【问题】
写了个antlr v3的语法:
grammar DDParserDemo;
options {
output = AST;
ASTLabelType = CommonTree; // type of $stat.tree ref etc...
}
ID : ('a'..'z'|'A'..'Z'|'_') ('a'..'z'|'A'..'Z'|'0'..'9'|'_')*
;
INT : '0'..'9'+;
;
FLOAT
: ('0'..'9')+ '.' ('0'..'9')* EXPONENT?
| '.' ('0'..'9')+ EXPONENT?
| ('0'..'9')+ EXPONENT
;
COMMENT
: '//' ~('\n'|'\r')* '\r'? '\n' {$channel=HIDDEN;}
| '/*' ( options {greedy=false;} : . )* '*/' {$channel=HIDDEN;}
;
WS : ( ' '
| '\t'
| '\r'
| '\n'
) {$channel=HIDDEN;}
;
STRING
: '"' ( ESC_SEQ | ~('\\'|'"') )* '"'
;
CHAR: '\'' ( ESC_SEQ | ~('\''|'\\') ) '\''
;
fragment
EXPONENT : ('e'|'E') ('+'|'-')? ('0'..'9')+ ;
fragment
HEX_DIGIT : ('0'..'9'|'a'..'f'|'A'..'F') ;
fragment
ESC_SEQ
: '\\' ('b'|'t'|'n'|'f'|'r'|'\"'|'\''|'\\')
| UNICODE_ESC
| OCTAL_ESC
;
fragment
OCTAL_ESC
: '\\' ('0'..'3') ('0'..'7') ('0'..'7')
| '\\' ('0'..'7') ('0'..'7')
| '\\' ('0'..'7')
;
fragment
UNICODE_ESC
: '\\' 'u' HEX_DIGIT HEX_DIGIT HEX_DIGIT HEX_DIGIT
;
NEWLINE : '\r'? '\n' ;
DECIMAL_VALUE
: (INT)+;
HEX_VALUE
: '0x' (HEX_DIGIT)+;
prog : identification+ ;
identification : definiton ','? (WS)* -> definiton
;
definiton : ID (' '|'\t')+ (DECIMAL_VALUE | HEX_VALUE)
;
结果出错了:
| [17:02:19] error(100): DDParserDemo.g:12:5: syntax error: antlr: MissingTokenException(inserted [@-1,0:0='<missing EOF>’,<-1>,12:4] at 😉 |
【解决过程】
1.才注意到,原来是INT后面多个冒号:
INT : '0'..'9'+;
;
去掉多余的冒号,即可:
INT : '0'..'9'+;
【总结】
虽然有antlr的语法错误,但是此处AntlrWorks却没有报错,显示的是绿色的:
所以,貌似此处属于语义错误?
否则,那就是antlrworks的bug了。
转载请注明:在路上 » 【已解决】error syntax error: antlr: MissingTokenException(inserted [@-1,0:0='<missing EOF>’,<-1>,12:4] at 😉