I've got a strange side effect of an antlr lexer rule and I've created an (almost) minimal working example to demonstrate it.
In this example I want to match the String [0..1]
for example. But when I debug the grammar the token stream that reaches the parser only contains [..1]
. The first integer, no matter how many digits it contains is always consumed and I've got no clue as to how that happens. If I remove the FLOAT
rule everything is fine so I guess the mistake lies somewhere in that rule. But since it shouldn't match anything in [0..1]
at all I'm quite puzzled.
I'd be happy for any pointers where I might have gone wrong. This is my example:
grammar min;
options{
language = Java;
output = AST;
ASTLabelType=CommonTree;
backtrack = true;
}
tokens {
DECLARATION;
}
declaration : LBRACEVAR a=INTEGER DDOTS b=INTEGER RBRACEVAR -> ^(DECLARATION $a $b);
EXP : 'e' | 'E';
LBRACEVAR: '[';
RBRACEVAR: ']';
DOT: '.';
DDOTS: '..';
FLOAT
: INTEGER DOT POS_INTEGER
| INTEGER DOT POS_INTEGER EXP INTEGER
| INTEGER EXP INTEGER
;
INTEGER : POS_INTEGER | NEG_INTEGER;
fragment NEG_INTEGER : ('-') POS_INTEGER;
fragment POS_INTEGER : NUMBER+;
fragment NUMBER: ('0'..'9');
The '0'
is discarded by the lexer and the following errors are produced:
line 1:3 no viable alternative at character '.'
line 1:2 extraneous input '..' expecting INTEGER
This is because when the lexer encounters '0.'
, it tries to create a FLOAT
token, but can't. And since there is no other rule to fall back on to match '0.'
, it produces the errors, discards '0'
and creates a DOT
token.
This is simply how ANTLR's lexer works: it will not backtrack to match an INTEGER
followed by a DDOTS
(note that backtrack=true
only applies to parser rules!).
Inside the FLOAT
rule, you must make sure that when a double '.'
is ahead, you produce a INTEGER
token instead. You can do that by adding a syntactic predicate (the ('..')=>
part) and produce FLOAT
tokens only when a single '.'
is followed by a digit (the ('.' DIGIT)=>
part). See the following demo:
declaration
: LBRACEVAR INTEGER DDOTS INTEGER RBRACEVAR
;
LBRACEVAR : '[';
RBRACEVAR : ']';
DOT : '.';
DDOTS : '..';
INTEGER
: DIGIT+
;
FLOAT
: DIGIT+ ( ('.' DIGIT)=> '.' DIGIT+ EXP?
| ('..')=> {$type=INTEGER;} // change the token here
| EXP
)
;
fragment EXP : ('e' | 'E') DIGIT+;
fragment DIGIT : ('0'..'9');
来源:https://stackoverflow.com/questions/10136324/antlr-lexer-rule-consumes-characters-even-if-not-matched