Replace Surprise

[Kazimir Majorinc]

I just discovered this unusual behaviour:

Code: Select all

(dolist (i '("first" "second" "third"))
  (println (replace "x" "hej, x" i)))

hej, first
hej, first
hej, first
"hej, first"

Is it bug or feature?

(If "hej, x" is replaced with (copy "hej, x") then everything works as expected. )

[Sammo]

Probably because replace is destructive and the 2nd and 3rd times through the loop, the argument to replace is "hej, first" instead of "hej, x" so that (replace "x" ...) doesn't find an "x" to replace.

[Kazimir Majorinc]

Yes, it has sense. This example is even simpler.

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(dotimes (i 101)
  (println (inc 0 i))) 
  
(dotimes (i 101)
  (println (inc (copy 0) i))) 

(exit)

I might even like it. But it is surprising.

[cormullion]

doesn't make sense to me... :) I expect this:

(replace "x" "hej, x" i)

to mean what it says - modification of a string 'constant', not for the string to be silently modified to something else because of a previous operation.

I know I'll never understand newLISP fully... :)

[xytroxon]

Well, you can always use the replace function's "second cousin" format ;)

Code: Select all

(dolist (i '("first" "second" "third"))
  (println (format (replace "x" "hej, x" "%s") i)))

Or more clearly and faster...

Code: Select all

(set 'fmtstr (replace "x" "hej, x" "%s"))
(dolist (i '("first" "second" "third"))
  (println (format fmtstr i)))

-- xytroxon