ANTLR: mismatched input - antlr

I couldn't understand a bug in my grammar. The file, Bug.g4, is:
grammar Bug;
text: TEXT;
WORD: ('a'..'z' | 'A'..'Z')+ ;
TEXT: ('a'..'z' | 'A'..'Z')+ ;
NEWLINE: [\n\r] -> skip ;
After running antlr4 and javac, I run
grun Bug text -tree
aa
line 1:0 mismatched input 'aa' expecting TEXT
(text aa)
But if I instead use text: WORD in the grammar, things are okay. What's wrong?

When two lexer rules each match the same string of text, and no other lexer rule matches a longer string of text, ANTLR assigns the token type according to the rule which appeared first in the grammar. In your case, a TEXT token can never be produced by the lexer rule because the WORD rule will always match the same text and the WORD rule appears before the TEXT rule in the grammar. If you were to reverse the order of these rules in the grammar, you would start to see TEXT tokens but you would never see a WORD token.

Related

Ambiguous Lexer rules in Antlr

I have an antlr grammar with multiple lexer rules that match the same word. It can't be resolved during lexing, but with the grammar, it becomes unambiguous.
Example:
conversion: NUMBER UNIT CONVERT UNIT;
NUMBER: [0-9]+;
UNIT: 'in' | 'meters' | ......;
CONVERT: 'in';
Input: 1 in in meters
The word "in" matches the lexer rules UNIT and CONVERT.
How can this be solved while keeping the grammar file readable?
When an input matches two lexer rules, ANTLR chooses either the longest or the first, see disambiguate. With your grammar, in will be interpreted as UNIT, never CONVERT, and the rule
conversion: NUMBER UNIT CONVERT UNIT;
can't work because there are three UNIT tokens :
$ grun Question question -tokens -diagnostics input.txt
[#0,0:0='1',<NUMBER>,1:0]
[#1,1:1=' ',<WS>,channel=1,1:1]
[#2,2:3='in',<UNIT>,1:2]
[#3,4:4=' ',<WS>,channel=1,1:4]
[#4,5:6='in',<UNIT>,1:5]
[#5,7:7=' ',<WS>,channel=1,1:7]
[#6,8:13='meters',<UNIT>,1:8]
[#7,14:14='\n',<NL>,1:14]
[#8,15:14='<EOF>',<EOF>,2:0]
Question last update 0159
line 1:5 missing 'in' at 'in'
line 1:8 mismatched input 'meters' expecting <EOF>
What you can do is to have only ID or TEXT tokens and distinguish them with a label, like this :
grammar Question;
question
#init {System.out.println("Question last update 0132");}
: conversion NL EOF
;
conversion
: NUMBER unit1=ID convert=ID unit2=ID
{System.out.println("Quantity " + $NUMBER.text + " " + $unit1.text +
" to convert " + $convert.text + " " + $unit2.text);}
;
ID : LETTER ( LETTER | DIGIT | '_' )* ; // or TEXT : LETTER+ ;
NUMBER : DIGIT+ ;
NL : [\r\n] ;
WS : [ \t] -> channel(HIDDEN) ; // -> skip ;
fragment LETTER : [a-zA-Z] ;
fragment DIGIT : [0-9] ;
Execution :
$ grun Question question -tokens -diagnostics input.txt
[#0,0:0='1',<NUMBER>,1:0]
[#1,1:1=' ',<WS>,channel=1,1:1]
[#2,2:3='in',<ID>,1:2]
[#3,4:4=' ',<WS>,channel=1,1:4]
[#4,5:6='in',<ID>,1:5]
[#5,7:7=' ',<WS>,channel=1,1:7]
[#6,8:13='meters',<ID>,1:8]
[#7,14:14='\n',<NL>,1:14]
[#8,15:14='<EOF>',<EOF>,2:0]
Question last update 0132
Quantity 1 in to convert in meters
Labels are available from the rule's context in the visitor, so it is easy to distinguish tokens of the same type.
Based on the info in your question, it's hard to say what the best solution would be - I don't know what your lexer rules are, for example - nor can I tell why you have lexer rules that are ambiguous at all.
In my experience with antlr, lexer rules don't generally carry any semantic meaning; they are just text that matches some kind of regular expression. So, instead of having VARIABLE, METHOD_NAME, etc, I'd just have IDENTIFIER, and then figure it out at a higher level.
In other words, it seems (from the little I can glean from your question) that you might benefit either from replacing UNIT and CONVERT with grammar rules, or just having a single rule:
conversion: NUMBER TEXT TEXT TEXT
and validating the text values in your ANTLR listener/tree-walker/etc.
EDIT
Thanks for updating your question with lexer rules. It's clear now why it's failing - as BernardK points out, antlr will always choose the first matching lexer rule. This means it's impossible for the second of two ambiguous lexer rules to match, and makes your proposed design infeasible.
My opinion is that lexer rules are not the correct layer to do things like unit validation; they excel at structure, not content. Evaluating the parse tree will be much more practical than trying to contort an antlr grammar.
Finally, you might also do something with embedded actions on parse rules, like validating the value of an ID token against a known set of units. It could work, but would destroy the reusability of your grammar.

