Virtual functions

[paul]

I was trying out the following sample code from Scotty Meyers:
Effective C++ book:

#include <iostream>

class B {
public:
virtual void f() const { std::cout << "B::f()" << std::endl; }
};

class D : public B {
public:
virtual void f() { std::cout << "D::f()" << std::endl; }
};

int main()
{
D dx;
B *pb = &dx;
pb->f(); // will call B::f()
return 0;
}

He says according to the standards this should be a compiler error but
some compilers allow this to work. I found Visual Studio and g++ 3.2
allow this.
I would like to know if this is true, if not what maybe the philosophy
behind allowing this ambigious behavior?

[Victor Bazarov]

Re: Virtual functions

paul wrote:

I was trying out the following sample code from Scotty Meyers:
Effective C++ book:
>
#include <iostream>
>
class B {
public:
virtual void f() const { std::cout << "B::f()" << std::endl; }
};
>
class D : public B {
public:
virtual void f() { std::cout << "D::f()" << std::endl; }
};
>
int main()
{
D dx;
B *pb = &dx;
pb->f(); // will call B::f()
return 0;
}
>
He says according to the standards this should be a compiler error

Does he say which compiler error it should be? What item from his 50
are you referring to?

but
some compilers allow this to work. I found Visual Studio and g++ 3.2
allow this.
I would like to know if this is true,

If *what* is true?

if not what maybe the philosophy
behind allowing this ambigious behavior?

In what way is this ambiguous?

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

[acehreli@gmail.com]

Re: Virtual functions

On Jul 16, 11:56 am, paul <pradhan.push.. .@gmail.comwrot e:

I was trying out the following sample code from Scotty Meyers:
Effective C++ book:
>
#include <iostream>
>
class B {
public:
        virtual void f() const { std::cout << "B::f()" << std::endl; }

There is a 'const' on that line.

>
};
>
class D : public B {
public:
        virtual void f() { std::cout << "D::f()" << std::endl; }

No 'const' on that line. Two compilers warn about that "name hiding": g
++, and Comeau online test compiler.

>
};
>
int main()
{
        D dx;
        B *pb = &dx;
        pb->f(); // will call B::f()

B has only one f() and D does not override it, so B::f() should be
called.

        return 0;
>
}
>
He says according to the standards this should be a compiler error but
some compilers allow this to work.

I can't see how it is a compiler error.

I found Visual Studio and g++ 3.2
allow this.
I would like to know if this is true, if not what maybe the philosophy
behind allowing this ambigious behavior?

I think name hiding was introduced to prevent certain types of hard to
detect errors. Without name hiding, a function in D might be in use
today through automatic type conversions on its parameter(s); but
then, introducing a function in B with the same name but with a better
matching parameter(s) could silently start using that function
instead, even though we're using D's interface. (Something like
that... :) )

Ali

[paul]

Re: Virtual functions

On Jul 16, 12:02 pm, Victor Bazarov <v.Abaza...@com Acast.netwrote:

paul wrote:

I was trying out the following sample code from Scotty Meyers:
Effective C++ book:

>

#include <iostream>

>

class B {
public:
   virtual void f() const { std::cout << "B::f()" << std::endl; }
};

>

class D : public B {
public:
   virtual void f() { std::cout << "D::f()" << std::endl; }
};

>

int main()
{
   D dx;
   B *pb = &dx;
   pb->f(); // will call B::f()
   return 0;
}

>

He says according to the standards this should be a compiler error

>
Does he say which compiler error it should be?  What item from his 50
are you referring to?

Item 48: Pay attention to compiler warnings. Actually I read again and
it doesn't say it's an error, just that some compilers will emit a
warning about name hiding.

 but
>

some compilers allow this to work. I found Visual Studio and g++ 3.2
allow this.
I would like to know if this is true,

>
If *what* is true?

I was confused, I read again and it says some compilers emit a warning
about name hiding and others don't.

 if not what maybe the philosophy
>

behind allowing this ambigious behavior?

>
In what way is this ambiguous?

I found this is as ambigious as having
virtual void f(int x);
virtual void f(char x);

[James Kanze]

Re: Virtual functions

On Jul 17, 1:22 am, paul <pradhan.push.. .@gmail.comwrot e:

On Jul 16, 12:02 pm, Victor Bazarov <v.Abaza...@com Acast.netwrote:

paul wrote:

I was trying out the following sample code from Scotty
Meyers: Effective C++ book:

First time I've heard him called Scotty:-).

#include <iostream>

class B {
public:
virtual void f() const { std::cout << "B::f()" << std::endl; }
};

class D : public B {
public:
virtual void f() { std::cout << "D::f()" << std::endl; }
};

int main()
{
D dx;
B *pb = &dx;
pb->f(); // will call B::f()
return 0;
}

He says according to the standards this should be a
compiler error

Does he say which compiler error it should be? What item
from his 50 are you referring to?

Item 48: Pay attention to compiler warnings. Actually I read
again and it doesn't say it's an error, just that some
compilers will emit a warning about name hiding.

Which is something completely different. There is no error in
the code, at least with regards to the standard. There is,
likely, an error with regards to what was wanted. Compilers are
free to warn about anything they feel like; in this case, the
warning is probably a good thing, since you probably meant to
make the derived function const as well (so that it would
override the function in the base class).

but

some compilers allow this to work. I found Visual Studio
and g++ 3.2 allow this. I would like to know if this is
true,

If *what* is true?

I was confused, I read again and it says some compilers emit a
warning about name hiding and others don't.

And it probably depends on the options with which you invoke the
compiler. Try using -Woverloaded-virtual with g++, for example.
(Don't know why this isn't included with -Wall, but it isn't.)

if not what maybe the philosophy

behind allowing this ambigious behavior?

In what way is this ambiguous?

I found this is as ambigious as having
virtual void f(int x);
virtual void f(char x);

I think your confusing ambigious with some other word. There's
not the slightest ambiguity here; the standard is clear as to
what is required. (Note that ambiguous can easily have two
different meanings in this context: the standard itself can be
ambiguous, in which case, we can't determine exactly what is
required; or a function call can be abiguous: the standard
(clearly?) says that the call is ambiguous, and requires the
compiler to issue a diagnostic. Neither applies here, however.)

The rule is somewhat surprising here, but is necessary in its
general form to prevent code from silently changing meaning when
presumably unrelated code (e.g. the private section of a base
class) evolves. It also turns out to be very coherent, working
over classes in the same way it works inside functions, etc.
Roughly speaking, when doing name lookup, the compiler
estabishes an ordered list of scopes, and stops at the first one
in which it finds the name.

If I could change something here (but it's way too late), I'd
require some sort of explicit statement as to 1) whether this
declaration is meant to override something in the base class or
not, and 2) whether the author wishes to allow this declaration
to be overridden. Something like:

void f() ;
// overrides nothing, cannot be overridden
overrides void f() ;
// overrides a virtual function in a
// base class, cannot be overridden
virtual void f() ;
// overrides nothing, but can be
// overridden.
overrides virtual void f() ;
// overrides a virtual function in a
// base class, and can be overriden in
// a derived class.

Name hiding would still work as it does now, but if you meant to
override, and made a mistake in the signature, you would get a
compiler error (and not just maybe a warning).

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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