Re: Map eats println output?

2011-10-17 Thread Walter van der Laan
That's a big relieve. Thanks for the answer! -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To un

Re: Map eats println output?

2011-10-17 Thread Baishampayan Ghose
> When I run this expression: > (map (fn [x] (println x) x) (range 5)) > > I expect to get (0 1 2 3 4) but instead I get (0 1 2 3 4 0 1 2 3 4) > It seems like the println output is included in the result list. > Am I missing something? The REPL is fooling you. (def *x (map (fn

Map eats println output?

2011-10-17 Thread Walter van der Laan
Hi, When I run this expression: (map (fn [x] (println x) x) (range 5)) I expect to get (0 1 2 3 4) but instead I get (0 1 2 3 4 0 1 2 3 4) It seems like the println output is included in the result list. Am I missing something? -- You received this message because you are subscribed to the

Re: println output

2008-12-28 Thread Meikel Brandmeyer
Hi, Am 28.12.2008 um 03:22 schrieb Mark Volkmann: Here's a related problem. I have a sequence of strings and I'd like to print each on a separate line. I know that this isn't the answer: (force (map println results)) I'm not sure what else to try. (doseq [x (range 3)] (println x)) (doseq [

Re: println output

2008-12-27 Thread Timothy Pratley
Meikel's example works fine for strings: user=> (doseq [s ["hi" "mum" "love u"]] (println s)) hi mum love u nil As does your map: user=> (dorun (map println ["hi" "mum" "love u"])) hi mum love u nil --~--~-~--~~~---~--~~ You received this message because yo

Re: println output

2008-12-27 Thread Mark Volkmann
Here's a related problem. I have a sequence of strings and I'd like to print each on a separate line. I know that this isn't the answer: (force (map println results)) I'm not sure what else to try. On Sat, Dec 27, 2008 at 10:12 AM, Meikel Brandmeyer wrote: > Hi, > > Am 26.12.2008 um 21:43 schrie

Re: println output

2008-12-27 Thread Meikel Brandmeyer
Hi, Am 26.12.2008 um 21:43 schrieb Mark Volkmann: (for [x (range 3)] (println x)) for is not a looping construct. for is a list comprehension. For side effects as above use doseq. (doseq [x (range 3)] (println x)) Sincerely Meikel smime.p7s Description: S/MIME cryptographic signature

Re: println output

2008-12-26 Thread Michael Wood
On Fri, Dec 26, 2008 at 10:43 PM, Mark Volkmann wrote: > > Why does > > (for [x (range 3)] (println x)) > > output > > (0 > nil 1 > nil 2 > nil) > > when run in the REPL instead of > > 0 > 1 > 2 This is because println returns nil every time it's run. user=> (println "test") test nil user=> Al

Re: println output

2008-12-26 Thread mehrheit
On Fri, 26 Dec 2008 14:43:23 -0600 "Mark Volkmann" wrote: > >Why does > >(for [x (range 3)] (println x)) > >output > >(0 >nil 1 >nil 2 >nil) > >when run in the REPL instead of > >0 >1 >2 > >and nothing at all when run from a script? > The seq of nils is the return value of `(for ...)'. It is pr

println output

2008-12-26 Thread Mark Volkmann
Why does (for [x (range 3)] (println x)) output (0 nil 1 nil 2 nil) when run in the REPL instead of 0 1 2 and nothing at all when run from a script? -- R. Mark Volkmann Object Computing, Inc. --~--~-~--~~~---~--~~ You received this message because you are s