I'm completely new to ANTLR and EBNF grammars to begin with, so this is probably a basic issue I'm simply not understanding.
I have a rule such as:
version_line : WS? 'VERS' WS? '=' WS? '1.0' WS? EOL ;
WS : ' '+ ;
EOL : '\r' | '\n' | '\r\n' | '\n\r' ;
that matches a statement in my input file that looks like this (with optional whitespace):
VERSION = 1.0
With the rule form above, I'm getting a successful match, although I get an exception with this form:
version_line : WS? 'VERS' WS? '=' WS? '1' '.0' WS? EOL ;
or this form:
vers开发者_StackOverflowion_line : WS? 'VERS' WS? '=' WS? DIGIT '.0' WS? EOL ;
DIGIT : '1' ;
Why is this different? I discovered this issue when trying to decompose the rule even more, hopefully ending up with something like this:
version_line : WS? 'VERS' WS? '=' WS? DIGIT '.' DIGIT WS? EOL ;
DIGIT : '0'..'9' ;
I see no problem, all four grammars produce the expected AST:
1
version_line : WS? 'VERSION' WS? '=' WS? '1.0' WS? EOL ;
WS : ' '+ ;
EOL : '\r' | '\n' | '\r\n' | '\n\r' ;
2
version_line : WS? 'VERSION' WS? '=' WS? '1' '.0' WS? EOL ;
WS : ' '+ ;
EOL : '\r' | '\n' | '\r\n' | '\n\r' ;
3
version_line : WS? 'VERSION' WS? '=' WS? DIGIT '.0' WS? EOL ;
DIGIT : '1' ;
WS : ' '+ ;
EOL : '\r' | '\n' | '\r\n' | '\n\r' ;
4
version_line : WS? 'VERSION' WS? '=' WS? DIGIT '.' DIGIT WS? EOL ;
DIGIT : '0'..'9' ;
WS : ' '+ ;
EOL : '\r' | '\n' | '\r\n' | '\n\r' ;
with input:
VERSION = 1.0
#
(Note that the #
in the input is a new line char!)
Tested with ANTLRWorks v1.3.1.
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