Why whould antlr rule won't making a nice parse tree? - antlr4

I'm trying to create a grammar that would help me parse a string like this:
[Hello:/c=0.3//a=hi/] [what:/c=0.4/] [are:/c=0.6//a=is/]
This is my grammar:
grammar MyGrammar;
WS: [ \t\r\n]+ -> skip; // skip spaces, tabs, newlines
sentence: WORD+;
WORD: '[' WORD_DESCRIPTOR ']';
WORD_DESCRIPTOR: WORD_IDENTIFIER ':' WORD_FEATURES_DESCRIPTORS;
WORD_IDENTIFIER: STRING;
WORD_FEATURES_DESCRIPTORS: WORD_FEATURE_DESCRIPTOR+;
WORD_FEATURE_DESCRIPTOR: '/' WORD_FEATURE_IDENTIFIER '=' WORD_FEATURE_VALUE '/';
WORD_FEATURE_IDENTIFIER:
C_FEATURE | A_FEATURE
;
C_FEATURE: 'c';
A_FEATURE: 'a';
WORD_FEATURE_VALUE: STRING | NUMBER;
fragment LETTER : LOWER | UPPER ;
fragment LOWER : 'a'..'z' ;
fragment UPPER : 'A'..'Z' ;
fragment DIGIT : '0'..'9' ;
fragment INTEGER: DIGIT+ ;
fragment NUMBER: INTEGER (DOT INTEGER)? ;
fragment STRING: LETTER+ ;
fragment DOT: '.' ;
The problem is that the parse tree has only one level.
What I'm doing wrong?

Your parse tree shows up the way it does because all tokens are leaf nodes, and all parser rules are internal nodes. Since you only have a single parser rule (sentence) and the rest are all tokens, this is the parse tree:
sentence
/ | | \
/ | | \
WORD WORD WORD WORD ...
You should see tokens as the atoms that your language is built from. Once you start creating tokens like TOKEN : TOKEN_A | TOKEN_B;, then that is often better defined as a parser rule: token : TOKEN_A | TOKEN_B;.
Try something like this instead:
sentence : word+ EOF;
word : '[' word_descriptor ']';
word_descriptor : word_identifier ':' word_feature_descriptors;
word_identifier : STRING;
word_feature_descriptors : word_feature_descriptor+;
word_feature_descriptor : '/' word_feature_identifier '=' word_feature_value '/';
word_feature_value : STRING | NUMBER;
word_feature_identifier : C_FEATURE | A_FEATURE;
C_FEATURE : 'c';
A_FEATURE : 'a';
NUMBER : INTEGER (DOT INTEGER)?;
STRING : LETTER+ ;
WS : [ \t\r\n]+ -> skip; // skip spaces, tabs, newlines
fragment LETTER : LOWER | UPPER;
fragment LOWER : [a-z];
fragment UPPER : [A-Z];
fragment DIGIT : [0-9];
fragment INTEGER : DIGIT+;
fragment DOT : '.';
which will create the following parse tree for your input:

Related

How to use the reserved words inside the string in ANTLR4?

