为什么Clojure没有？还是什么？功能？

Question

似乎我必须缺少一些明显常用的习语,但缺少这两个功能似乎有点令我感到困惑.有some但它返回nil而不是False,为什么不是一个any?功能？

Answer 1

some被认为与存在时相同any?.有一个密切命名的功能not-any?,只是some在引擎盖下调用:

(source not-any?)
(def
...
not-any? (comp not some))

你可以简单地写任何:

(def any? (comp boolean some))

补丁欢迎:)只需填写并首先在贡献者协议中邮寄.

考虑到not-any?从1.0开始包含该函数,关于命名的观点尤其正确

(defn any? [pred col] (not (not-any? pred col)))
(any? even? [1 2 3])
true
(any? even? [1  3])
false

我想没有人来补丁发送？(提示暗示轻推)

当使用基于的任何代码some(not-any?在引擎盖下调用一些)时要小心匹配pred和col的类型或使用捕获类型异常的pred

(if (some odd? [2 2 nil 3]) 1 2)
No message.
  [Thrown class java.lang.NullPointerException]

ps:这个例子来自clojure 1.2.1

Answer 2

nil评估为假. (if nil 1 2)评估为2.

some返回满足谓词的第一个元素.任何没有nil或false评估为真的东西.所以(if (some odd? [2 2 3]) 1 2)评估为1.

Answer 3

您正在寻找some-fn，尽管您必须通过它生成的谓词的返回值进行线程化boolean。

user> (doc every-pred)
-------------------------
clojure.core/every-pred
([p] [p1 p2] [p1 p2 p3] [p1 p2 p3 & ps])
  Takes a set of predicates and returns a function f that returns true if all of its
  composing predicates return a logical true value against all of its arguments, else it returns
  false. Note that f is short-circuiting in that it will stop execution on the first
  argument that triggers a logical false result against the original predicates.
nil
user> (doc some-fn)
-------------------------
clojure.core/some-fn
([p] [p1 p2] [p1 p2 p3] [p1 p2 p3 & ps])
  Takes a set of predicates and returns a function f that returns the first logical true value
  returned by one of its composing predicates against any of its arguments, else it returns
  logical false. Note that f is short-circuiting in that it will stop execution on the first
  argument that triggers a logical true result against the original predicates.
nil

这些函数的相似性与every?和some相似。