antlr4

问题 The input content: The grammar: grammar test; p : EOF; Char : [a-z]; fragment Tab : '\t'; fragment Space : ' '; T1 : (Tab|Space)+ ->skip; T2 : '#' T1+ Char+; The matching result is this: [@0,0:6='# abc',<T2>,1:0] <<<<<<<< PLACE 1 [@1,7:6='<EOF>',<EOF>,1:7] line 1:0 extraneous input '# abc' expecting <EOF> Please ignore the error in the last line. I am wondering why the token matched at PLACE 1 is T2 . In the grammar file, the T2 lexer rule goes after the T1 lexer rule. So I expect T1 rule

问题 I'm trying to create a Beginning-Of-Line token: lexer grammar ScriptLexer; BOL : {getCharPositionInLine() == 0;}; // Beginning Of Line token But the above emits the error The name 'getCharPositionInLine' does not exist in the current context As it creates this code: private void BOL_action(RuleContext _localctx, int actionIndex) { switch (actionIndex) { case 0: getCharPositionInLine() == 0; break; } } Where the getCharPositionInLine() method doesn't exist... 回答1: Simplest approach is to just

问题 I'm trying to adapt to JS target the Expr.g4 of the book. In this example, actions are directly in the grammar. They include utility functions defined in @parser::members , that are called in the rules. The example works well in Java, but in its JS translation I have 2 problem: - getting the action function visible by the action rule - getting the tokens recognized in the function. I finally manage to get this working formulation: @parser::members { myeval = function(left, op, right) { switch

问题 Here is the example. This ($type) is not recognised by ANTLR4. Number //options { backtrack=true; } : IntegerLiteral { $type = IntegerLiteral; } | FloatLiteral { $type = FloatLiteral; } | IntegerLiteral { $type = IntegerLiteral; } ; What could this be replaced by? Thank you. 回答1: In ANTLR v4, do: Number : IntegerLiteral {setType(IntegerLiteral);} | ... 回答2: In ANTLR 4, this is the new syntax: Foo : Bar -> type(SomeType) | ... ; However, for the rule you have above you should just remove the

问题 I'm trying to use antlr4 version 4.4 and the python2 runtime. The grammar is from the antlr4 book, page 6, file: Hello.g4 : grammar Hello; r : 'hello' ID ; ID : [a-z]+ ; WS : [ \t\r\n]+ -> skip ; and I generate lexer and parser with command antlr4 -Dlanguage=Python2 Hello.g4 the files HelloLexer.py , HelloParser.py and HelloListener.py among other, are then generated. I make a main program test.py to test the generated python parser: from antlr4 import * from HelloLexer import HelloLexer from

问题 I tried this solution but it didn't seem to work for me Here's an excerpt from my grammer: module : BEGIN MODULE IDENT STRING module_element* END MODULE ; module_element : element_1 | element_2 | element_3 | ... ; There is a bigger tree below each element. Now when a RecognitionException occurs I want to consume tokens until either the next module_element matches or the parent END MODULE matches. Any hints on how to do this inside a class inheriting from DefaultErrorStrategy? edit: Here is a

问题 I'm trying to convert my grammar from v3 to v4 and having some trouble. In v3 I have rules like this: dataspec[DataLayout layout] returns [DataExtractor extractor] @init { DataExtractorBuilder builder = new DataExtractorBuilder(layout); } @after { extractor = builder.create(); } : first=expr { builder.addAll(first); } (COMMA next=expr { builder.addAll(next); })* ; expr returns [List<ValueExtractor> ext] ... However, with rules in v4 returning these custom context objects instead of what I

问题 I'm new to ANTLR and trying to write grammar in ANTLR4 without any prior brush with the previous version. I'm following the book ' The Definitive ANTLR 4 Reference '. I use Eclipse and installed ANTLR4 IDE as given in here. I wrote the following grammar in Expr.g4: grammar Expr; import Common; options{ language = Java; } prog: stat+; stat: expr NEWLINE | ID '=' expr NEWLINE | NEWLINE; expr: expr ('/'|'*') expr | expr ('+'|'-') expr | INT | ID | '('expr')'; The Common.g4 contains the following

问题 I am writing a grammar to handle scalar and vector expressions. The grammar below is simplified to show the problem I have where a scalar expression can be derived from a vector and a vector can be derived from a scalar. For example, a vector could be a literal [1, 2, 3] or the product of a scalar and a vector 2 * [1, 2, 3] (equivalent to [2, 4, 6] ). A scalar could be a literal 2 or an index into a vector [1, 2, 3][1] (equivalent to 2 ). grammar LeftRecursion; Integer : [0-9]+ ; WhiteSpace :

问题 I am new to Scala and I am trying to parse Scala files with the use of Scala Grammar and ANTLR. Below is the code for Scala Grammar which I got from the git hub link: https://github.com/antlr/grammars-v4/tree/master/scala There are chances of repo to be moved so I am pasting the Scala grammar code here: grammar Scala; literal : '-'? IntegerLiteral | '-'? FloatingPointLiteral | BooleanLiteral | CharacterLiteral | StringLiteral | SymbolLiteral | 'null' ; qualId : Id ('.' Id)* ; ids : Id (',' Id