Antlr规则优先级

编程入门行业动态更新时间:2024-10-12 03:26:28

本文介绍了Antlr规则优先级的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！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.