如何为此功能添加arity？

Question

我想要这个功能

(defn ret-lowest-str-len
    "Computes the lengths of two strings. Returns default length -- len -- 
     if len is <= the length of str-1 and len is <= length of str-2.

     Else, returns smaller of length of str-1 and str-2."

    [str-1 str-2 len]

    (let [l1 (count str-1)
          l2 (count str-2)]

        (if (and (<= len l1) (<= len l2))
            len
            (if (< l1 l2)
                l1
                l2))))

能够有两个参数签名.该示例显示了str-1 str-2和len(固定长度).这样做了,所以如果一个字符串小于15的固定默认值,将返回一个不会导致溢出异常的长度值.

我希望能够在没有str-2的情况下传递str-1和len,我不太清楚如何做到这一点.

我知道如果没有传入l2,代码将不得不改变.我想知道如何设置arity.任何例子将不胜感激.

谢谢.

Answer 1

defn有两种语法

(defn foo [] (expression) (expression))

和

(defn foo
  ([] (exp1) (exp2)) ; arity 0
  ([x] (exp1) (exp2)) ;arity 1
  ([x y] (exp1) (exp2)) ; arity 2

一个常见的模式是让较小的arity版本填入默认值并调用更高的arity情况:

(defn ret-lowest-str-len
([str-1 len] (ret-lowest-str-len str-1 "" len))
([str-1 str-2 len]
  (let [l1 (count str-1)
        l2 (count str-2)]
      (if (and (<= len l1) (<= len l2))
          len
          (if (< l1 l2)
              l1
              l2)))))

在旁注中你可以使用可变数量的参数编写一个类似的函数:

user> (defn ret-lowest-str-len [& args] (reduce min (map count args)))
#'user/ret-lowest-str-len
user> (ret-lowest-str-len "a" "aaaa" "aaaaaaaaa")
1