On Sun 08 Nov 2015 11:23, <to...@tuxteam.de> 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