I am a newbie to ANTLR4 and language compilers. I am working on building a language compiler using ANTLR4 Java. I have a small problem with parsing strings. The reserved words/ Tokens are getting matched instead of string. For eg: IF is a keyword token in my lexer but how to use "if" as a string?
Lexer file:
lexer grammar testgrammar;
IF : I F;
ENDIF : E N D I F;
ELSE : E L S E;
CASE : C A S E;
ENDCASE : E N D C A S E;
BREAK : B R E A K;
SWITCH : S W I T C H;
SUBSTRING : S U B S T R I N G;
COMMA : ',' ;
SEMI : ';' ;
COLON : ':' ;
LPAREN : '(' ;
RPAREN : ')' ;
DOT : '.' ;// ('.' {$setType(DOTDOT);})? ;
LCURLY : '{' ;
RCURLY : '}' ;
AND : '&&' ;
OR : '||' ;
DOUBLEQUOTES : '"' ;
COMPARATOR : '=='| '>=' | '>' | '<' | '<=' | '!=' ;
SYMBOLS : '§' | '$' | '%' | '/' | '=' | '?' | '#' | '_' | '#' | '€';
LETTER : [A-Za-z\u00e4\u00c4\u00d6\u00f6\u00dc\u00fc\u00df];
NUMERICVALUE : NUMBER ('.' NUMBER)?;
STRING_LITERAL : '\'' ('\'\'' | ~('\''))* '\'';
NOTCONDITION : NOT;
OPERATORS : OPERATOR;
COMMENT : (('/*' .*? '*/') | ('//' ~[\r\n]*)) -> skip;
WS : (' ' | '\t' | '\r' | '\n')+ -> skip;
fragment A:('a'|'A');
fragment B:('b'|'B');
fragment C:('c'|'C');
fragment D:('d'|'D');
fragment E:('e'|'E');
fragment F:('f'|'F');
fragment G:('g'|'G');
fragment H:('h'|'H');
fragment I:('i'|'I');
fragment J:('j'|'J');
fragment K:('k'|'K');
fragment L:('l'|'L');
fragment M:('m'|'M');
fragment N:('n'|'N');
fragment O:('o'|'O');
fragment P:('p'|'P');
fragment Q:('q'|'Q');
fragment R:('r'|'R');
fragment S:('s'|'S');
fragment T:('t'|'T');
fragment U:('u'|'U');
fragment V:('v'|'V');
fragment W:('w'|'W');
fragment X:('x'|'X');
fragment Y:('y'|'Y');
fragment Z:('z'|'Z');
fragment NUMBER:[0-9]+;
fragment OPERATOR: ('+'|'-'|'&'|'*'|'~');
fragment NOT: ('!');
grammar:
parser grammar testParser;
symbolCharacters: (SYMBOLS | operators) ;
word:
( symbolCharacters | LETTER )+
;
wordList:
word+
;
I am not supposed share full grammar. But i have shared enough information i guess. I can understand that the words are formed from LETTERS and Symbol characters. One workaround i can do is making word rule like:
word:
( symbolCharacters | LETTER | IF | SWITCH | CASE | ELSE | BREAK )+
;
I have a lot of tokens. I dont want to add everything individually. Is there any other nice way to accomplish this?
Valid expression
Error expression
How to make the parser ignore the keywords inside the string?
Your same grammar does not have the problem you describe:
➜ antlr4 testgrammar.g4
➜ javac *.java
➜ echo "if 'if' endif" | grun testgrammar tokens -tokens
[#0,0:1='if',<IF>,1:0]
[#1,3:6=''if'',<STRING_LITERAL>,1:3]
[#2,8:12='endif',<ENDIF>,1:8]
[#3,14:13='<EOF>',<EOF>,2:0]
(perhaps you have inadvertently "corrected" the problem as you trimmed your grammar down, so I'll elaborate a bit.)
In short, during the lexing/tokenization phase of ANTLR parsing your input, ANTLR will, naturally, attempt to match you Lexer rules. If ANTLR finds a match of multiple rules for the current characters of your input stream, it follows two rules to determine a "winner".
If a rule matches a longer sequence of input characters, then that rule will be used.
If two rules match the same number of input characters, then the rule appearing first in your grammar will be used.
In your case, neither really comes into play as the grammar, when it reaches the ', will attempt to complete the STRING_LITERAL rule, and will find a match for the characters 'if'. It will never even attempt to match you IF lexer rule.
BTW, I did have to correct the symbolCharacters parser rule to be
symbolCharacters: (SYMBOLS | OPERATORS);

