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From: mleblanc@athena.mit.edu (Marc LeBlanc)
Newsgroups: comp.lang.c++
Subject: PolyMorphism
Message-ID: <1990Apr10.112351.24301@athena.mit.edu>
Date: 10 Apr 90 11:23:51 GMT
Sender: news@athena.mit.edu (News system)
Organization: Massachusetts Institute of Technology
Lines: 47

>    If we use a Base pointer to construct a Derived object, will the
>Derived destructer be used when been deleted? I have used the
>following code to explore:
>--------------------------
>// test derived constructer and destructer

>#include <stream.h>

>class Base
>{
>  public:
>    Base() { cout << "This is base constructer\n";};
>    ~Base() { cout << "This is base destructer\n";};
>};

>class Derived : public Base
>{
>  public:
>    Derived() { cout << "This is derived constructer\n";};
>    ~Derived() { cout << "This is derived destructer\n";};
>};

>main()
>{
>    Base* test;

>    test = new Derived;
>    delete test;
>}
>----------------------------
>And the output as follow:

>This is base constructer
>This is derived constructer
>This is base destructer

>So the derived destructer won't be called if we construct the Derived
>object by Base pointer. But isn't it more natural to call both? If I
>misunderstood the C++, is there anyway you can call Derived class
>destructer?

I think this would qualify as an "undocumented feature."  If you
declare the base destructor as virtual, you get the desired effect. (I
tried it).  Shouldn't this be the default?

------------------------------------------------------------------------
Marc LeBlanc                        		mleblanc@athena.mit.edu