Antlr规则优先级 [英] Antlr rule priorities
问题描述
首先,我知道这种语法没有意义,但是它是为测试ANTLR规则优先级行为而创建的
Firstly I know this grammar doesn't make sense but it was created to test out the ANTLR rule priority behaviour
grammar test;
options
{
output=AST;
backtrack=true;
memoize=true;
}
rule_list_in_order :
(
first_rule
| second_rule
| any_left_over_tokens)+
;
first_rule
:
FIRST_TOKEN
;
second_rule:
FIRST_TOKEN NEW_LINE SECOND_TOKEN NEW_LINE;
any_left_over_tokens
:
NEW_LINE
| FIRST_TOKEN
| SECOND_TOKEN;
FIRST_TOKEN
: 'First token here'
;
SECOND_TOKEN
: 'Second token here';
NEW_LINE
: ('\r'?'\n') ;
WS : (' '|'\t'|'\u000C')
{$channel=HIDDEN;}
;
当我将此语法输入此处的第一个标记\ n此处的第二个标记"时,它与second_rule相匹配.
When I give this grammar the input 'First token here\nSecond token here', it matches the second_rule.
我希望它与第一个规则匹配,然后与any_left_over_tokens匹配,因为first_rule出现在Rule_order_list的第二个规则之前,后者是起点.谁能解释为什么会这样?
I would have expected it to match the first rule then any_left_over_tokens because the first_rule appears before the second_rule in the rule_order_list which is the start point. Can anyone explain why this happens?
欢呼
推荐答案
首先,ANTLR的词法分析器将从上到下对输入进行标记化.因此,首先定义的令牌优先于其下面的令牌.如果规则的标记重叠,则匹配最多字符的规则将优先(贪婪匹配).
First of all, ANTLR's lexer will tokenize the input from top to bottom. So tokens defined first have a higher precedence than the ones below it. And in case rule have overlapping tokens, the rule that matches the most characters will take precedence (greedy match).
相同的原则在解析器规则中也适用.首先定义的规则也将首先匹配.例如,在规则foo
中,子规则a
将首先在b
之前尝试:
The same principle holds within parser rules. Rules defined first will also be matched first. For example, in rule foo
, sub-rule a
will first be tried before b
:
foo
: a
| b
;
请注意,在您的情况下,2 nd 规则不匹配,但尝试匹配,但由于没有尾随换行符而失败,并产生错误:
Note that in your case, the 2nd rule isn't matched, but tries to do so, and fails because there is no trailing line break, producing the error:
line 0:-1 mismatched input '<EOF>' expecting NEW_LINE
所以,根本没有匹配的东西.但是那很奇怪.由于您已经设置了backtrack=true
,因此它至少应该回溯并匹配:
So, nothing is matched at all. But that is odd. Because you've set the backtrack=true
, it should at least backtrack and match:
-
first_rule
(此处的第一个令牌") -
any_left_over_tokens
(换行符") -
any_left_over_tokens
(此处是第二个令牌")
first_rule
("First token here")any_left_over_tokens
("line-break")any_left_over_tokens
("Second token here")
如果不匹配first_rule
首先,甚至不尝试匹配second_rule
.
if not match first_rule
in the first place and not even try to match second_rule
to begin with.
在手动执行谓词(并在选项{...} 部分中禁用backtrack
)时,快速演示如下:
A quick demo when doing the predicates manually (and disabling the backtrack
in the options { ... } section) would look like:
grammar T;
options {
output=AST;
//backtrack=true;
memoize=true;
}
rule_list_in_order
: ( (first_rule)=> first_rule {System.out.println("first_rule=[" + $first_rule.text + "]");}
| (second_rule)=> second_rule {System.out.println("second_rule=[" + $second_rule.text + "]");}
| any_left_over_tokens {System.out.println("any_left_over_tokens=[" + $any_left_over_tokens.text + "]");}
)+
;
first_rule
: FIRST_TOKEN
;
second_rule
: FIRST_TOKEN NEW_LINE SECOND_TOKEN NEW_LINE
;
any_left_over_tokens
: NEW_LINE
| FIRST_TOKEN
| SECOND_TOKEN
;
FIRST_TOKEN : 'First token here';
SECOND_TOKEN : 'Second token here';
NEW_LINE : ('\r'?'\n');
WS : (' '|'\t'|'\u000C') {$channel=HIDDEN;};
可以通过以下类进行测试:
which can be tested with the class:
import org.antlr.runtime.*;
public class Main {
public static void main(String[] args) throws Exception {
String source = "First token here\nSecond token here";
ANTLRStringStream in = new ANTLRStringStream(source);
TLexer lexer = new TLexer(in);
CommonTokenStream tokens = new CommonTokenStream(lexer);
TParser parser = new TParser(tokens);
parser.rule_list_in_order();
}
}
产生预期的输出:
first_rule=[First token here]
any_left_over_tokens=[
]
any_left_over_tokens=[Second token here]
请注意,使用以下内容无关紧要:
Note that it doesn't matter if you use:
rule_list_in_order
: ( (first_rule)=> first_rule
| (second_rule)=> second_rule
| any_left_over_tokens
)+
;
或
rule_list_in_order
: ( (second_rule)=> second_rule // <--+--- swapped
| (first_rule)=> first_rule // <-/
| any_left_over_tokens
)+
;
,两者都会产生预期的输出.
, both will produce the expected output.
所以,我的猜测是您可能已经找到了一个错误.
So, my guess is that you may have found a bug.
如果您想要一个明确的答案(Terence Parr在这里的访问频率比他在这里更高的频率),则可以尝试ANTLR邮件列表.
Yout could try the ANTLR mailing-list, in case you want a definitive answer (Terence Parr frequents there more often than he does here).
祝你好运!
PS.我用ANTLR v3.2进行了测试
PS. I tested this with ANTLR v3.2
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