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From: Loryn Jenkins <loryn@s054.aone.net.au>
Subject: Re: Software landmines (loops)
Date: 1998/09/01
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Date: 1998-09-01T09:36:15+00:00
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Matthew Heaney wrote:
> 
> Loryn Jenkins <loryn@s054.aone.net.au> writes:
> 
> > How is this more complicated?
> >
> > equal (l,r: LIST): BOOLEAN is
> >       require
> >               l.count = r.count
> >       do
> >               Result := l.first /= r.first
> >               if Result then
> >                       from
> >                               l.start; r.start
> >                       until
> >                               not Result or l.off
> >                       loop
> >                               Result := l.item /= r.item
> >                               l.forth; r.forth
> >                       end
> >               end
> >       end
> 
> There are a few things I don't like about this:
> 
> 1) You have to test a flag every iteration of the loop.  That adds (a
> marginal amount of) inefficiency.

Oh well. I can live with that. (Unless my profiler tells me it is in a
hot spot ... which doesn't happen too often.)

> 2) The loop predicate has 4 possible values.  The original example had
> only 2.

And you had two exit points, and I have one. (Probably because I'm not
familiar with your style) I actually have to work a bit harder than
reading my version.

> 3) There's more nesting.

Sorry, I don't see that. You have one for loop and an if statement. I
have one if statement and a loop. Seems the same amount of nesting to
me.

> So yes, I feel the above implementation is more complex than the
> original implementation that used muliple returns.

Oh well, we'll have to agree to disagree then.

> That's what the check L.Top /= R.Top means: if the number of items is
> different, then you know immediately that the stacks can't be equal.
> When you reach the loop, you know the stack depths are the same.

Ah! For some reason, I thought you were testing the first item on the
stack. So, my amended code, to match your semantics is thus:

 equal (l,r: LIST): BOOLEAN is
       require
               l /= Void and r /= Void
       do
               Result := l.count /= r.count
               if Result then
                       from
                               l.start; r.start
                       until
                               not Result or l.off
                       loop
                               Result := l.item /= r.item
                               l.forth; r.forth
                       end
               end
       end


> The way I look at this problem, is that it's like searching for a
> specific house on an unfamiliar block.  As you're driving, when you find
> the house, you stop immediately and declare success.  You don't have to
> drive to the end of the block, and then say where the house was.

Yeah, yeah. Same here, given the amended code.

Loryn Jenkins

PS: I really would agree with you if the only counter examples were Tim
McDermott's and Jim Cochrane's ones.