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使用ANTLR4和Netbeans平台我创建了一个语法,在该语法中我在Netbeans平台应用程序中实现了语法高亮显示。一切工作正常,当我试图改变我的代码。IlegalStateException当我将未完成的代码键入支持语法高亮的JEditorPane中时
grammar MyRule;
my_rule : '(' my_rule ')'
| binary
| binary_hyst
| my_rule (AND my_rule)+
| my_rule (OR my_rule)+
;
binary : '(' binary ')'
| unary (EQ | NE) unary
;
binary_hyst : '(' binary_hyst ')'
| unary hyst? (GT | LT | GTE | LTE) unary hyst?
;
hyst : (HYST '(' unary ')');
unary : '(' unary ')'
| NUMBER
| STRING
| (MOV | DEC | INC) '(' unary ',' NUMBER ')'
| unary '*' unary
| unary '/' unary
| unary '+' unary
| unary '-' unary
;
// TOKENS
HYST : ('hyst');
MOV : ('mov');
INC : ('inc');
DEC : ('dec');
AND : ('&&');
OR : ('||');
EQ : ('==');
NE : ('!=');
GT : ('>');
LT : ('<');
GTE : ('>=');
LTE : ('<=');
NUMBER
: '-'? INT '.' [0-9]+ // 1.35, 0.3, -4.5
| '-'? INT // -3, 45
;
fragment INT : '0' | [1-9] [0-9]* ; // no leading zeros
STRING : '"' (ESC | .)*? '"' ;
ID : [a-zA-Z_]+;
fragment ESC: '\\' [btnr"\\] ;
WS : [ \t\n\r]+ -> channel(HIDDEN) ;
此语法允许这样的代码:("some_string" >= 20)
。但是,当我试图改变我的代码(即("some_id" >= 20) && ("some_other_id < 10)
我收到一个IllegalStateException:
java.lang.IllegalStateException: Lexer de.nor[email protected]181848
returned null token but lexerInput.readLength()=2
lexer-state: null
tokenStartOffset=20, readOffset=22, lookaheadOffset=23
Chars: "&\n" - these characters need to be tokenized.
Fix the lexer to not return null token in this state.
at org.netbeans.lib.lexer.LexerInputOperation.checkLexerInputFinished(LexerInputOperation.java:457)
at org.netbeans.lib.lexer.LexerInputOperation.nextToken(LexerInputOperation.java:217)
at org.netbeans.lib.lexer.inc.TokenListUpdater.relex(TokenListUpdater.java:627)
at
...
line 1:1 token recognition error at: '&\n'
这种情况权当我输入第一个“&”为“& &”令牌看来,Netbeans的要分析我的。代码和词法分析器认识到这不是一个有效的标记,这是正确的。如果我简单地复制并粘贴一个'& &'它工作正常,并且Netbeans不会引发异常所以这似乎来自我键入如何解决此问题?
完美地工作,谢谢! – padde 2014-09-30 06:36:27