On Wed, Aug 24, 2011 at 5:58 PM, octopusgrabbus <octopusgrab...@gmail.com> wrote: > (defn f1 > [in-seq] > (loop [new-seq [] cur-seq in-seq] > (if (nil? (first cur-seq)) > new-seq > (if-not (nil? (x-in-seq (first cur-seq) new-seq)) > (recur (conj new-seq (first cur-seq)) (rest cur-seq))))))
You don't have a second branch for that last if, so the whole thing evaluates to nil if there are any duplicates, which there are. You need an else clause of (recur new-seq (rest cur-seq)). Or you could just use clojure.core/distinct :) -- Protege: What is this seething mass of parentheses?! Master: Your father's Lisp REPL. This is the language of a true hacker. Not as clumsy or random as C++; a language for a more civilized age. -- 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