ANTLR4 Grammar Issue with Decimal Numbers

I'm new to ANTLR and using ANTLR4 (4.7.2 Jar file). I'm currently working on Oracle Parser.
I'm having issues with Decimal numbers. I have kept only the relevant parts.
My grammar file is as below.
Now when I parse the below statement it is fine. ".1" is a valid number in my case.
BEGIN a NUMBER:=.1; END;
I haven't shown the grammar but the below are valid cases for me in Oracle.
a NUMBER:= .1; // with Space after operator
a NUMBER:=1.1; // without Space after operator
a NUMBER:=1; // without Space after operator
a NUMER:= 3; // with Space after operator
Now I need to create a tablespace as below.
CREATE TABLESPACE tbs_01 DATAFILE +DATA/BR/CONTROLFILE/Current.260.750;
Here the Digits 260 & 750 are tokenized along with the DOT (as per the definition of NUMERIC_LITERAL). I would want this to be 2 separate digits separated by DOT (and assigned to filenumber and incarnation_number resp as shown in the grammar).
How do I do this?
I have tried using _input.LA(-1)!='.'}? etc but was not working correctly for me.
I tried many other steps mentioned (most solutions were for ANTLR3 and not working in ANTLR4). Is there a simple way to do this in LEXER? I do not want to write a Parser rule to split the decimal digits.
grammar Oracle;
parse
: ( sql_statements | error )* EOF
;
error
: UNEXPECTED_CHAR
{
throw new RuntimeException("UNEXPECTED_CHAR=" + $UNEXPECTED_CHAR.text);
}
;
sql_statements
: 'CREATE' 'TABLESPACE' tablespace_name 'DATAFILE' fully_qualified_file_name ';'
| 'BEGIN' var1 'NUMBER' ':=' num1 ';' 'END' ';'
;
tablespace_name : IDENTIFIER;
fully_qualified_file_name : K_PLUS_SIGN diskgroup_name K_SOLIDUS db_name K_SOLIDUS file_type K_SOLIDUS file_type_tag '.' filenumber '.' incarnation_number;
diskgroup_name : IDENTIFIER;
db_name : IDENTIFIER;
file_type : IDENTIFIER;
file_type_tag : IDENTIFIER;
filenumber : NUMERIC_LITERAL;
incarnation_number : NUMERIC_LITERAL;
var1 : IDENTIFIER;
num1 : NUMERIC_LITERAL;
IDENTIFIER : [a-zA-Z_] ([a-zA-Z] | '$' | '_' | '#' | DIGIT)* ;
K_PLUS_SIGN : '+';
K_SOLIDUS : '/';
NUMERIC_LITERAL
: DIGIT+ ( '.' DIGIT+ )? ( E ('+'|'-')? DIGIT+ )? ('D' | 'F')?
| '.' DIGIT+ ( E ('+'|'-')? DIGIT+ )? ('D' | 'F')?
;
SPACES : [ \u000B\t\r\n] -> skip;
WS : [ \t\r\n]+ -> skip;
UNEXPECTED_CHAR : . ;
fragment DIGIT : [0-9];
fragment A : [aA];
fragment B : [bB];
fragment C : [cC];
fragment D : [dD];
fragment E : [eE];
fragment F : [fF];
fragment G : [gG];
fragment H : [hH];
fragment I : [iI];
fragment J : [jJ];
fragment K : [kK];
fragment L : [lL];
fragment M : [mM];
fragment N : [nN];
fragment O : [oO];
fragment P : [pP];
fragment Q : [qQ];
fragment R : [rR];
fragment S : [sS];
fragment T : [tT];
fragment U : [uU];
fragment V : [vV];
fragment W : [wW];
fragment X : [xX];
fragment Y : [yY];
fragment Z : [zZ];
Your Dsl has a natural ambiguity: in some instances, numbers are integers and in others, decimals.
If the Dsl provides sufficient guard conditions, Antlr modes can be used to isolate the instances. For example, in the given Dsl, decimal numbers appear to always occur between := and ; guards.
...
K_ASSIGN : ':=' -> pushMode(Decimals);
K_SEMI : ';' ;
NUMERIC_LITERAL : DIGIT+ ;
...
mode Decimals;
D_SEMI : ';' -> type(K_SEMI), popMode ;
NUMERIC:
DIGIT+ ( '.' DIGIT+ )? ( E ('+'|'-')? DIGIT+ )? 'D'
| 'F')?
| '.' DIGIT+ ( E ('+'|'-')? DIGIT+ )? ('D' | 'F')?
-> type(NUMERIC_LITERAL);

