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Message from discussion Why is (eqv? g g) unspecified in R6RS when g is a procedure?
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John Cowan  
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 More options May 9 2012, 3:02 pm
Newsgroups: comp.lang.scheme
From: John Cowan <johnwco...@gmail.com>
Date: Wed, 9 May 2012 12:02:39 -0700 (PDT)
Local: Wed, May 9 2012 3:02 pm
Subject: Re: Why is (eqv? g g) unspecified in R6RS when g is a procedure?
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On Wednesday, May 9, 2012 2:29:47 AM UTC-4, Michael Sperber wrote:
> (eqv? (lambda (x) (+ x 1)) (lambda (x) (+ x 1)))

> This is the kind of program transformation we wanted to allow in R6RS.
> Since one expression has become two, you also generally get two
> evaluations of `lambda' expressions, and, hence, two objects that are
> not `eqv?'.

As I pointed out, this bites both ways: given this expression, a compiler
may notice that the two expressions are both pure and identical, and reduce
the whole thing to just #t.

> >     (memv cdr (list car cdr cons)) => #f rather than a list of two elements

> This example, I believe, specifically used simple primitives,

I actually invented this example myself, so I used primitives only to keep
it short.  But perhaps I chose better than I knew.

> Does this help?

Well, it explains the motivation well enough.  Unfortunately, we are still
stuck with the issue: procedures are supposed to be first-class in Scheme,
but when you put them into a data structure, you can't be sure of getting
them out again.  In the Lambda Order, that's not supposed to happen.

> Historical aside: Prior to R6RS, the ML people had ridiculed us for
> years for the old behavior of `eqv?'.

Ah, language design by competitive insults!  If I'd known that was the
proper procedure, it would have made my R7RS efforts so much easier....

 
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