相等字符串之间的差异

Question

我正在尝试Racket使用parser-tools.编写一个简单的解析器.我得到了一个我无法解释的行为(我是一个球拍新手,也许这是微不足道的).

请考虑以下代码:

#lang racket
(require parser-tools/yacc
         parser-tools/lex
         (prefix-in : parser-tools/lex-sre))


(define-tokens value-tokens   ;;token which have a value
  (STRING-VALUE ))
(define-empty-tokens op-tokens ;;token without a values
  (EOF))
(define-lex-abbrevs ;;abbreviation 
  [STRING (:+ (:or (:/ "a" "z") (:/ "A" "Z") (:/ "0" "9") "." "_" "-"))]
  )


(define lex-token
  (lexer
   [(eof) 'EOF]
   ;; recursively call the lexer on the remaining input after a tab or space.  Returning the "1+1")
   ;; result of that operation.  This effectively skips all whitespace.
   [(:or #\tab #\space #\newline)
    (lex-token input-port)]
   [(:seq STRING) (token-STRING-VALUE lexeme)]
   ))

(define test-parser
  (parser
   (start query)
   (end EOF)
   (tokens value-tokens op-tokens)
   (error (?(ok? name value) (printf "Couldn't parse: ~a\n" name)))

   (grammar
    (query [(STRING-VALUE)      $1])
    )))

(define s (open-input-string 
           "abcd123"))

(define res
  (test-parser (lambda () (lex-token s))))

(define str "abcd123")

在这些定义之后,res是一个字符串:

> (string? res)
#t

等等str.

如果我尝试与"abcd123"字符串执行比较,我会得到两个不同的结果:

> (eq? res "abcd123")
#f

> (eq? str "abcd123")
#t

这是为什么？我在这里错过了什么？

Answer 1

你应该比较字符串equal?.