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 (lambda (k) ...) evaluated, value of k is (* 2 []), then I call (k 0) ; <---- When does the control flow come back here? (k 0) ---jump to--> (* 2 []), and then? What happens after (*2 0) ? What is the continuation of (*2 0)? On Friday, October 26, 2018 at 2:11:43 AM UTC+8, Joao Pedro Abreu De Souza wrote: > > 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 > functions), we will do something like (+ 1 (prompt (* 2 (call/cc (lambda > (k) (set! x k) 2))))) > > this will set x with (* 2 []), because delimited capture from prompt. > call/cc in racket works in a similar way, if around the most extern s-exp > exists a prompt. > > 2) well, when I dont know exactly why begin dont work and lambda do, but I > have a clue : if you do (syntax->datum (expand '(begin (print "hi") (print > "hello")))) in drracket, you will receive > '(begin (#%app print '"hi") (#%app print '"hello")) > > and if you do (syntax->datum (expand '((lambda () (print "hi") (print > "hello"))))) you will receive '(#%app (lambda () (#%app print '"hi") (#%app > print '"hello"))) > > Probably, as begin dont is involved in a #%app, he dont is invoked with a > prompt. But this is just speculation of my part. > > Em qui, 25 de out de 2018 às 14:13, <serioa...@gmail.com <javascript:>> > escreveu: > >> 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> 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 >>>> >>>> -- >>>> 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 racket-users...@googlegroups.com. >>>> For more options, visit https://groups.google.com/d/optout. >>>> >>> -- >> 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 racket-users...@googlegroups.com <javascript:>. >> For more options, visit https://groups.google.com/d/optout. >> > -- You received this message because you are subscribed to the Google Groups "Racket Users" group. 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