> (for [a (r 2), b (r 3)] [a b]) > ; produces: > <range 2> > (<range 3> > [0 0] [0 1] <range 3> > [0 2] [1 0] [1 1] [1 2])
Why does (r 2) get evaluated before the seq is needed ? => (def k (for [a (r 2), b (r 3)] [a b]) ) <range 2> #'a/k - Thanks -- 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 unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
