Re: What's the different betteen pure virtual function and virtual function

From:
"Daniel T." <daniel_t@earthlink.net>
Newsgroups:
comp.lang.c++
Date:
Tue, 03 Jun 2008 20:41:21 -0400
Message-ID:
<daniel_t-232E13.20412103062008@earthlink.vsrv-sjc.supernews.net>
In article
<daniel_t-0BA75B.06391803062008@earthlink.vsrv-sjc.supernews.net>,
 "Daniel T." <daniel_t@earthlink.net> wrote:

Jack <Jack.L.China@gmail.com> wrote:

I meet a question with it ,

I did not get clear the different betteen them,


First, pure virtual:

class Base1 {
public:
   virtual void pure() = 0;
};

class Derived1 { }; // will not compile

class Derived2 {
public:
   void pure() { cout << "pure\n"; }
};

int main() {
   Base1 b; // will not compile
}

Now virtual:

class Base1 {
public:
   virtual void vert();
};

class Derived1 { }; // will compile

int main() {
   Base1 b; // will compile
}

That's the difference.


Well, I posted the above without much due diligence! Sorry about that...

class Base {
public:
   virtual void pure() = 0; // whether Base1::pure() is defined or not
};

class Derived1 : public /*or protected or private*/ Base1 { };

class Derived2 : public Base1 {
public:
   void pure() { }
};

int main() {
   Base b; // will not compile
   Derived1 d; // will not compile
   Derived2 d2; // will compile
}

now virtual:

class Base {
public:
   virtual void virt() { }
};

class Derived1 : public /*or protected or private*/ Base1 { };

class Derived2 : public Base1 {
public:
   void virt() { }
};

int main() {
   Base b; // will compile
   Derived1 d; // will compile
   Derived2 d2; // will compile
}

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