Clojure - Calculate with big numbers

Question

I want to calculate !1000 in clojure, how can I do this without getting a integer-overflow exception?

My factorial code is right now: (reduce * (range 1 1001)).

score 24 · Accepted Answer · answered Oct 10 '12 at 19:29

24

You could use the *' operator which supports arbitrary precision by automatically promoting the result to BigInt in case it would overflow:

(reduce *' (range 1 1001))

answered Oct 10 '12 at 19:29

mtyaka

1

Thanks! That was a bit easier and cleaner. – Sawny Oct 10 '12 at 19:43
1

What are the pros / cons of doing it this way instead of @Hamza's way? – Sawny Oct 10 '12 at 19:45

score 6 · Answer 2 · answered Oct 10 '12 at 19:24

6

Put N at the end of the number which makes it a bigint,

(reduce * (range 1N 1001N))

answered Oct 10 '12 at 19:24

Hamza Yerlikaya

score 2 · Answer 3 · answered Oct 10 '12 at 19:59

2

Coerce the parameters to clojure.lang.BigInt

(reduce * (range (bigint 1) (bigint 1001)))

I.e. if you are working with an third-party library that doesn't use *'

(defn factorial' [n]
   (factorial (bigint n)))

answered Oct 10 '12 at 19:59

noahlz

3 Answers3