通过匹配给定字符串中的模式来获取字符串数组(Get the array of strings by matching the pattern in the given string)

 您的代码有两个问题： 
 
 第一个是你在when语句中检查(rest coll)而不是(rest x) ，导致无限循环。 
 
 第二个是map返回一个惰性序列，这意味着没有发生对connections的调用。 这可以通过使用doall将调用包装到map来修复，但是由于您对副作用而不是返回值感兴趣，所以应该使用doseq 。 
 
 这个版本的函数（多余的loop也已被删除）应该工作： 
 
(defn make-conts [coll]
    (when (not (empty? (rest coll)))
      (doseq [endpoint (rest coll)]
        (connections (first coll) endpoint))
      (recur (rest coll))))
 
 您可以使用clojure.math.combinatorics的组合，而不是自己编写代码来生成组合： 
 
(require '[clojure.math.combinatorics :as combi])

(defn make-conts [coll]
   (doseq [[p1 p2] (combi/combinations coll 2)]
     (connections p1 p2)))

There are two problems with your code: 
 
The first is that you are checking (rest coll) instead of (rest x) in the when statement, leading to an infinite loop. 
 
The second is that map returns a lazy sequence, meaning that the calls to connections is not taking place. This can be fixed by wrapping the call to map with doall, but as you are interested in the side effects, not the return values, you should use doseq instead. 
 
This version of the function (where the superfluous loop has also been removed) should work: 
 
(defn make-conts [coll]
    (when (not (empty? (rest coll)))
      (doseq [endpoint (rest coll)]
        (connections (first coll) endpoint))
      (recur (rest coll))))
 
Instead of writing the code to generate the combinations yourself, you can use combinations from clojure.math.combinatorics:
 
(require '[clojure.math.combinatorics :as combi])

(defn make-conts [coll]
   (doseq [[p1 p2] (combi/combinations coll 2)]
     (connections p1 p2)))