Virtual << operator?

[Rob McDonald]

I would like to force all the classes in my hierarchy to implement the
<< operator for testing purposes. My base class is a pure virtual
class.

I started out by working with operator overloading in the derived
class. I have been trying to use what I learned there to create an
appropriate virtual class to force overloading.
class Base{

// Can't define operator with two arguments inside Base class
virtual std::ostream& operator<<(std: :ostream& s, Base& b) = 0;

// When I did this for the concrete derived class, it failed because
the
// compiler doesn't seem to 'find' the implementation of << for Base
virtual std::ostream& operator<<(std: :ostream& s) = 0;
}

// For the concrete derived class, I got this two-argument approach to
work.
// However, you can't declare virtual functions outside a class.
virtual std::ostream& operator<<(std: :ostream& s, Base& b) = 0;

Any suggestions are appreciated.

Rob

[Frank Birbacher]

Hi!

Why not use an oridnary virtual function instead of an operator for
printing?

Rob McDonald schrieb:

Any suggestions are appreciated.

struct Printable
{
virtual void print(std::ostr eam&) const =0;
};

static inline std::ostream& operator << (
std::ostream& stream, Printable const& p
)
{
if(stream)
p.print(stream) ;
return stream;
}

Regards,
Frank

[Rob McDonald]

On Jun 14, 3:08 pm, Frank Birbacher <bloodymir.c... @gmx.netwrote:

Hi!

Why not use an oridnary virtual function instead of an operator for
printing?

Probably because that is far too simple and logical.

Worked great.

Thanks,

Rob

[AnonMail2005]

On Jun 14, 6:25*pm, Rob McDonald <rob.a.mcdon... @gmail.comwrote :

On Jun 14, 3:08 pm, Frank Birbacher <bloodymir.c... @gmx.netwrote:

Hi!

Why not use an oridnary virtual function instead of an operator for
printing?

Probably because that is far too simple and logical.

Worked great.

Thanks,

Rob

BTW, this exact scenario is covered in The C++ Programming Language.

[Rob McDonald]

On Jun 14, 5:49 pm, "AnonMail2...@g mail.com" <AnonMail2...@g mail.com>
wrote:

>
BTW, this exact scenario is covered in The C++ Programming Language.

I didn't see it there when I was trying to figure this out. Is it in
the operator overloading section, or the streams section, or someplace
else.

Thanks,

Rob

[=?utf-8?Q?David_C=C3=B4me?=]

On Sat, 14 Jun 2008 22:48:17 +0200, Rob McDonald
<ro************ @gmail.comwrote :

I would like to force all the classes in my hierarchy to implement the
<< operator for testing purposes. My base class is a pure virtual
class.

I started out by working with operator overloading in the derived
class. I have been trying to use what I learned there to create an
appropriate virtual class to force overloading.
class Base{

// Can't define operator with two arguments inside Base class
virtual std::ostream& operator<<(std: :ostream& s, Base& b) = 0;

// When I did this for the concrete derived class, it failed because
the
// compiler doesn't seem to 'find' the implementation of << for Base
virtual std::ostream& operator<<(std: :ostream& s) = 0;
}

// For the concrete derived class, I got this two-argument approach to
work.
// However, you can't declare virtual functions outside a class.
virtual std::ostream& operator<<(std: :ostream& s, Base& b) = 0;

Any suggestions are appreciated.

Rob

To be virtual a function must be membre of a classe. Or any function
member receive
implicitly this* for first argument. To operator<<, the fisrt argument
must be an stream.
So you can't have operator<< like a function member, specially virtual.

[AnonMail2005]

On Jun 14, 9:33*pm, Rob McDonald <rob.a.mcdon... @gmail.comwrote :

On Jun 14, 5:49 pm, "AnonMail2...@g mail.com" <AnonMail2...@g mail.com>
wrote:

BTW, this exact scenario is covered in The C++ Programming Language.

I didn't see it there when I was trying to figure this out. *Is it in
the operator overloading section, or the streams section, or someplace
else.

In section 21.2.3.1 Virtual Output Functions of C++PL Special Edition.
It has a non-member operator<< function which takes as the second
argument
a base class, but the function just calls a virtual function to do
it's
work.

HTH