Don Saklad wrote:
> What does this mean?...
>
> (let ((C call-with-current-continuation)) (apply (lambda (x y) (x y)) (map
> ((lambda (r) ((C C) (lambda (s) (r (lambda l (apply (s s) l)))))) (lambda
> (f) (lambda (l) (if (null? l) C (lambda (k) (display (car l)) ((f (cdr l))
> (C k))))))) '((#\J #\d #\D #\v #\s) (#\e #\space #\a #\i #\newline)))))
Your pretty printer is broken?
You never read "A Short History of the Confederate States of America"?
Will
Jens Axel S�gaard wrote:
> Run it in a Scheme.
(To the OP:) And if you want to try to understand what's going on try to
understand the following stuff ([xx] is difficulty in Knuth's
logarithmic scale):
- [25] lambda calculus
- [30] fixed point combinator(s)
- [40] continuations and continuation passing style
(WARNING: May take more than 5 minutes. More than a day is also likely.)
Don Saklad <·······@nestle.csail.mit.edu> writes:
> What does this mean?...
>
> (let ((C call-with-current-continuation)) (apply (lambda (x y) (x y)) (map
> ((lambda (r) ((C C) (lambda (s) (r (lambda l (apply (s s) l)))))) (lambda
> (f) (lambda (l) (if (null? l) C (lambda (k) (display (car l)) ((f (cdr l))
> (C k))))))) '((#\J #\d #\D #\v #\s) (#\e #\space #\a #\i #\newline)))))
It means you need to learn some more scheme.
--
__Pascal Bourguignon__ http://www.informatimago.com/
"Specifications are for the weak and timid!"
Don Saklad <·······@nestle.csail.mit.edu> writes:
> What does this mean?...
>
> (let ((C call-with-current-continuation)) (apply (lambda (x y) (x y)) (map
> ((lambda (r) ((C C) (lambda (s) (r (lambda l (apply (s s) l)))))) (lambda
^^^
> (f) (lambda (l) (if (null? l) C (lambda (k) (display (car l)) ((f (cdr l))
> (C k))))))) '((#\J #\d #\D #\v #\s) (#\e #\space #\a #\i #\newline)))))
As far as I know, that's not a valid program in any language.
Thomas F. Burdick wrote:
> Don Saklad <·······@nestle.csail.mit.edu> writes:
>
>> What does this mean?...
>>
>> (let ((C call-with-current-continuation)) (apply (lambda (x y) (x y)) (map
>> ((lambda (r) ((C C) (lambda (s) (r (lambda l (apply (s s) l)))))) (lambda
> ^^^
>> (f) (lambda (l) (if (null? l) C (lambda (k) (display (car l)) ((f (cdr l))
>> (C k))))))) '((#\J #\d #\D #\v #\s) (#\e #\space #\a #\i #\newline)))))
>
> As far as I know, that's not a valid program in any language.
(lambda args ...) in Scheme is like (lambda (&rest args) ...) in Common
Lisp.
Pascal
--
3rd European Lisp Workshop
July 3 - Nantes, France - co-located with ECOOP 2006
http://lisp-ecoop06.bknr.net/
Pascal Costanza <··@p-cos.net> writes:
> Thomas F. Burdick wrote:
> > Don Saklad <·······@nestle.csail.mit.edu> writes:
> >
> >> What does this mean?...
> >>
> >> (let ((C call-with-current-continuation)) (apply (lambda (x y) (x y)) (map
> >> ((lambda (r) ((C C) (lambda (s) (r (lambda l (apply (s s) l)))))) (lambda
> > ^^^
> >> (f) (lambda (l) (if (null? l) C (lambda (k) (display (car l)) ((f (cdr l))
> >> (C k))))))) '((#\J #\d #\D #\v #\s) (#\e #\space #\a #\i #\newline)))))
> > As far as I know, that's not a valid program in any language.
>
> (lambda args ...) in Scheme is like (lambda (&rest args) ...) in
> Common Lisp.
Yuck, that's right. I think I forgot that on purpose.