Access a class as if it were POD

[Simon Elliott]

I've got a requirement to implement a C++ class which should be usable
as if it were an unsigned long:

Foo my_foo;

my_foo = 1;

if (my_foo < 1000)
{
// do this
}

etc.

Is there a simple way I can do this with references, or do I need to
overload every operator under the sun?

--
Simon Elliott http://www.ctsn.co.uk

[Mike Wahler]

"Simon Elliott" <Simon at ctsn.co.uk> wrote in message
news:41***********************@news.gradwell.net.. .

I've got a requirement to implement a C++ class which should be usable
as if it were an unsigned long:

Foo my_foo;

my_foo = 1;

if (my_foo < 1000)
{
// do this
}

etc.

Is there a simple way I can do this
Yes.
with references,
You don't need references (but you can use them if you like).
or do I need to
overload every operator under the sun?

No.

Hints:

Foo::Foo(unsigned long){}
Foo::operator unsigned long(){}

This may or may not be 'cheating', depending upon more
specific defintion of your requirements.

-Mike

[Simon Elliott]

On 08/10/2004, Mike Wahler wrote:

Hints:

Foo::Foo(unsigned long){}
Foo::operator unsigned long(){}

This may or may not be 'cheating', depending upon more
specific defintion of your requirements.

Thanks for this. I'd expect I'll also need a
Foo::operator=(unsigned long){}
--
Simon Elliott http://www.ctsn.co.uk

[Mike Wahler]

"Simon Elliott" <Simon at ctsn.co.uk> wrote in message
news:41***********************@news.gradwell.net.. .

On 08/10/2004, Mike Wahler wrote:

Hints:

Foo::Foo(unsigned long){}
Foo::operator unsigned long(){}

This may or may not be 'cheating', depending upon more
specific defintion of your requirements.

Thanks for this. I'd expect I'll also need a
Foo::operator=(unsigned long){}

Not necessarily. Remember what a default compiler synthesized
assignment operator will do. It's issues like this that caused
my 'possible cheat' remark above -- we're simply delegating the work
to a built-in type, and depending upon already defined operations.
IOW a 'wrapper'.

-Mike