Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!wuarchive!udel!rochester!kodak!uupsi!sunic!ugle.unit.no!hanche
From: hanche@imf.unit.no (Harald Hanche-Olsen)
Newsgroups: comp.lang.scheme
Subject: Re: eqness of procedures
Message-ID: <HANCHE.91May11164038@hufsa.imf.unit.no>
Date: 11 May 91 14:40:38 GMT
References: <9105080039.aa09228@mc.lcs.mit.edu>
	<1991May8.063427.25012@Think.COM>
	<ALAN.91May9100139@august.lcs.mit.edu>
	<CARLTON.91May10135234@husc10.harvard.edu>
	<ALAN.91May10232758@august.lcs.mit.edu>
Sender: news@ugle.unit.no
Organization: The Norwegian Institute of Technology, Trondheim, Norway.
Lines: 20
In-Reply-To: alan@lcs.mit.edu's message of 10 May 91 21:27:58 GMT

In article <ALAN.91May10232758@august.lcs.mit.edu> alan@lcs.mit.edu (Alan Bawden) writes:

   Well, but certainly

     (eq? (cons 1 2) (cons 1 2))

   is required to be false.  I can imagine a programmer thinking it would be
   convenient if every evaluation of a LAMBDA-expression resulted in a unique
   (in the sense of EQ?) closure.

If I have understood it right, that form is required to be false
because of the existence of procedures like set-car! and set-cdr!
which could change equal-looking objects to look different.  You
cannot change a closure in an analogous way.  So what is so convenient
about (lambda (x) x) not being eq? to (lambda (x) x) ?

- Harald Hanche-Olsen <hanche@imf.unit.no>
  Division of Mathematical Sciences
  The Norwegian Institute of Technology
  N-7034 Trondheim, NORWAY