是否可以撰写例如:
(defn- multiple-of-three? [n] (zero? (mod n 3))
(defn- multiple-of-five? [n] (zero? (mod n 5))
成:
multiple-of-three-or-five?
所以我可以用它来过滤:
(defn sum-of-multiples [n]
(->> (range 1 n)
(filter multiple-of-three-or-five?)
(reduce +)))
另外我不想这样定义:
(defn- multiple-of-three-or-five? [n]
(or (multiple-of-three? n)
(multiple-of-five? n)))
例如,使用Javascript模块Ramda,它将实现为:http://ramdajs.com/docs/#either
const multipleOfThreeOrFive = R.either(multipleOfThree, multipleOfFive)
答案 0 :(得分:7)
当然,在Clojure中,这是some-fn
。
(def multiple-of-three-or-five?
(some-fn multiple-of-three? multiple-of-five?))
(multiple-of-three-or-five? 3) ; => true
(multiple-of-three-or-five? 4) ; => false
(multiple-of-three-or-five? 5) ; => true