Because applying the continuation captured by Racket's
call-with-current-continuation will *replace* existing continuations.
Therefore the exception handler is removed. To obtain a continuation
that will only *extend* the current continuation, use
call-with-composable-continuation.
(Another way
Thanks Shu-hung.
So here is my understanding:
#lang racket
(begin
(define saved-k #f) ; wrapped in "prompt 1"
(+ 1 (call/cc
(lambda (k)
(set! saved-k k)
0))) ; wrapped in "prompt 2"
(println 'hello) ; wrapped in "prompt 3"
(saved-k 100) ; wrapped in
Shu-hung, very cool information.
Em qui, 25 de out de 2018 às 16:57, Shu-Hung You <
shu-hung@eecs.northwestern.edu> escreveu:
> FWIW here's the part in the document that could be helpful:
>
> 1. The section in the Guide that talks about the concept of prompts
> and their usage in the
FWIW here's the part in the document that could be helpful:
1. The section in the Guide that talks about the concept of prompts
and their usage in the continuation in Racket:
https://docs.racket-lang.org/guide/prompt.html
The prompt is kind of like a mark on the continuations that lets the
About (+ 1 (prompt (* 2 (call/cc (lambda (k) (set! x k) 2)
1. "prompt" form does not exist in racket.
It is a procedure application or special form ?
2. assume this procedure or special form exists to construct a prompt.
(+ 1 (prompt (* 2 (call/cc (lambda (k) (set! x k) 2)
After the
well, let's go by parts :
1) call/cc in principle will capture the complete continuation of a
expression, right? Delimited continuation will capture ... welll, delimited
continuations. But delimited by what? By a prompt.
In a delimited contninuation style(not really, but I dont want to mix
Thank Joao
I change my code like this:
#lang racket
((lambda ()
(define saved-k #f)
(println (+ 1 (call/cc
(lambda (k) ; k is the captured continuation
(set! saved-k k)
0
(println 'hello)
(saved-k 100)
))
Now it works as I expected.
But why?
Well, call/cc is like (in racket) delimited continuation, and have a
implicit prompt around a s-exp, so, as begin is a macro, he don't create a
prompt. The continuation captured is (+ 1 []) in your example.
If you change the begin to a let, this works, because let expand to a
application of a
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