Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!usc!apple!sun-barr!newstop!sun!basselope!shiffman
From: shiffman%basselope@Sun.COM (Hank Shiffman)
Newsgroups: comp.lang.lisp
Subject: Re: what does "destructively sorted" mean in KCL?
Message-ID: <119605@sun.Eng.Sun.COM>
Date: 4 Aug 89 20:08:47 GMT
References: <3916@phri.UUCP>
Sender: news@sun.Eng.Sun.COM
Reply-To: shiffman@sun.UUCP (Hank Shiffman)
Organization: Sun Microsystems, Mountain View
Lines: 45

In article <3916@phri.UUCP> roy@phri.UUCP (Roy Smith) writes:
>	I'm trying to figure out how the sort function works in KCL, but I
>think I'm getting hung up on what "Common Lisp: The Language" means when it
>says in section 14.5:  "The sequence is destructively sorted".  I took
>"destructively sorted" to mean "sorted in-place", but it looks like it
>really means "trashed, but a sorted copy is returned".  For example:
>
>>(setq foo '(1 3 9 4 -2 0 9))
>(1 3 9 4 -2 0 9)
>
>>foo
>(1 3 9 4 -2 0 9)
>
>>(sort foo #'>)
>(9 9 4 3 1 0 -2)
>
>>foo
>(1 0 -2)
>
>	Sort returns a sorted list as its value, but the argument gets
>trashed.  Is this really the way it's supposed to work?  Do I have to do
>(setq foo (sort foo #'>)) to get it to work?

Yes, this is exactly the way it's supposed to work.  Notice that in
your example, FOO points to a CONS whose CAR is 1 both before and
after the sort.  When CLtl says that a function on a list operates
destructively, it means that the CONS cells of the original list may
be reused.  FOO still points to the CONS cell which has a CAR of 1.
It's just not the first CONS cell in the sorted list.

Keep in mind that since SORT is a function, (sort foo #'>) causes sort
to be handled the *value of FOO*, not the symbol itself.  So SORT (as
it's currently defined) couldn't put sorted list back into FOO even if
it wanted to.


-- 
Hank Shiffman                                     (415) 336-4658
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