Path: utzoo!utgpu!water!watmath!clyde!att!osu-cis!tut.cis.ohio-state.edu!cs.utexas.edu!wasatch!utah-gr!uplherc!sp7040!obie!wsccs!dharvey
From: dharvey@wsccs.UUCP (David Harvey)
Newsgroups: comp.lang.lisp
Subject: Re: Question about eval
Summary: Redefing EVAL is nasty NO-NO!
Message-ID: <747@wsccs.UUCP>
Date: 24 Oct 88 02:22:48 GMT
References: <13733@iuvax.cs.indiana.edu>
Lines: 41

In article <13733@iuvax.cs.indiana.edu>, dfried@iuvax.cs.indiana.edu (Dan Friedman) writes:
> Are there any LISP implementations, especially of Common LISP,
> which get 5 instead of 1 as the result of the following run?
> 
> >>> (defun eval (x) 1)
> eval
> 
> >>> 5
> 1
> 

	Yes, VAX' Common Lisp will return the value of 5.  It will
also not let you so cavalierly redefine this system primitive.  It
sqwawks about it in no uncertain terms before allowing you to hang
yourself.  5 returns 5, and (eval anyentity) returns 1.  If you do
an explicit eval of anthing over or less than 1 arg it will complain
now.

> Does this change to eval ever change what load does?
> 
> Does (setf (symbol-function 'eval) #'(lambda (x) 1))
> do exactly the same thing as (defun eval (x) 1) in all 
> Common Lisp implementations?
> 
> ... Dan

	Unfortunately, VAX Common Lisp will not sqwawk at you when you
redefine in this manner.  It accomplishes the same results as above.
Although I have not tried this with any other Common Lisp, the results
of the setf and the defun should be the same on all machines, since
the defun is nothing more than a macro that expands to the setf.  Oops,
check that.  This assumes that setf and defun are both found in package
Lisp.  Where else would they be?

dharvey		aliases:	Lispin' Dave
				Hacker Horribulus

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and Nobody is responsible for me.
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