate an implicit destructor for derived classes, if the class's author does not do so, any derived classes will not be abstract. Therefore having the pure virtual destructor in the base class will not make any difference for the derived classes. It will only make the base class abstract (thanks for @kappa's comment).

One may also assume that every deriving class would probably need to have specific clean-up code and use the pure virtual destructor as a reminder to write one but this seems contrived (and unenforced).

Note: The destructor is the only method that even if it is pure virtual has to have an implementation in order to instantiate derived classes (yes pure virtual functions can have implementations).

struct foo {
    virtual void bar() = 0;

void foo::bar() { /* default implementation */ }

class foof : public foo {
    void bar() { foo::bar(); } // have to explicitly call default implementation.
5/23/2017 11:47:20 AM

All you need for an abstract class is at least one pure virtual function. Any function will do; but as it happens, the destructor is something that any class will have—so it's always there as a candidate. Furthermore, making the destructor pure virtual (as opposed to just virtual) has no behavioral side effects other than to make the class abstract. As such, a lot of style guides recommend that the pure virtual destuctor be used consistently to indicate that a class is abstract—if for no other reason than it provides a consistent place someone reading the code can look to see if the class is abstract.

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