这是ClojureScript中re-seq实现的一个bug。
(defn re-seq
"Returns a lazy sequence of successive matches of re in s."
[re s]
(let [match-data (re-find re s)
match-idx (.search s re)
match-str (if (coll? match-data) (first match-data) match-data)
post-idx (+ match-idx (max 1 (count match-str)))
post-match (subs s post-idx)]
(when match-data (lazy-seq (cons match-data (when (<= post-idx (count s)) (re-seq re post-match)))))))
问题出现在它递归到re-seq，使用字符串的剩余部分，这样做意味着^[a-f]将再次匹配这个新，更短的字符串。
一个解决方案是让你的正则表达式具有黏性
(js/RegExp. #"^." "y")
这使得后续使用你的正则表达式会注意到之前的匹配，请注意，你需要小心地放置此代码，因为它需要在正确的位置创建，它不能是全局的！如果您全局使用，您将遇到类似这样的奇怪状态问题
(let [re (js/RegExp. #"^." "y")]
[(re-seq re "cccc")
(re-seq re "abbb")])
;; => [("c" "c") nil]
（我完全无法解释！）
re-seq的另一种实现可能会为您进行此初始克隆
(defn re-seq2
"Returns a lazy sequence of successive matches of re in s."
[re s]
(let [re-seq* (fn re-seq* [re s]
(let [match-data (re-find re s)
match-idx (.search s re)
match-str (if (coll? match-data) (first match-data) match-data)
post-idx (+ match-idx (max 1 (count match-str)))
post-match (subs s post-idx)]
(when match-data (lazy-seq (cons match-data (when (<= post-idx (count s)) (re-seq* re post-match)))))))]
(re-seq* (js/RegExp. re "y") s)))
(let [re #"^."]
[(re-seq2 re "cccc")
(re-seq2 re "abbb")])
;; => [("c") ("a")]
