Functional algorithm to reverse a number

Question

I've been playing with Clojure lately and I can't get this algorithm to work:

(defn reverse-number [number reversed]
   (if (= number 0)
     reversed
     (reverse-number (/ number 10) 
                     (+ (rem number 10) (* reversed 10)))))

This is how I should call it (reverse-number 123 0) and the result I expect is: 321. When I run this, the REPL just hangs.

Can someone explain me, please, what is happening, what I did wrong and how to get this function working?

Note: I know I can use string functions to reverse a number. Actually, I already did this, but I'm not interested in this solution. All I want is to make the leap to functional languages. That's why I try multiple approaches.

Using string functions:

(defn reverse-number [n]
  (Integer. (clojure.string/reverse (str n))))

(reverse-number 123)  ; --> 321

I don't like this version since it feels like cheating by using the string version of reverse.

@Gumbo: 123, why? This is the answer I expect. I certainly wouldn't expect 0123 since that wouldn't be a correct number... — Igor Popov
– Igor Popov, Commented Dec 30, 2012 at 17:08

soulcheck · Accepted Answer · 2012-12-30 13:39:56Z

5

You should use quot instead of /.

/ in clojure will give you a fraction so number will never be 0 (unless it's 0 from the beginning), while quot will give you "integer division".

Examples:

user=> (/ 123 10)                  
123/10
user=> (quot 123 10)
12

answered Dec 30, 2012 at 13:39

soulcheck

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1 Comment

D.Shawley Over a year ago

+1 - I was just about to ask how Clojure implemented truncating division.

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