Re: take! 0==1?

[William ML Leslie <william.leslie.ttg@gmail.com>]

On 12 July 2013 17:14, Jan Nieuwenhuizen <janneke@gnu.org> wrote:
> Hi,
>
> Reading the documentation of take!
>
>     -- Scheme Procedure: take lst i
>     -- Scheme Procedure: take! lst i
>         Return a list containing the first I elements of LST.
>
>         `take!' may modify the structure of the argument list LST in order
>         to produce the result.
>
> its behaviour surpsises me.
>
> For list LST, (take! lst 0) leaves LST in the same state
> as (take! lst 1) does.  Worse, the return value suggests
> that it worked

Well, there's no pair that can be mutated in (take! lst 0), and (take!
lst n) where n is the length of the list is just like setting the nth
cdr to nil, which it already is.  Have I understood this correctly?

>     scheme@(guile-user)> (use-modules (srfi srfi-1))
>     scheme@(guile-user)> (define lst '(a))
>     scheme@(guile-user)> (take! lst 0)
>     $4 = ()
>     scheme@(guile-user)> lst
>     $5 = (a)
>     scheme@(guile-user)> (take! lst 1)
>     $6 = (a)
>     scheme@(guile-user)> lst
>     $7 = (a)
>     scheme@(guile-user)>
>
> How are you doing such things?  Is anyone using take! at all to reduce a
> list to n elements?  Using list-cdr-set! and set!/list-set! also seems a bit
> clumsy.
>
> Greetings, Jan
>
> --
> Jan Nieuwenhuizen <janneke@gnu.org> | GNU LilyPond http://lilypond.org
> Freelance IT http://JoyofSource.com | Avatar®  http://AvatarAcademy.nl

--
William Leslie

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