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From: Matthew Heaney <matthew_heaney@acm.org>
Subject: Re: Software landmines (loops)
Date: 1998/09/01
Message-ID: <m3yas4q514.fsf@mheaney.ni.net>#1/1
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NNTP-Posting-Date: Mon, 31 Aug 1998 20:12:50 PDT
Newsgroups: comp.lang.eiffel,comp.object,comp.software-eng,comp.lang.ada
Date: 1998-09-01T00:00:00+00:00
List-Id: <comp.lang.ada>

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.

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

3) There's more nesting.

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


> Sorry for the language change ... it's the one I'm familiar with.

AOK, I understood it OK.  (I also read the 1st ed of Bertrand's book.)
 
> By the way, I'm not sure whether it's a problem in your Ada code, but my
> Eiffel code could fail if r has greater or fewer items than l. Hence the
> precondition.

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.


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.

Likewise for the example I provided.  Once you determine that the stacks
are unequal, you quit and go home.  You don't need go all the way to the
end of the subprogram to declare victory (unless of course the stacks
happen to really be equal).