Re: Streams in Ada

[Robert A Duff <bobduff@world.std.com>]

Brian Gaffney <b_gaffney@deja.com> writes:

> Robert A Duff wrote:
> > 
> > Florian Weimer <Florian.Weimer@RUS.Uni-Stuttgart.DE> writes:
> > 
> > > Ted Dennison<dennison@telepath.com> writes:
> > >
> > > > I think you meant, "all possible Ada implementations", which I would
> > > > disagree with.
> > >
> > > Do impossible Ada implementations exist? ;-)
> > 
> > No, but some possible Ada implementations don't exist.  ;-)
> > 
> Are you sure?

Yes, I'm quite sure that some possible Ada implementations do not (yet?)
exist.

>....  Isn't there a current _very_ long thread about a current 
> 'impossible' Ada implementation :-)?  Oops, that's an Ada compiler/tool
> set, sorry.
> 
> Seriously, shouldn't the Storage_Unit refer to the addressable unit on 
> the machine?

Yes.

>..  For example, on a machine with 32-bit addressing, does the
> potential address space cover 2**32 8-bit bytes, 32-bit words, or
> 256-bit 
> cache-words?

I'm not sure what you mean by "32-bit addressing".  If an address points
to an 8-bit quantity, then Storage_Unit = 8.  If it points to a 32-bit
quantity, then Storage_Unit = 32.  It doesn't matter how big an address
is.

The cache has nothing to do with it.  It's the machine architecture that
matters.

>..  I would think most machines are byte-addressable,
> therefore
> Storage_Unit should be 8.

True on most machines, these days.

>...  On machines we use, we have a 16-bit address 
> space and can reference 2**16 16-bit half-words.  Therefore Storage_Unit
> would be 16 (assuming an Ada compiler existed, HA!), even though 8-bit 
> bytes and 32-bit words can be referenced.

How are 8-bit bytes referenced?  If by "load high/low byte pointed to by
address", then yes, Storage_Unit should be 16.

- Bob