我知道通过使用Xeger，我们可以获得指定模式的随机值。
String regex = "[0-9]{2}";
Xeger generator = new Xeger(regex);
String result = generator.generate();
我想知道有没有办法返回指定正则表达式的所有有效字符串。例如，对于 pattern： ，我们可以从 到 获取所有值。[0-9]{2}0099
谢谢
编辑：
在这里，我们不考虑无限的输出，如+和*;我们如何获取有限正则表达式的所有值？
上次编辑：
谢谢大家！最后，我不考虑所有可能的值，因为可能有数千个。我限制一个特定的数字作为值的数量来减少数量。
由于正则表达式是由有限状态机定义的，我想知道是否有一些东西能够在这样的机器上自动推理，并且非常适合重新用于这项工作......和 clojure.core.logic 交付
所以，我看了正则表达式语法的这个定义（不幸的是，它缺少{}量词，但它们应该很容易添加到我的代码中）将其调整为java转义，并制定了这个110行长的clojure程序：
(ns regexp-unfolder.core
(:require [instaparse.core :as insta])
(:require [clojure.core.logic :as l])
(:require [clojure.set :refer [union difference]])
(:gen-class :methods [#^{:static true} [unfold [String] clojure.lang.LazySeq]])
)
(def parse-regexp (insta/parser
"re = union | simple-re?
union = re '|' simple-re
simple-re = concat | base-re
concat = simple-re base-re
base-re = elementary-re | star | plus
star = elementary-re '*'
plus = elementary-re '+'
elementary-re = group | char | '$' | any | set
any = '.'
group = '(' re ')'
set = positive-set | negative-set
positive-set = '[' set-items ']'
negative-set = '[^' set-items ']'
set-items = set-item*
set-item = range | char
range = char '-' char
char = #'[^\\\\\\-\\[\\]]|\\.'" ))
(def printables (set (map char (range 32 127))))
(declare fns handle-first)
(defn handle-tree [q qto [ type & nodes]]
(if (nil? nodes)
[[q [""] qto]]
((fns type handle-first) q qto nodes)))
(defn star [q qto node &]
(cons [q [""] qto]
(handle-tree q q (first node))))
(defn plus [q qto node &]
(concat (handle-tree q qto (first node))
(handle-tree qto qto (first node))))
(defn any-char [q qto & _] [[q (vec printables) qto]] )
(defn char-range [[c1 _ c2]]
(let [extract-char (comp int first seq second)]
(set (map char (range (extract-char c1) (inc (extract-char c2)))))))
(defn items [nodes]
(union (mapcat
(fn [[_ [type & ns]]]
(if (= type :char)
#{(first ns)}
(char-range ns)))
(rest (second nodes)))))
(defn handle-set [q qto node &] [[q (vec (items node)) qto]])
(defn handle-negset [q qto node &] [[q (vec (difference printables (items node))) qto]])
(defn handle-range [q qto & nodes] [[q (vec (char-range nodes)) qto]])
(defn handle-char [q qto node &] [[q (vec node) qto]] )
(defn handle-concat [q qto nodes]
(let [syms (for [x (rest nodes)] (gensym q))]
(mapcat handle-tree (cons q syms) (concat syms [qto] ) nodes)
))
(defn handle-first [q qto [node & _]] (handle-tree q qto node))
(def fns {:concat handle-concat, :star star, :plus plus, :any any-char, :positive-set handle-set, :negative-set handle-negset, :char handle-char})
(l/defne transition-membero
[state trans newstate otransition]
([_ _ _ [state trans-set newstate]]
(l/membero trans trans-set)))
(defn transitiono [state trans newstate transitions]
(l/conde
[(l/fresh [f]
(l/firsto transitions f)
(transition-membero state trans newstate f))]
[(l/fresh [r]
(l/resto transitions r)
(transitiono state trans newstate r))])
)
(declare transitions)
;; Recognize a regexp finite state machine encoded in triplets [state, transition, next-state], adapted from a snippet made by Peteris Erins
(defn recognizeo
([input]
(recognizeo 'q0 input))
([q input]
(l/matche [input] ; start pattern matching on the input
(['("")]
(l/== q 'ok)) ; accept the empty string if we are in an accepting state
([[i . nput]]
(l/fresh [qto]
(transitiono q i qto transitions) ; assert it must be what we transition to qto from q with input symbol i
(recognizeo qto nput)))))) ; recognize the remainder
(defn -unfold [regex]
(def transitions
(handle-tree 'q0 'ok (parse-regexp regex)))
(map (partial apply str) (l/run* [q] (recognizeo q))))
使用core.logic编写，应该很容易将其调整为正则表达式匹配器
我将可打印字符从 32 个限制为 126 个 ascii，否则处理诸如 的正则表达式会太麻烦，但您可以轻松扩展它...另外，我还没有实现联合，可选模式以及字符类的\w，\s等转义[^c]
到目前为止，这是我在clojure中写的最大的东西，但基础知识似乎被很好地覆盖了......一些例子：
regexp-unfolder.core=> (-unfold "ba[rz]")
("bar" "baz")
regexp-unfolder.core=> (-unfold "[a-z3-7]")
("a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s" "t" "u" "v" "w" "x" "y" "z" "3" "4" "5" "6" "7")
regexp-unfolder.core=> (-unfold "[a-z3-7][01]")
("a0" "a1" "b0" "b1" "c0" "c1" "d0" "d1" "e0" "e1" "f0" "f1" "g0" "g1" "h0" "h1" "i0" "i1" "j0" "j1" "k0" "k1" "l0" "l1" "m0" "m1" "n0" "n1" "o0" "o1" "p0" "p1" "q0" "q1" "r0" "r1" "s0" "s1" "t0" "t1" "u0" "u1" "v0" "v1" "w0" "w1" "x0" "x1" "y0" "y1" "z0" "z1" "30" "31" "40" "41" "50" "51" "60" "70" "61" "71")
regexp-unfolder.core=> (-unfold "[^A-z]")
(" " "@" "!" "\"" "#" "$" "%" "&" "'" "(" ")" "*" "+" "," "-" "." "/" "0" "1" "2" "3" "4" "5" "6" "7" "8" "9" ":" ";" "{" "<" "|" "=" "}" ">" "~" "?")
regexp-unfolder.core=> (take 20 (-unfold "[abc]*"))
("" "a" "b" "c" "aa" "ab" "ac" "ba" "ca" "aaa" "bb" "cb" "aab" "bc" "cc" "aac" "aba" "aca" "baa" "caa")
regexp-unfolder.core=> (take 20 (-unfold "a+b+"))
("ab" "aab" "abb" "abbb" "aaab" "abbbb" "aabb" "abbbbb" "abbbbbb" "aabbb" "abbbbbbb" "abbbbbbbb" "aaaab" "aabbbb" "aaabb" "abbbbbbbbb" "abbbbbbbbbb" "aabbbbb" "abbbbbbbbbbb" "abbbbbbbbbbbb")
自从我开始以这种方式开始以来，我还实现了无限的输出:)
如果有人有兴趣，我在这里上传
显然，这里有一个如何从普通的旧Java调用的例子：unfold
import static regexp_unfolder.core.unfold;
public class UnfolderExample{
public static void main(String[] args){
@SuppressWarnings("unchecked")
Iterable<String> strings = unfold("a+b+");
for (String s : strings){
System.out.println(s);
}
}
}