antlr4 two lexer rule match the same string

I'm currently using antlr4 to build a parser, but I encountered a problem which I tried my best but didn't figure out. Can you help me to explaain and solve it ?
# grammer file : PluginDoc.g4:
grammer PluginDoc
pluginDef : pluginName | pluginDesc;
pluginName : PluginName IDENTIFIER;
pluginDesc : PluginDesc TEXT;
PluginName '#pluginName'
PluginDesc '#pluginDesc'
IDENTIFIER : [a-zA-Z_]+;
TEXT : ~( ' ' | '\n' | '\t' )+;
input content is:
#pluginName kafka
#pluginDesc abc
If I put IDENTIFIER before TEXT, I will get "mismatched input 'abc' expecting TEXT"
If I put TEXT before IDENTIFIER, I will get "mismatched input 'kafka' expecting IDENTIFIER"
Looks like both IDENTIFIER and TEXT are matched, how can I only match IDENTIFIER in pluginName and only match TEXT in pluginDesc ?
First of all, you have several errors in the grammar that you posted:
The header of the file should specify grammar, not grammer. Your Lexer tokens PluginName and PluginDesc do not have a colon in front of them and semicolon to terminate them. It is also an (unwritten?) rule to write your parser rules as all lower-case and your lexer rules as all upper-case.
grammar PluginDoc;
pluginDef : pluginName | pluginDesc;
pluginName : PLUGIN_NAME IDENTIFIER;
pluginDesc : PLUGIN_DESC TEXT;
PLUGIN_NAME : '#pluginName';
PLUGIN_DESC : '#pluginDesc';
IDENTIFIER : [a-zA-Z_]+;
TEXT : ~( ' ' | '\n' | '\t' )+;
Some of the problems that I encountered while testing your grammar were due to the unhandled whitespace. First of all, you should include a Lexer rule to skip the whitespace at the end of the file after all of the other Lexer rules.
WS: [ \n\t\r]+ -> skip;
Next, there is a problem with your TEXT and IDENTIFIER clashing with each other. When the character stream is tokenized by the Lexer, kafka and abc can be both IDENTIFIER and TEXT token. Since the Lexer lexes in a top-down fashion, they are both tokenized as whateve Lexer rule comes first in your grammar. This causes the error that you encounter - whatever you define as the second rule cannot be matched in the parser because it was not sent in as a token.
As suggested by Lucas, you should probably match both of these as TEXT and do the subsequent checking for validity of the input in your Listener/Visitor.
grammar PluginDoc;
pluginDef : (pluginName | pluginDesc)* EOF;
pluginName : PLUGIN_NAME TEXT;
pluginDesc : PLUGIN_DESC TEXT;
PLUGIN_NAME: '#pluginName';
PLUGIN_DESC: '#pluginDesc';
TEXT : ~[ \r\n\t]+;
WS: [ \r\n\t]+ -> skip;
I also changed the pluginDef Parser rule to
pluginDef : (pluginName | pluginDesc)* EOF;
since it was my impression that you want to input both #pluginName X and #pluginDesc Y at once and identify them. If this is not the case, feel free to change back to what you had before.
The resulting AST produced by the modified grammar above onyour sample input:
You can also run this with a text file as an input.

