OSCL/cl-ppcre
2
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Report a syntax error on "(?<".

Browse Source 
Instead of failing due to NIL not being a character in a call to ALPHA-CHAR-P.
This commit is contained in:
Stas Boukarev
2016-12-27 19:33:27 +03:00
parent 0f6fe804f8
commit 0f295337d9
1 changed files with 32 additions and 30 deletions
Download Patch File Download Diff File Expand all files Collapse all files

62
lexer.lisp
View File

@ -672,37 +672,39 @@ closing #\> will also be consumed."
;; might be a look-behind assertion or a named group, so ;; might be a look-behind assertion or a named group, so
;; check next character ;; check next character
(let ((next-char (next-char-non-extended lexer))) (let ((next-char (next-char-non-extended lexer)))
(if (alpha-char-p next-char) (cond ((and next-char
(progn (alpha-char-p next-char))
;; we have encountered a named group ;; we have encountered a named group
;; are we supporting register naming? ;; are we supporting register naming?
(unless *allow-named-registers* (unless *allow-named-registers*
(signal-syntax-error* (1- (lexer-pos lexer)) (signal-syntax-error* (1- (lexer-pos lexer))
"Character '~A' may not follow '(?<'." "Character '~A' may not follow '(?<' (because ~a = NIL)"
next-char)) next-char
;; put the letter back '*allow-named-registers*))
(decf (lexer-pos lexer)) ;; put the letter back
;; named group (decf (lexer-pos lexer))
:open-paren-less-letter) ;; named group
(case next-char :open-paren-less-letter)
((#\=)
;; positive look-behind
:open-paren-less-equal)
((#\!)
;; negative look-behind
:open-paren-less-exclamation)
((#\))
;; Perl allows "(?<)" and treats
;; it like a null string
:void)
((nil)
;; syntax error
(signal-syntax-error "End of string following '(?<'."))
(t (t
;; also syntax error (case next-char
(signal-syntax-error* (1- (lexer-pos lexer)) ((#\=)
"Character '~A' may not follow '(?<'." ;; positive look-behind
next-char )))))) :open-paren-less-equal)
((#\!)
;; negative look-behind
:open-paren-less-exclamation)
((#\))
;; Perl allows "(?<)" and treats
;; it like a null string
:void)
((nil)
;; syntax error
(signal-syntax-error "End of string following '(?<'."))
(t
;; also syntax error
(signal-syntax-error* (1- (lexer-pos lexer))
"Character '~A' may not follow '(?<'."
next-char )))))))
(otherwise (otherwise
(signal-syntax-error* (1- (lexer-pos lexer)) (signal-syntax-error* (1- (lexer-pos lexer))
"Character '~A' may not follow '(?'." "Character '~A' may not follow '(?'."