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Comment by danneu | Hacker News Reader

Clojure's namespace declaration:

    (ns my-app.my-namespace
      (:require [some-library.some-namespace :as foo])

is pretty liberating for me after working with Ruby for so long.

Composition is tedious to reason about in Ruby when your only clue is a menagerie of `require`s at the top of the file. Or, more commonly, when your only clue is a Gemfile because all the gems are auto-required and omnipresent.

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An offtopic clojure question.

Today I was seeing this -- https://en.wikipedia.org/wiki/Knuth-Morris-Pratt_algorithm

How can I begin to implement it in clojure ?

Specifically, how do I _ingeneral_ deal with assignments and while loops.

In common-lisp, for example I would use the `loop` macro and defvar-sefq combo, (beginning lisper).

In ruby, = and while itself.

Clojure has most of the semantics you're used to in other languages.

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

    (loop [n 0]
      (when (< n 10)
        (print n)
        (recur (inc n))))

    ;=> 0123456789

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

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 ?

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.

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

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