list-set not working in Racket

Question

Why is list-set not working in following function in Racket:

(define (trimlist sl)
  (let ((num 0) (ss "") )
    (set! num (length sl))
    (for ((i num))
         (println i)
         (set! ss (list-ref sl i))
         (println ss)
         (set! ss (string-trim ss))
         (println ss)
         (list-set sl i ss )  )
     sl ))

>  (trimlist (list " a" "b " " c "))
0
" a"
"a"
1
"b "
"b"
2
" c "
"c"
'(" a" "b " " c ")

The println statements show that each element is getting trimmed properly but the returned list still has untrimmed items. The trimmed elements are not replacing original elements in the list.

The documentation of list-set is here: https://docs.racket-lang.org/reference/pairs.html#%28def._%28%28lib._racket%2Flist..rkt%29._list-set%29%29

Where is the problem and how can I solve it?

score 2 · Accepted Answer · answered Jul 13 '16 at 13:21

2

This is because list-set is not destructive (if it was, it's name would be "list-set!"). You can fix that with

(set! sl (list-set sl i ss))

Anyway, consider also a more schemish solution, like:

(define (trim-list sl) (map string-trim sl))

answered Jul 13 '16 at 13:21

dercz

962
6
9

Perfect solutions. Does Scheme code always runs in Racket? – rnso Jul 13 '16 at 13:37
I would consider `(define (trim-list s1) (map string-trim s1))` to be both scheme-y and racket-y. – Alex Knauth Jul 13 '16 at 14:03
My question here is whether all (or most) Scheme code runs in Racket without modification? – rnso Jul 13 '16 at 14:05
1

@rnso http://stackoverflow.com/questions/3345397/how-is-racket-different-from-scheme – dercz Jul 13 '16 at 14:23
Thanks for a very useful link. – rnso Jul 13 '16 at 14:55

list-set not working in Racket

1 Answers1