bug#21855: eq?

[Andy Wingo]

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