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From: jinx@zurich.ai.mit.edu (Guillermo J. Rozas)
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Subject: Re: eqness of procedures
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Date: 11 May 91 22:10:04 GMT
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In-reply-to: hanche@imf.unit.no's message of 11 May 91 14:40:38 GMT

In article <HANCHE.91May11164038@hufsa.imf.unit.no> hanche@imf.unit.no (Harald Hanche-Olsen) writes:

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   From: hanche@imf.unit.no (Harald Hanche-Olsen)
   Newsgroups: comp.lang.scheme
   Date: 11 May 91 14:40:38 GMT
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   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

You cannot? What about things like 

(let ((loc 'unknown))
  (lambda (new)
    (let ((old loc))
      (set! loc new)
      old)))

Should two such closures be EQ?/EQV? when they have different
environments?  Should such a closure not be EQ?/EQV? to itself?  Such
closures behave very much like structures built out of mutable pairs,
and should behave very much like them as far as EQ?/EQV? are
concerned.

In general, the intent was that LAMBDA is very much like CONS, and
every time a LAMBDA expression is evaluated, you get a different
object (in the sense of EQ?/EQV?).  Note that the formal semantics
represents procedure objects as pairs of locations and functions, and
models this behavior exactly.

This restriction was relaxed somewhat in order to allow compilers to
merge procedure objects that it can prove will always operate
indistinguishably.  In other words, it is spurious to make procedures
distinct when only EQ?/EQV? would be able to tell them apart.

Since determining function equivalence is an undecidable problem, you
cannot expect implementations to do it perfectly, and rather than
imposing precise rules that might be improved later, we left the
decision to the implementors.

Thus

  (eqv? (lambda (x) x) (lambda (y) y))

may be either true or false.

Similarly for

  (define (make-new-procedure)
    (lambda (y) y))

  (eqv? (make-new-procedure) (make-new-procedure))