Re: Easy question about Character manipulation

["(see below)" <yaldnif.w@blueyonder.co.uk>]

On 30/1/07 22:05, in article wccwt346uhg.fsf@shell01.TheWorld.com, "Robert A
Duff" <bobduff@shell01.TheWorld.com> wrote:

>     type T = 1..10;
>     procedure P(X: var T);
>     var A : 1..10;
>     type S = 1..10;
>     var B : S;
>     var C : 1..12345;
> 
> is it legal to pass A, B, and/or C to P's X, in Pascal?
> 
>> Value parameters are in effect assigned their values,
>> and the permissive semantics of assignment compatibility apply;
>> var parameters must be of equivalent types, and type-equivalence in
>> ISO Pascal is name-equivalence.
> 
> ...which seems to say that B and C are illegal above.

Yes, and A as well.

>>> ...  I claim that we're just arguing over
>>> terminology: saying "so-and-so are different types, but there are all
>>> kinds of implicit conversions" amounts to roughly the same thing as
>>> "so-and-so are the same type".
>> 
>> It is unfortunate that the examples being used are subranges,
>> because the assignment compatibility rules for subranges are
>> the most lax. This complexity in the semantics of Pascal
>> is necessary because it lacks the concept of a subtype, and
>> is forced to smuggle in something of the facility by rather
>> ad hoc special rules. So to a small extent your claim is valid.
> 
> That is, "small extent" means "true for integers, but not true for
> records".  Right?

Not quite.
It is true for subranges of any discrete base type,
i.e. integer, char, boolean, user-defined enumerated types;
it is arguable for set types; and is definitely false for all other types.

>> One of the nicest things about Ada is how it cleared this mess up.
> 
> Right.  I think the Ada "subtype S is Integer range 1..10" is pretty
> close to Pascal's "type S = 1..10", but Pascal has no rough equivalent
> to Ada's "type T is range 1..10".  True?

Yes, because Pascal has no way of getting the equivalent of:

   type T is new ...;

-- 
Bill Findlay
<surname><forename> chez blueyonder.co.uk