Problems with ANTLR4 grammar

I have a very simple grammar file, which looks like this:
grammar Wort;
// Parser Rules:
word
: ANY_WORD EOF
;
// Lexer Rules:
ANY_WORD
: SMALL_WORD | CAPITAL_WORD
;
SMALL_WORD
: SMALL_LETTER (SMALL_LETTER)+
;
CAPITAL_WORD
: CAPITAL_LETTER (SMALL_LETTER)+
;
fragment SMALL_LETTER
: ('a'..'z')
;
fragment CAPITAL_LETTER
: ('A'..'Z')
;
If i try to parse the input "Hello", everything is OK, BUT if if modify my grammar file like this:
...
// Parser Rules:
word
: CAPITAL_WORD EOF
;
...
the input "Hello" is no longer recognized as a valid input. Can anybody explain, what is going wrong?
Thanx, Lars
The issue here has to do with precedence in the lexer grammar. Because ANY_WORD is listed before CAPITAL_WORD, it is given higher precedence. The lexer will identify Hello as a CAPITAL_WORD, but since an ANY_WORD can be just a CAPITAL_WORD, and the lexer is set up to prefer ANY_WORD, it will output the token ANY_WORD. The parser acts on the output of the lexer, and since ANY_WORD EOF doesn't match any of its rules, the parse fails.
You can make the lexer behave differently by moving CAPITAL_WORD above ANY_WORD in the grammar, but that will create the opposite problem -- capitalized words will never lex as ANY_WORDs. The best thing to do is probably what Mephy suggested -- make ANY_WORD a parser rule.

ANTLR and Empty Strings Contradictory Behavior

I fundamentally don't understand how antlr works. Using the following grammar:
blockcomment : '/\*' ANYCHARS '\*/';
ANYCHARS : ('a'..'z' | '\n' | 'r' | ' ' | '0'..'9')* ;
I get a warning message when I compile the grammar file that says:
"non-fragment lexer rule 'ANYCHARS' can match the empty string"
Fine. I want it to be able to match empty strings as: "/\*\*/" is perfectly valid. But when I run "/\*\*/" in the TestRig I get:
missing ANYCHARS at '*/'
Obviously I could just change it so that '/**/' is handled as a special case:
blockcomment : '/\*' ANYCHARS '\*/' | '/**/';
But that doesn't really address the underlying issue. Can someone please explain to me what I am doing wrong? How can ANTLR raise a warning about matching empty strings and then not match them at the same time?
add "fragment" to ANYCHARS? It will then do what you want.
"non-fragment lexer rule 'ANYCHARS' can match the empty string"
The error message hints you to make ANYCHARS fragment.
Empty string cannot be matched as a token, that would end up with infinitely many empty tokens anywhere in the source.
You want to make the ANYCHARS part of the BLOCKCOMMENT token, rather than a separate token.
That is basically what fragments are good for - they simplify the lexer rules, but don't produce tokens.
BLOCKCOMMENT : '/*' ANYCHARS '*/';
fragment ANYCHARS : ('a'..'z' | '\n' | 'r' | ' ' | '0'..'9')* ;
EDIT: switched parser rule blockcomment to lexer rule BLOCKCOMMENT to enable fragment usage

How to consume text until newline in ANTLR?

How do you do something like this with ANTLR?
Example input:
title: hello world
Grammar:
header : IDENT ':' REST_OF_LINE ;
IDENT : 'a'..'z'+ ;
REST_OF_LINE : ~'\n'* '\n' ;
It fails, with line 1:0 mismatched input 'title: hello world\n' expecting IDENT
(I know ANTLR is overkill for parsing MIME-like headers, but this is just at the top of a more complex file.)
It fails, with line 1:0 mismatched input 'title: hello world\n' expecting IDENT
You must understand that the lexer operates independently from the parser. No matter what the parser would "like" to match at a certain time, the lexer simply creates tokens following some strict rules:
try to match tokens from top to bottom in the lexer rules (rules defined first are tried first);
match as much text as possible. In case 2 rules match the same amount of text, the rule defined first will be matched.
Because of rule 2, your REST_OF_LINE will always "win" from the IDENT rule. The only time an IDENT token will be created is when there's no more \n at the end. That is what's going wrong with your grammars: the error messages states that it expects a IDENT token, which isn't found (but a REST_OF_LINE token is produced).
I know ANTLR is overkill for parsing MIME-like headers, but this is just at the top of a more complex file.
You can't just define tokens (lexer rules) you want to apply to the header of a file. These tokens will also apply to the rest of the more complex file. Perhaps you should pre-process the header separately from the rest of the file?
antlr parsing is usually done in 2 steps.
1. construct your ast
2. define your grammer
pseudo code (been a few years since I played with antlr) - AST:
WORD : 'a'..'z'+ ;
SEPARATOR : ':';
SPACE : ' ';
pseudo code - tree parser:
header: WORD SEPARATOR WORD (SPACE WORD)+
Hope that helps....