I am trying to write a simple parser in Racket, using the parser-tools. I got a behaviour which I could not explain (I am a Racket newbie, perhaps that is trivial).
Consider the following code:
#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 (λ(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")
After those definitions, res is a string:
> (string? res)
#t
and so is str.
If I try to execute a comparison with the "abcd123" string I get two different results:
> (eq? res "abcd123")
#f
> (eq? str "abcd123")
#t
Why is that? What am I missing here?
