5 ms·

Clojure has most of the semantics you're used to in other languages. http://clojuredocs.org/clojure_core/clojure.core/let http://clojuredocs.org/clojure_core/c

by danneu 13y ago

Clojure has most of the semantics you're used to in other languages. http://clojuredocs.org/clojure_core/clojure.core/let http://clojuredocs.org/clojure_core/clojure.core/let (let [username "prollyignored" karma 7] (println username "has" karma "karma points.")) ; prollyignored has 7 karma points. http://clojuredocs.org/clojure_core/1.2.0/clojure.core/loop http://clojuredocs.org/clojure_core/1.2.0/clojure.core/loop (loop [n 0] (when (< n 10) (print n) (recur (inc n)))) ;=> 0123456789 http://clojuredocs.org/clojure_core/clojure.core/for http://clojuredocs.org/clojure_core/clojure.core/for (for [x [0 1 2 3 4 5] :let [y (* x 3)] :when (even? y)] y) ;=> (0 6 12) Or just: (filter even? (map #(* % 3) [0 1 2 3 4 5])) ;=> (0 6 12) Browse the Clojure cheatsheet: http://clojure.org/cheatsheet http://clojure.org/cheatsheet

prollyignored 13y ago
Aha ! Now in, while m+i is less than the length of S, do: if W[i] = S[m + i], if i equals the (length of W)-1, return m let i ← i + 1 .... How would I implement the `return m` part ?
- dizzystar 13y ago
  There is no explicit return value because the functions aren't functions: they are values, thus the result of the algorithm is bound to the value that is bound to the function. When you write: (defn square [x] (* x x)) You are really writing: (def square (fn [x] (* x x))) and x * x is bound to the value of square. There would be no logical reason to have an explicit return value here.
- danneu 13y ago
  It's pretty much the same as this: (loop [n 0] (if (= n 10) n ; true case: return n (recur (inc n)))) ; false case: loop with n++ ;=> 10

by danneu 13y ago

prollyignored 13y ago

Aha ! Now in, while m+i is less than the length of S, do: if W[i] = S[m + i], if i equals the (length of W)-1, return m let i ← i + 1 .... How would I implement the `return m` part ?

dizzystar 13y ago
There is no explicit return value because the functions aren't functions: they are values, thus the result of the algorithm is bound to the value that is bound to the function. When you write: (defn square [x] (* x x)) You are really writing: (def square (fn [x] (* x x))) and x * x is bound to the value of square. There would be no logical reason to have an explicit return value here.
danneu 13y ago
It's pretty much the same as this: (loop [n 0] (if (= n 10) n ; true case: return n (recur (inc n)))) ; false case: loop with n++ ;=> 10