[英]Represent regular expression as context free grammar
我正在为一个简单的正则表达式引擎编写解析器。
引擎支持a .. z
|
*
以及串联和括号
这是我制作的CFG:
exp = concat factor1
factor1 = "|" exp | e
concat = term factor2
factor2 = concat | e
term = element factor3
factor3 = * | e
element = (exp) | a .. z
等于
S = T X
X = "|" S | E
T = F Y
Y = T | E
F = U Z
Z = *| E
U = (S) | a .. z
对于交替和关闭,我可以轻松地处理它们,方法是向前看,然后根据令牌选择产品。 但是,由于它是隐式的,因此无法通过向前看来处理串联。
我想知道如何处理串联或语法有问题吗?
这是我的OCaml代码进行解析:
type regex =
| Closure of regex
| Char of char
| Concatenation of regex * regex
| Alternation of regex * regex
(*| Epsilon*)
exception IllegalExpression of string
type token =
| End
| Alphabet of char
| Star
| LParen
| RParen
| Pipe
let rec parse_S (l : token list) : (regex * token list) =
let (a1, l1) = parse_T l in
let (t, rest) = lookahead l1 in
match t with
| Pipe ->
let (a2, l2) = parse_S rest in
(Alternation (a1, a2), l2)
| _ -> (a1, l1)
and parse_T (l : token list) : (regex * token list) =
let (a1, l1) = parse_F l in
let (t, rest) = lookahead l1 in
match t with
| Alphabet c -> (Concatenation (a1, Char c), rest)
| LParen ->
(let (a, l1) = parse_S rest in
let (t1, l2) = lookahead l1 in
match t1 with
| RParen -> (Concatenation (a1, a), l2)
| _ -> raise (IllegalExpression "Unbalanced parentheses"))
| _ ->
let (a2, rest) = parse_T l1 in
(Concatenation (a1, a2), rest)
and parse_F (l : token list) : (regex * token list) =
let (a1, l1) = parse_U l in
let (t, rest) = lookahead l1 in
match t with
| Star -> (Closure a1, rest)
| _ -> (a1, l1)
and parse_U (l : token list) : (regex * token list) =
let (t, rest) = lookahead l in
match t with
| Alphabet c -> (Char c, rest)
| LParen ->
(let (a, l1) = parse_S rest in
let (t1, l2) = lookahead l1 in
match t1 with
| RParen -> (a, l2)
| _ -> raise (IllegalExpression "Unbalanced parentheses"))
| _ -> raise (IllegalExpression "Unknown token")
对于LL语法,FIRST集是允许作为规则的第一个标记的标记。 可以迭代地构造它们,直到达到固定点为止。
从第1步开始,然后重复第2步和第3步,直到FIRST设置达到固定点(不要更改)。 现在,您已经有了语法的真正第一集,并且可以使用前瞻性来确定每个规则。
注意:在您的代码中,parse_T函数与FIRST(T)集不匹配。 例如,如果您查看“ a | b”,则输入parse_T,而“ a”与parse_F调用匹配。 然后,前瞻为“ |” 在您的语法中匹配epsilon,但在您的代码中不匹配。
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