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?
"begin" special form packages a sequence of expressions into a single
expression. After compiled, it should be continuation-passing style.
"lambda" special form also have a sequence of expressions in it.
They are essentially the same, except the lambda can be applied.
What is prompt you mentioned?
I have not heard of this concept before.
Can you tell me something about the prompt?
Thanks.
On Friday, October 26, 2018 at 12:40:37 AM UTC+8, Joao Pedro Abreu De Souza
wrote:
>
> 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 lambda, so create prompt as you expected
>
> Em quinta-feira, 25 de outubro de 2018, <serioa...@gmail.com <javascript:>>
> escreveu:
>
>> Dear all,
>>
>> I am learning call/cc in racket, so I wrote some experiment code:
>>
>> #lang racket
>>
>> (begin
>>
>> (define saved-k #f)
>>
>> (+ 1 (call/cc
>> (lambda (k) ; k is the captured continuation
>> (set! saved-k k)
>> 0)))
>>
>> (println 'hello)
>>
>> (saved-k 100)
>>
>> ;; why 'hello not print more than once??
>>
>> )
>>
>> The execution result of this codeis:
>>
>> 1
>> hello
>> 101
>>
>>
>> It does not meet my expectation, my expectation is :
>>
>>
>> 1
>> hello
>> 101
>> hello
>> 101
>> ....loop
>>
>>
>> About continuation, for example:
>>
>> assume exp1 is:
>> (+ 1 (call/cc
>> (lambda (k) ; k is the captured continuation
>> (set! saved-k k)
>> 0)))
>>
>> exp2 is:
>> (println 'hello)
>>
>> Whether exp2 is the continuation of exp1
>>
>> Are these two expressions equivalent?
>>
>> (begin
>> (exp1)
>> (exp2))
>>
>>
>> (exp1 (lambda (v)
>> (exp2 continuation-of-exp2)))
>>
>>
>> So exp2 is exp1's continuation?
>>
>> It makes me confused.
>>
>> Excuse me, this question may be stupid, forgive my ignorance...
>>
>> Thanks.
>>
>> Br,
>> serioadamo97
>>
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