Clojure ::获取列表的单个元素,如果列表包含多个元素,则抛出异常

Question

我知道在我的代码中的某个点,列表只有一个元素,所以我用它来获取它

(first alist)

但是如果列表有多个元素来提醒我错误的情况,我也希望代码能够破解.在Clojure中实现这一目标的惯用方法是什么？

Answer 1

替换first为only(或其他诗歌命名的)函数,其中包含您要进行断言的前提条件:

(defn only [x] {:pre [(nil? (next x))]} (first x))

(only [1])
=> 1

(only [1 2])
=> AssertionError Assert failed: (nil? (next x))  user/only (NO_SOURCE_FILE:1)

Answer 2

这将在包含一个元素以外的任何元素的集合上爆炸。在惰性序列上也能正常工作。

(defn only
 "Gives the sole element of a sequence"
 [coll]
 (if (seq (rest coll))
   (throw (RuntimeException. "should have precisely one item, but had at least 2"))
   (if (seq coll)
     (first coll)
     (throw (RuntimeException. "should have precisely one item, but had 0")))))