Tail recursion in clojure

Question

This is a lisp code that uses tail recursion.

(defun factorial (f n)
    (if (= n 1)
        f
        (factorial (* f n) (- n 1))))

I translate this into clojure code expecting the same tail recursion optimization.

(defn fact [f n]
    (if (= n 1)
        f
        (fact (* f n) (dec n))))

However I got this integer overflow (not stack overflow) even with small number such as (fact 1 30).

ArithmeticException integer overflow  clojure.lang.Numbers.throwIntOverflow (Numbers.java:1374)

I tried with recur, but got the same error.

(defn factorial [f n]
    (if (= n 1)
        f
        (recur (* f n) (dec n))))

What's wrong with the clojure code?

Also it's worth noting that Clojure, because of limitations of the JVM, doesn't support automatic tail call optimization. `recur` is indeed the way to go for a recursive idiom in this case. — JohnJ, May 20 '13 at 18:40
Where in the clojure docs can I find examples of using recur without loop? The way you have used it here. — SultanLegend, Jun 09 '17 at 17:01

A. Webb · Accepted Answer · 2013-05-20T18:22:43.457

18

Nothing, just use BigInts:

(factorial 1N 30N) ;=> 265252859812191058636308480000000N

The arguments may be small, but the result is not!

Note that ticked versions of the arithmetic operators are also available, which support arbitrary precision:

(reduce *' (range 1 31)) ;=> 265252859812191058636308480000000N

edited May 20 '13 at 18:22

answered May 20 '13 at 18:10

A. Webb

1 Answers1