I've read a lot about call/cc, and each time wind up just moving on. So
this is an early New Year's resolution: getting a better understanding of
it.
According to Wikipedia's page on continuations, the continuation of the
statement:
((call/cc f) e2)
is:
(lambda (c) (c e2))
#lang racket
(define ret #f)
(define ret2 (lambda (c) (add1 c)))
(add1
(call/cc
(lambda (k)
(set! ret k) ;; Now ret should be equivalent to ret2.
2)))
(ret 9) ;; Why does this print twice?
(ret2 5) ;; This only prints once, like I would have expected.
--
You received this message because you are subscribed to the Google Groups
"Racket Users" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/racket-users/CACgrOx%2BnXZpQv_MbDwdAZJSGpMiD%2BE1gB4YH8Rsb_y3%3D6RaAnQ%40mail.gmail.com.
