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Subject: |
eq? |
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Date: |
Sat, 07 Nov 2015 13:58:48 +0100 |
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User-agent: |
Notmuch/0.21 (http://notmuchmail.org) Emacs/24.5.1 (x86_64-unknown-linux-gnu) |
So I wanted to try out gnu guix and thus make myself more familiar with
guile first. While running some tests I encountered a problem/bug with eq?:
$ guile -v
guile (GNU Guile) 2.1.1
$ guile
scheme@(guile-user)>
(define (multirember a lat)
(cond
((null? lat) '())
((eq? (car lat) a) (multirember a (cdr lat)))
(else (cons (car lat) (multirember a (cdr lat))))))
scheme@(guile-user)> (multirember '(a b) '(x y (a b) z (a b)))
$1 = (x y z)
So why does guile return (x y z)? I expected (x y (a b) z (a b)). I know
eq? should only be used with symbols (and thus this example is more
theoretical) but nevertheless the return value is not right, since (eq?
'(a b) '(a b)) returns #f (Btw same in guile 2.0.11).
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--- Begin Message ---
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Subject: |
Re: bug#21855: eq? |
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Date: |
Fri, 24 Jun 2016 17:31:11 +0200 |
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User-agent: |
Gnus/5.13 (Gnus v5.13) Emacs/24.5 (gnu/linux) |
On Sun 08 Nov 2015 11:23, <address@hidden> writes:
> On Sat, Nov 07, 2015 at 01:58:48PM +0100, Atticus wrote:
>> So I wanted to try out gnu guix and thus make myself more familiar with
>> guile first. While running some tests I encountered a problem/bug with eq?:
>>
>> $ guile -v
>> guile (GNU Guile) 2.1.1
>>
>> $ guile
>> scheme@(guile-user)>
>> (define (multirember a lat)
>> (cond
>> ((null? lat) '())
>> ((eq? (car lat) a) (multirember a (cdr lat)))
>> (else (cons (car lat) (multirember a (cdr lat))))))
>>
>> scheme@(guile-user)> (multirember '(a b) '(x y (a b) z (a b)))
>> $1 = (x y z)
>>
>> So why does guile return (x y z)? I expected (x y (a b) z (a b)). I know
>> eq? should only be used with symbols (and thus this example is more
>> theoretical) but nevertheless the return value is not right, since (eq?
>> '(a b) '(a b)) returns #f (Btw same in guile 2.0.11).
>
> Hm. As far as I know (eq? '(a b) '(a b)) is not *guaranteed* to evaluate
> to #f. The implementation might be free to re-use things it "knows" to be
> constant (I might be wrong, though).
Tomas is correct; within one compilation unit, constant literals will be
deduplicated. That means that within one compilation unit, (eq? '(a b)
'(a b)) will indeed be #t.... yarggghhhh.... but:
scheme@(guile-user)> (eq? '(a b) '(a b))
$1 = #f
scheme@(guile-user)> ,optimize (eq? '(a b) '(a b))
$2 = #f
Evidently the optimizer is doing the compare at compile-time, which it
is allowed to do, and at compile-time the values are actually distinct.
I will see if I can fix that. However Tomas' logic is impeccable :)
Closing as things are all working fine, I think.
Cheers,
Andy
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