Why does this work - haskell mysteries?

[Petter Egesund]

Thanks,

the Scheme-version made it even clearer!

Cheers,

Petter

On Saturday 04 October 2003 13:53, you wrote:
> On Sun, 5 Oct 2003 11:02:37 +0000
>
> Petter Egesund <pegesund at online.no> wrote:
> > Hi & thanks for answering.
> >
> > I think I got it - the chaning of the functions lies in the last part
> > of
> >
> >  (\w -> if v==w then n else sto w)
> >
> > I am used to higher ordered functions from Scheme, but it was the
> > delayed evaluation which played me the trick here. This function is
> > built when updating, and not executed before asking value 'x'?!
>
> If by delayed evaluation you mean lazy evaluation then that has nothing
> to do with it.  Obviously the function isn't executed before asking the
> value of 'x' because no function can run without it's argument(s). The
> same representation will behave exactly the same in Scheme.
>
> (define init-store (lambda (key) 0))
> (define (lookup-store store key) (store key))
> (define (update-store store key value)
>    (lambda (lookup-key)
>       (if (equal? key lookup-key)
>           value
>           (lookup-store store lookup-key))))
>
> Obviously, this will make the function taking lookup-key when
> update-store is called (just like Haskell) and (just like Haskell) it
> will only be executed when applied to a key to lookup.