使用 Scheme 解析字符串

Question

我正在尝试编写一个简单的解析器，它从字符串 e.xml 创建一个 sxml 表达式。 G。

"This is a [Test]" ===> (item "This is a" (subitem "Test"))

任何想知道给定示例中的方括号的人都可以查看所谓的 Leiden conventions .

这是我到目前为止编写的代码:

(define my-sequence '("this" "[" "is" "a" "]" "test"))

(define (left-square-bracket? item)
  (or (equal? item "[")
      (eq? item #\x005b)))

(define (right-square-bracket? item)
  (or (equal? item "]")
      (eq? item #\x005d)))

(define (parse-sequence sequence)
  (cond ((null? sequence) '())
        ((left-square-bracket? (car sequence))
         (let ((subsequence (get-subsequence (cdr sequence))))
           (list subsequence)))
        (else
         (cons (car sequence)
               (parse-sequence (cdr sequence))))))

(define (get-subsequence sequence)
  (if (right-square-bracket? (car sequence))
      '()
      (cons (car sequence)
            (get-subsequence (cdr sequence)))))

正在评估 (parse-sequence my-sequence) yield ("this" ("is" "a")) .已创建嵌套表达式，但程序在没有计算最后一项的情况下完成 "test" .问题是，我如何从 get-subsequence 返回至 parse-sequence ?

任何帮助表示赞赏，非常感谢提前! :)

Answer 1

要解决您最初的问题，如何返回多个值:使用“值”表单。这是一个示例实现，其中内部过程返回要处理的剩余列表和到目前为止的结果。它在左括号上递归。

(define (parse-sequence lst)

  (define (parse-seq lst)
    (let loop ((lst lst) (res null))
      (cond
        ((null? lst) (values null res))
        ((string=? (car lst) "[")
         (let-values ([(lst2 res2) (parse-seq (cdr lst))])
           (loop lst2 (append res (list res2)))))
        ((string=? (car lst) "]")
         (values (cdr lst) res))
        (else
          (loop (cdr lst) (append res (list (car lst))))))))

  (let-values ([(lst res) (parse-seq lst)])
    res))

然后

(parse-sequence '("this" "is" "a" "test"))
(parse-sequence '("this" "[" "is" "a" "]" "test"))
(parse-sequence '("this" "[" "is" "[" "a" "]" "]" "test"))

会屈服

'("this" "is" "a" "test")
'("this" ("is" "a") "test")
'("this" ("is" ("a")) "test")