Antlr4语法歧义 [英] Antlr4 grammar ambiguity

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本文介绍了Antlr4语法歧义的处理方法,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

我有以下语法(对于SO来说是最小的)

I have the following grammar ( minimized for SO)

grammar Hello;

odataIdentifier  : identifierLeadingCharacter identifierCharacter*; 


identifierLeadingCharacter : Alpha| UNDERSCORE;
identifierCharacter :  identifierLeadingCharacter | Digit;
identifierUnreserved    : identifierCharacter | (MINUS | DOT | TILDE);

Digit  : ZERO_TO_FIVE |[6-9];

ONEHUNDRED_TO_ONEHUNDREDNINETYNINE : '1' Digit Digit;            // 100-199
TWOHUNDRED_TO_TWOHUNDREDFOURTYNINE : '2' ZERO_TO_FOUR Digit;     // 200-249
TWOHUNDREDFIFTY_TO_TWOHUNDREDFIFTYFIVE : '25' ZERO_TO_FIVE;      // 250-255
TEN_TO_NINETYNINE : ONE_TO_NINE Digit;                           // 10-99

ZERO_TO_ONE : [0-1];
ZERO_TO_TWO : ZERO_TO_ONE | [2];
ZERO_TO_THREE : ZERO_TO_TWO | [3];
ZERO_TO_FOUR : ZERO_TO_THREE | [4];
ZERO_TO_FIVE : ZERO_TO_FOUR | [5];

ONE_TO_TWO  : [1-2];
ONE_TO_THREE  : ONE_TO_TWO | [3];
ONE_TO_FOUR  : ONE_TO_THREE | [4]; 
ONE_TO_NINE  : ONE_TO_FOUR | [5-9]; 

Alpha  : [a-zA-Z];

MINUS : [-];
DOT : '.';
UNDERSCORE : '_';
TILDE : '~';

WS  :  (' '|'\r'|'\t'|'\u000C'|'\n') -> skip
    ;

对于输入c9来说,它工作正常,但是当我有2位数字时,例如c10,它说:

for input c9 it works fine, but when i have 2 digits for example c10 it says:

extraneous input '92' expecting {<EOF>, Digit, Alpha, '_'}

所以我想它解析9和解析2,但不知道这应该是TEN_TO_NINETYNINE还是2 Digit Digit. 我对此很菜鸟,所以想知道我的分析是否正确,以及如何减轻这种情况...

so i guess it parses 9 and parses 2 and doesn't know if this should be TEN_TO_NINETYNINE or 2 Digit Digit. i am a noob to this, so wondering if my analysis is right and how could i alleviate this ...

推荐答案

您的输入将生成一个Alpha令牌,后跟一个TEN_TO_NINETYNINE令牌.尽管解析器规则identifierLeadingCharacter确实允许使用Alpha令牌,但identifierCharacter规则不能与TEN_TO_NINETYNINE令牌匹配.

Your input is resulting in an Alpha token followed by a TEN_TO_NINETYNINE token. While the parser rule identifierLeadingCharacter does allow the Alpha token, the identifierCharacter rule cannot match a TEN_TO_NINETYNINE token.

输入10总是产生一个TEN_TO_NINETYNINE令牌,而不是两个Digit令牌,因为前者匹配更多的输入,并且词法分析器规则很贪婪.

The input 10 will always produce a TEN_TO_NINETYNINE token rather than two Digit tokens, because the former matches more of the input and lexer rules are greedy.

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