user=> (defrecord Foo [a])user.Foouser=> (def a (Foo. 1))#'user/
auser=> (def b (Foo. 2))#'user/b
user=> (defrecord Other [b])
user.Other
user=> (def q (Other. 123))
#'user/q
user=> (def p (Other. 456))
#'user/puser=> (filter #(= (class %) user.Foo) (map var-get (vals (ns-
publics 'user))))
(#:user.Foo{:a 2} #:user.Foo{:a 1})Cheers, -M On Nov 5, 4:03 am, Colin Taylor <[email protected]> wrote: > Hi, > > given a namespace sources > > with [simplified definitions] > > (defrecord Source [name url]) > > (def google (source "Google" "google.com")) > (def bbc (source "BBC" "bbc.co.uk")) > (def nbc ... > etc. > > what would be the idiomatic way to implement > > (defn get-all-sources) > > cheers > Colin -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to [email protected] Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/clojure?hl=en
