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From: jtc@dimensional.com (Jim Cochrane)
Subject: Re: Software landmines (loops)
Date: 1998/08/31
Message-ID: <6sf4gl$hb6@flatland.dimensional.com>#1/1
X-Deja-AN: 386563995
References: <slrn6sopmg.4fg.jdege@jdege.visi.com> <35f51e53.48044143@
 <m3af4mq7f4 <m31zpxqutn.fsf@mheaney.ni.net>
Organization: Dimensional Communications
NNTP-Posting-Date: Mon, 31 Aug 1998 15:27:54 MDT
Newsgroups: comp.lang.eiffel,comp.object,comp.software-eng,comp.lang.ada
Date: 1998-08-31T00:00:00+00:00
List-Id: <comp.lang.ada>

How about the following version, without multiple exit points (written in
Eiffel because it is a good language for this style [it actually doesn't
allow the multiple return style] and because I'm familiar with Eiffel.)

class STACK [G] ...
is_equal (other: STACK [G]): BOOLEAN is
        -- Are all items in Current equal to all item in other?
    require
        other /= Void
    local
        i1, i2: STACK_ITERATOR
    do
        !!i1.make (Current); !!i2.make (other)
        from
            i1.start; i2.start
        until
            i1.after or i2.after or else i1.item /= i2.item
        loop
            i1.forth; i2.forth
        end
        Result := i1.after and i2.after
    end

I don't think this is particularly hard to understand or maintain, plus it
is simpler than the algorithm below - it eliminates the if statement at the
beginning.  (STACK_ITERATOR behavior is, hopefully, obvious - i.item means
the value of item at the current cursor position of i.)  I threw this together
just for this post, so apologies if there are any bugs (and bonus points to
those that find them :-) ).

In article <m31zpxqutn.fsf@mheaney.ni.net>,
Matthew Heaney  <matthew_heaney@acm.org> wrote:
>ahussey@it.uq.edu.au (Andrew Hussey) writes:
>
>> >Using an exit from the middle avoids the headaches (literally)
>> >engendered by using an extra flag in the predicate.  When you want to
>> >exit, you just say that you want to exit, directly.  No mental
>> >gymnastics are required in order to determine whether you'll "really"
>> >exit, as would be the case using the flag approach.
>> 
>> That's brilliant, now your code is much easier to write!
>> Now let's see who has an easier time *testing* their code.
>> I think you'll find the control-flow errors you introduce
>> in the spaghetti you produce will more than make up for
>> any gain you have from rapid coding.  
>
>Here's is something I whipped up for another post recently.  It's an
>equality operator for a bounded stack.  
>
>The implementation of the function has multiple returns.
>
>Does this implementation fit your definition of spaghetti code?
>
>Would the implementation be better by not using multiple returns?
>
>
>function "=" (L, R : Stack_Type) return Boolean is
>begin
>
>   if L.Top /= R.Top then
>      return False;
>   end if;
>
>   for Index in Positive range 1 .. L.Top loop
>      if L.Items (Index) /= R.Items (Index) then
>         return False;
>      end if;
>   end loop;
>
>   return True;
>
>end "=";
>
>
>My feeling is that trying to implement this operation using only a
>single return would just make it more complicated.
>
>


-- 
Jim Cochrane
jtc@dimensional.com