ANTLR词法分析器规则即使不匹配也会消耗字符? [英] ANTLR lexer rule consumes characters even if not matched?
问题描述
我有一个反语言词典规则的奇怪副作用,并且我创建了一个(几乎)最小的工作示例来演示它.
在此示例中,我想例如匹配字符串[0..1]
.但是,当我调试语法时,到达解析器的令牌流仅包含[..1]
.第一个整数,无论它包含多少个数字,总是会消耗掉,而且我不知道它是如何发生的.如果我删除FLOAT
规则,一切都很好,所以我猜错误出在该规则的某处.但是由于它根本不匹配[0..1]
中的任何内容,所以我很困惑.
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.
对于任何可能出错的指针,我都会感到高兴.这是我的示例:
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 $b);
EXP : 'e' | 'E';
LBRACEVAR: '[';
RBRACEVAR: ']';
DOT: '.';
DDOTS: '..';
FLOAT
: INTEGER DOT POS_INTEGER
| INTEGER DOT POS_INTEGER EXP INTEGER
| INTEGER EXP INTEGER
;
INTEGER : POS_INTEGER | NEG_INTEGER;
fragment NEG_INTEGER : ('-') POS_INTEGER;
fragment POS_INTEGER : NUMBER+;
fragment NUMBER: ('0'..'9');
推荐答案
'0'
被词法分析器丢弃,并产生以下错误:
The '0'
is discarded by the lexer and the following errors are produced:
line 1:3 no viable alternative at character '.'
line 1:2 extraneous input '..' expecting INTEGER
这是因为当词法分析器遇到'0.'
时,它将尝试创建FLOAT
令牌,但不能.并且由于没有其他规则可以依赖以匹配'0.'
,因此它会产生错误,丢弃'0'
并创建DOT
令牌.
This is because when the lexer encounters '0.'
, it tries to create a FLOAT
token, but can't. And since there is no other rule to fall back on to match '0.'
, it produces the errors, discards '0'
and creates a DOT
token.
这就是ANTLR的词法分析器的工作方式:不会回溯匹配INTEGER
后跟DDOTS
的内容(请注意,backtrack=true
仅适用于解析器规则!).
This is simply how ANTLR's lexer works: it will not backtrack to match an INTEGER
followed by a DDOTS
(note that backtrack=true
only applies to parser rules!).
在FLOAT
规则内,必须确保在双精度'.'
前面时,产生的是INTEGER
令牌.您可以通过添加语法谓词(('..')=>
部分)并仅在单个'.'
后跟数字(('.' DIGIT)=>
部分)时产生FLOAT
令牌来做到这一点.请参见以下演示:
Inside the FLOAT
rule, you must make sure that when a double '.'
is ahead, you produce a INTEGER
token instead. You can do that by adding a syntactic predicate (the ('..')=>
part) and produce FLOAT
tokens only when a single '.'
is followed by a digit (the ('.' DIGIT)=>
part). See the following demo:
declaration
: LBRACEVAR INTEGER DDOTS INTEGER RBRACEVAR
;
LBRACEVAR : '[';
RBRACEVAR : ']';
DOT : '.';
DDOTS : '..';
INTEGER
: DIGIT+
;
FLOAT
: DIGIT+ ( ('.' DIGIT)=> '.' DIGIT+ EXP?
| ('..')=> {$type=INTEGER;} // change the token here
| EXP
)
;
fragment EXP : ('e' | 'E') DIGIT+;
fragment DIGIT : ('0'..'9');
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