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?
On Wednesday, May 9, 2012 2:29:47 AM UTC-4, Michael Sperber wrote:As I pointed out, this bites both ways: given this expression, a compiler
> (eqv? (lambda (x) (+ x 1)) (lambda (x) (+ x 1)))
> This is the kind of program transformation we wanted to allow in R6RS.
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 elementsI actually invented this example myself, so I used primitives only to keep
> This example, I believe, specifically used simple primitives,
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 forAh, language design by competitive insults! If I'd known that was the
> years for the old behavior of `eqv?'.
proper procedure, it would have made my R7RS efforts so much easier....
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