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50 lines
1.4 KiB
Racket
50 lines
1.4 KiB
Racket
#lang racket/base
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(require brag/examples/0n1n
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brag/support
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rackunit)
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(define (lex ip)
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(port-count-lines! ip)
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(lambda ()
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(define next-char (read-char ip))
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(cond [(eof-object? next-char)
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(token eof)]
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[(char=? next-char #\0)
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(token "0" "0")]
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[(char=? next-char #\1)
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(token "1" "1")])))
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;; The only rule in the grammar is:
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;;
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;; rule-0n1n: ["0" rule-0n1n "1"]
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;;
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;; It makes use of the "maybe" pattern. The result type of the
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;; grammar rule is:
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;;
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;; rule-0n1n: (U #f
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;; (list "0" rule-0n1n "1"))
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(check-equal? (syntax->datum (parse #f (lex (open-input-string "0011"))))
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'(rule-0n1n "0" (rule-0n1n "0" (rule-0n1n) "1") "1"))
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(check-equal? (syntax->datum (parse #f (lex (open-input-string "01"))))
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'(rule-0n1n "0" (rule-0n1n) "1"))
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(check-equal? (syntax->datum (parse #f (lex (open-input-string ""))))
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'(rule-0n1n))
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(check-equal? (syntax->datum (parse #f (lex (open-input-string "000111"))))
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'(rule-0n1n "0" (rule-0n1n "0" (rule-0n1n "0" (rule-0n1n) "1") "1") "1"))
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(check-exn exn:fail:parsing?
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(lambda () (parse #f (lex (open-input-string "0001111")))))
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(check-exn exn:fail:parsing?
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(lambda () (parse #f (lex (open-input-string "0001110")))))
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(check-exn exn:fail:parsing?
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(lambda () (parse #f (lex (open-input-string "10001110")))))
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