Re: Anybody in US using ADA ? One silly idea..

[Dmitry A. Kazakov <mailbox@dmitry-kazakov.de>]

On Mon, 03 Feb 2003 08:26:33 -0500, Hyman Rosen <hyrosen@mail.com>
wrote:

>Dmitry A. Kazakov wrote:
>> Then you should explain what "is a" does mean. It is definitely not
>> mathematical =, as memcpy shows.
>
>Memcpy is not a "mathematical =" but a procedure which copies
>a bunch of bytes in memory. You won't like my explanation of
>"is a" becuase it's purely practical - "B is a A" means that
>a pointer or reference to A may hold a pointer or reference to
>an object of type B (or a type which "is a B") and may call
>methods of A using that pointer or reference.

But this is an unsatisfactory definition. It involves pointers and
references, you have to define what they are. You have to admit that
your "is a" does not work if there are no pointers. Is an integer in a
register "is a" an integer in the memory, on a remote host? What does
it mean "may call". In FORTRAN-IV you may call a function passing
INTEGER*4 instead of REAL.*4!

It is much simplier just to say an instance of B is substitutable for
a parameter where A is required. With further clarification what
"substitutable" means.

>> If you define a type, say, "field".
>
>But we are not talking about defining a type, but about
>defining a procedure.

But each procedure is related to the type of its parameters. Then a
type is a set of values and operations on them. So to define a type is
to define these operations as well.

>This procedure defines an interface
>implicitly by virtue of the operations it uses.

This clearly violates DbC principle. The only contract of the
procedure is that it takes an object of some known type [+ possibly
pre- and postconditions]. The type of the object determines all
operations the procedure may use. If you restrict this set of
operation and want to expose this fact, then it is *another* type.

>You then
>claim that this implicit interface needs to be augmented
>with operations that the procedure doesn't need, and then
>that types upon which the procedure operates need to have
>these useless operations, and that this somehow achieves
>safety! I am just baffled.

They are not useless, they constitute the type. That's the point.

---
Regards,
Dmitry Kazakov
www.dmitry-kazakov.de