virtual function declaration

[newbie]

class AbstractBox {
public:
virtual double area() const = 0;
}

class BoxA : public AbstractBox {
virtual double area() { return 0.1; }
}

class BoxB : public AbstractBox {
double area() { return 0.2; }
}

Very dump example, but I hope you get my question: does the 'virtual'
keyword in BoxA take any effect? I don't think so, but I do see code
like BoxA, but not like BoxB.

Thanks

[osmium]

"newbie" writes:

class AbstractBox {
public:
virtual double area() const = 0;
}

class BoxA : public AbstractBox {
virtual double area() { return 0.1; }
}

class BoxB : public AbstractBox {
double area() { return 0.2; }
}

Very dump example, but I hope you get my question: does the 'virtual'
keyword in BoxA take any effect? I don't think so, but I do see code
like BoxA, but not like BoxB.

It has no effect, once virtual always virtual. I see it simply as good
manners for a coder to pass on what information he already knows to the poor
slob who must maintain this nightmare he has come up with.

[pmouse]

Virtual is only used on base classes, i.e., classes that might be
inherited by others. There is a huge difference in this case, but none
in yours. (Suppose BoxB is inherited from BoxA, then yes, virtual
should be declared on A, otherwise it's just a good manner to keep the
class extend-able)

Regards,

PQ

On May 1, 9:11 pm, newbie <mitbb...@yahoo .comwrote:

class AbstractBox {
public:
virtual double area() const = 0;

}

class BoxA : public AbstractBox {
virtual double area() { return 0.1; }

}

class BoxB : public AbstractBox {
double area() { return 0.2; }

}

Very dump example, but I hope you get my question: does the 'virtual'
keyword in BoxA take any effect? I don't think so, but I do see code
like BoxA, but not like BoxB.

Thanks

[Richard Powell]

On May 1, 9:26 pm, pmouse <pmo...@cogeco. cawrote:

Virtual is only used on base classes, i.e., classes that might be
inherited by others. There is a huge difference in this case, but none
in yours. (Suppose BoxB is inherited from BoxA, then yes, virtual
should be declared on A, otherwise it's just a good manner to keep the
class extend-able)

So if a function is virtual in the base class, and an inherited class
has the same function, then the inherited function is virtual too.
But what if the function overloads the arguments, is it still virtual:

class AbstractBox {
public:
virtual double area() const = 0;
};

class BoxA : public AbstractBox {
virtual double area() { return 0.1; }
};

class BoxB : public AbstractBox {
double area() { return 0.2; } /* this is virtual too */
};

class BoxC : public AbstractBox {
double area(int a) { return a*0.2; } /* is this virtual too? */
};

[Victor Bazarov]

Richard Powell wrote:

On May 1, 9:26 pm, pmouse <pmo...@cogeco. cawrote:

>Virtual is only used on base classes, i.e., classes that might be
inherited by others. There is a huge difference in this case, but
none in yours. (Suppose BoxB is inherited from BoxA, then yes,
virtual should be declared on A, otherwise it's just a good manner
to keep the class extend-able)

So if a function is virtual in the base class, and an inherited class
has the same function, then the inherited function is virtual too.
But what if the function overloads the arguments, is it still virtual:

class AbstractBox {
public:
virtual double area() const = 0;
};

class BoxA : public AbstractBox {
virtual double area() { return 0.1; }

This function does not override 'area' in the base class. It hides
the other function since the signature is different. And since the
'AbstractBox::a rea' is pure, this class (and others) are still
abstract.

};

class BoxB : public AbstractBox {
double area() { return 0.2; } /* this is virtual too */

No, it isn't virtual. The signature of this function is not the
same as 'area' in the base class 'AbstractBox'. 'const' is missing.

};

class BoxC : public AbstractBox {
double area(int a) { return a*0.2; } /* is this virtual too? */

No, it's not, for the same reason: the signature is different.

};

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Jim Langston]

"Victor Bazarov" <v.********@com Acast.netwrote in message
news:f1******** **@news.datemas .de...

Richard Powell wrote:

>On May 1, 9:26 pm, pmouse <pmo...@cogeco. cawrote:

>>Virtual is only used on base classes, i.e., classes that might be
inherited by others. There is a huge difference in this case, but
none in yours. (Suppose BoxB is inherited from BoxA, then yes,
virtual should be declared on A, otherwise it's just a good manner
to keep the class extend-able)

So if a function is virtual in the base class, and an inherited class
has the same function, then the inherited function is virtual too.
But what if the function overloads the arguments, is it still virtual:

class AbstractBox {
public:
virtual double area() const = 0;
};

class BoxA : public AbstractBox {
virtual double area() { return 0.1; }

This function does not override 'area' in the base class. It hides
the other function since the signature is different. And since the
'AbstractBox::a rea' is pure, this class (and others) are still
abstract.

>};

class BoxB : public AbstractBox {
double area() { return 0.2; } /* this is virtual too */

No, it isn't virtual. The signature of this function is not the
same as 'area' in the base class 'AbstractBox'. 'const' is missing.

>};

class BoxC : public AbstractBox {
double area(int a) { return a*0.2; } /* is this virtual too? */

No, it's not, for the same reason: the signature is different.

>};

The compilation of this test program did throw some suprises for me.

#include <iostream>

class Base
{
public:
virtual ~Base() {}
virtual int Foo() { return 1; }
virtual int Bar() { return 2; }
};

class Derived: public Base
{
public:
int Foo() { return 10; }
int Bar( int x ) { return x; }
};

int main()
{
Base* Inst = new Derived;
std::cout << Inst->Foo() << "\n";
std::cout << Inst->Bar() << "\n";
std::cout << Inst->Bar( 20 ) << "\n"; // Won't compile
std::cout << dynamic_cast<De rived*>( Inst )->Bar() << "\n"; // Won't
compile
std::cout << dynamic_cast<De rived*>( Inst )->Bar( 20 ) << "\n";
}

[Thomas J. Gritzan]

Jim Langston wrote:

The compilation of this test program did throw some suprises for me.

#include <iostream>

class Base
{
public:
virtual ~Base() {}
virtual int Foo() { return 1; }
virtual int Bar() { return 2; }
};

class Derived: public Base
{
public:
int Foo() { return 10; }
int Bar( int x ) { return x; }
};

int main()
{
Base* Inst = new Derived;
std::cout << Inst->Foo() << "\n";
std::cout << Inst->Bar() << "\n";
std::cout << Inst->Bar( 20 ) << "\n"; // Won't compile

Inst is a Base* and Base doesn't have a function Bar(int).

std::cout << dynamic_cast<De rived*>( Inst )->Bar() << "\n"; // Won't
compile

You try to call Bar(void) on a Derived* but the declaration of Bar(int)
shadows Bar(void) from the base class. You can:

1) override Bar(void) in Derived -or-
2) insert "using Base::Bar;" in Derived to make the Bar(void) function part
of the Derived class too.

std::cout << dynamic_cast<De rived*>( Inst )->Bar( 20 ) << "\n";
}

--
Thomas
http://www.netmeister.org/news/learn2quote.html