Re: [racket-users] Racket Solution of String Recursive Replace

2015年9月17日木曜日 2時26分58秒 UTC+9 Jens Axel Søgaard: > > (regexp-replace "B" "BBB" "A") > "ABB" > > (regexp-replace* "B" "BBB" "A") > "AAA" > > > Notice the star. > > > > > 2015-09-16 19:21 GMT+02:00 : > Hi, > > > > I aim to write a racket script like a bash. > > > > $

[racket-users] About Web-Server API

Hi, I want to build my web server in internet and make a simple blog page. Is the racket's web server API useful as php or python, or experimental or for teaching? -- You received this message because you are subscribed to the Google Groups Racket Users group. To unsubscribe from this group

Re: [racket-users] About Web-Server API

Thank you, Jay. I'll try it. 2015年8月17日月曜日 23時45分07秒 UTC+9 Jay McCarthy: There's nothing experimental or especially teaching oriented in the API. It is as useful as PHP and Python's. I personally think it is more useful because of some special Racket-y things it does, but I'm obviously

Re: [racket-users] Re: is this a bug?

2015年8月8日土曜日 15時40分15秒 UTC+9 Michael Titke: These procedure were necessary to represent questions like IS THERE ANY stack on the table onto which I could put this card?, Is the top card of ALL foundations a king?. If so then you have won! :-) You are right, Michael. But I'm not willing to

[racket-users] Re: is this a bug?

Dear crafty schemer, Micheal Perhaps may you mistake quantifier and identifier? I don't know about lisp, at least in C, we call the [a-zA-Z][0-9a-zA-Z]* identifier. And in general, quantifier is any and exists in math. -- You received this message because you are subscribed to the Google

[racket-users] Re: is this a bug?

Andrew I just want to know it. ormap is cool. Alex Umm, I'm thinking everyday about I change racket's (default) syntax. But I'm afraid of lack of docs or supports. Matthias Thanks for correcting my mistake. -- You received this message because you are subscribed to the Google Groups

[racket-users] Re: is this a bug?

In racket, the proc apply-or is needed. (or #f 1 2 #f 3 4 #f) 1 (apply or (list #f 1 2 #f 3 4 #f)) . or: bad syntax in: or (define (apply-or lst) (foldr or #t lst)) . or: bad syntax in: or (define (apply-or lst) (foldr (lambda (x y) (or x y)) #t lst)) (apply-or (list #f 1 2 #f 3 4 #t)) 1

[racket-users] is this a bug?

Dear racket maintainers, Maybe, I found a bug. (foldr (lambda (x y) (list x y)) '() '(a1 a2 a3 a4 a5)) '(a1 (a2 (a3 (a4 (a5 ()) (foldl (lambda (x y) (list x y)) '() '(a1 a2 a3 a4 a5)) '(a5 (a4 (a3 (a2 (a1 ()) Thanks, I hope racket language is more popular. Taro -- You received this

[racket-users] Re: is this a bug?

You are right. ; proc, init and lst are from the racket reference. (define (my-foldl proc init lst) (define (rec res rest1) (if (null? rest1) res (rec (proc res (first rest1)) (rest rest1 (rec init lst)) (my-foldl (lambda (x y) (list x y)) '() '(a1 a2 a3 a4 a5))

[racket-users] Re: is this a bug?

Sorry Stephen, I expected, (foldl (lambda (x y) (list x y)) () '(list a1 a2 a3 a4 a5)) (list (list (list (list (list a1 a2) a3) a4) a5). it may be a problem in non-commutative operations. Taro -- You received this message because you are subscribed to the Google Groups Racket Users group.

[racket-users] Re: is this a bug?

http://stackoverflow.com/questions/8778492/why-is-foldl-defined-in-a-strange-way-in-racket Oh, I see. Racket's holdl is non-regular type for calculation efficiency. Thank you, Stephen. -- You received this message because you are subscribed to the Google Groups Racket Users group. To

Re: [racket-users] Question : Local Definition

Thanks for very good answer. -- 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+unsubscr...@googlegroups.com. For more options, visit

[racket-users] Question : Local Definition

Nice to meet you. I'm a racket beginner and I'm not a English speaker. Please forgive my poor skills. In racket tutorials or HtDP, it uses some local definitions, as follows: (define (f x) (local [((define g x) (add1 x))] (f (g x. I'm not sure about the difference from the code: