Re: unpacking a list

[David Kastrup]

Urs Liska <address@hidden> writes:

> Hi all,
>
> I don't seem to find a way to "unpack" a list. I think I won't explain
> the background, but basically what I need is that
>     (list (list 1 2 3) 4)
> becomes
>     (list 1 2 3 4)
>
> This is a mock-up of my real code:
>
> #(define (func-a)
>    (list 1 2 3))
>
> #(define (func-b)
>    (list
>     (func-a)
>     4))
>
> #(display (func-b))
>
> What I need is that the call to (func-a) doesn't evaluate to a list but
> to its elements.
> I have "solved" it by actually using something like
>     (append (func-a) (list f4))
> but I think that's not really clean.

It most definitely _is_ clean.

> I'd prefer directly "unpacking" the list in situ.
>
> Any suggestions?

You can write

#(define (func-b)
   `(,@(func-a) 4))

but this will not result in anything significantly different in the
ultimate code.

You can also write a generic unpacker like

#(define (list-elts x)
   (cond ((null? x) x)
         ((pair? x) (append! (list-elts (car x)) (list-elts (cdr x))))
         (else (list x))))

And then use something like (list-elts (cons (func-a) 4))

But that in no way is cleaner than the call to append when you know your
original structures anyway.

-- 
David Kastrup