Why does (reduce * []) evaluate to 1?

Question

In clojure, calling reduce * on en empty collection returns 1. This is quite surprising.

I made this discovery while creating a factorial function, defined as follow :

(defn factorial [n] (reduce * (take-while #(> % 0) (iterate dec n))))

(factorial 0) is correctly returning 1, without me having to write a special case for zero. How come ?

Answer 1

Checking the code for * and + shows that these two functions implement the 0-arity case by returning the identity for the operation. In the case of * the code is (with dosctring and metadata removed):

(defn *
  ([] 1)
  ([x] (cast Number x))
  ([x y] (. clojure.lang.Numbers (multiply x y)))
  ([x y & more]
     (reduce1 * (* x y) more)))