ANTLR4 ambiguity - how to solve

I would like to solve the following ambiguity:
grammar test;
WS : (' ' | '\t' | '\n' | '\r' | '\f')+ -> skip;
program
:
input* EOF;
input
: '%' statement
| inputText
;
inputText
: ~('%')+
;
statement
: Identifier '=' DecimalConstant ';'
;
DecimalConstant
: [0-9]+
;
Identifier
: Letter LetterOrDigit*
;
fragment
Letter
: [a-zA-Z$##_.]
;
fragment
LetterOrDigit
: [a-zA-Z0-9$##_.]
;
Sample input:
%a=5;
aa bbbb
As soon as I put a space after "aa" with values like "bbbb" an ambiguity is created.
In fact I want inputText to contain the full string "aa bbbb".
There is no ambiguity. The input aa bbbb will always be tokenised as 2 Identifier tokens. No matter what any parser rule is trying to match. The lexer operates independently from the parser.
Also, the rule:
inputText
: ~('%')+
;
does not match one or more characters other than '%'.
Inside parser rules, the ~ negates tokens, not characters. So ~'%' inside a parser rule will match any token, other than a '%' token. Inside the lexer, ~'%' matches any character other than '%'.
But creating a lexer rule like this:
InputText
: ~('%')+
;
will cause your example input to be tokenised as a single '%' token, followed by a large 2nd token that'd match this: a=5;\naa bbbb. This is how ANTLR's lexer works: match as much characters as possible (no matter what the parser rule is trying to match).
I found the solution:
grammar test;
WS : (' ' | '\t' | '\n' | '\r' | '\f')+ -> skip;
program
:
input EOF;
input
: inputText ('%' statement inputText)*
;
inputText
: ~('%')*
;
statement
: Identifier '=' DecimalConstant ';'
;
DecimalConstant
: [0-9]+
;
Identifier
: Letter LetterOrDigit*
;
fragment
Letter
: [a-zA-Z$##_.]
;
fragment
LetterOrDigit
: [a-zA-Z0-9$##_.]
;

Can I make my ANTLR4 Lexer discard a character from the input stream?

I'm working on parsing PDF streams. In section 7.3.4.2 on literal string objects, the PDF Reference says that a backslash within a literal string that isn't followed by an end-of-line character, one to three octal digits, or one of the characters "nrtbf()\" should be ignored. Is there a way to get the recover method in my lexer to ignore a backslash in this situation?
Here is my simplified parser:
parser grammar PdfStreamParser;
options { tokenVocab=PdfSteamLexer; }
array: LBRACKET object* RBRACKET ;
dictionary: LDOUBLEANGLE (NAME object)* RDOUBLEANGLE ;
string: (LITERAL_STRING | HEX_STRING) ;
object
: NULL
| array
| dictionary
| BOOLEAN
| NUMBER
| string
| NAME
;
content : stat* ;
stat
: tj
;
tj: ((string Tj) | (array TJ)) ; // Show text
Here's the lexer. (Based on the advice in this answer I'm not using a separate string mode):
lexer grammar PdfStreamLexer;
Tj: 'Tj' ;
TJ: 'TJ' ;
NULL: 'null' ;
BOOLEAN: ('true'|'false') ;
LBRACKET: '[' ;
RBRACKET: ']' ;
LDOUBLEANGLE: '<<' ;
RDOUBLEANGLE: '>>' ;
NUMBER: ('+' | '-')? (INT | FLOAT) ;
NAME: '/' ID ;
// A sequence of literal characters enclosed in parentheses.
LITERAL_STRING: '(' ( ~[()\\]+ | ESCAPE_SEQUENCE | LITERAL_STRING )* ')' ;
// Escape sequences that can occur within a LITERAL_STRING
fragment ESCAPE_SEQUENCE
: '\\' ( [\r\nnrtbf()\\] | [0-7] [0-7]? [0-7]? )
;
HEX_STRING: '<' [0-9A-Za-z]+ '>' ; // Hexadecimal data enclosed in angle brackets
fragment INT: DIGIT+ ; // match 1 or more digits
fragment FLOAT: DIGIT+ '.' DIGIT* // match 1. 39. 3.14159 etc...
| '.' DIGIT+ // match .1 .14159
;
fragment DIGIT: [0-9] ; // match single digit
// Accept all characters except whitespace and defined delimiters ()<>[]{}/%
ID: ~[ \t\r\n\u000C\u0000()<>[\]{}/%]+ ;
WS: [ \t\r\n\u000C\u0000]+ -> skip ; // PDF defines six whitespace characters
I can override the recover method in the PdfStreamLexer class and get notified when the LexerNoViableAltException occurs, but I'm not sure how to (or if it's possible to) ignore the backslash and continue on with the LITERAL_STRING tokenization.
To be able to skip part of the string, you'll need to use lexical modes. Here's a quick demo:
lexer grammar DemoLexer;
STRING_OPEN
: '(' -> pushMode(STRING_MODE)
;
SPACES
: [ \t\r\n] -> skip
;
OTHER
: .
;
mode STRING_MODE;
STRING_CLOSE
: ')' -> popMode
;
ESCAPE
: '\\' ( [nrtbf()\\] | [0-7] [0-7] [0-7] )
;
STRING_PART
: ~[\\()]
;
NESTED_STRING_OPEN
: '(' -> type(STRING_OPEN), pushMode(STRING_MODE)
;
IGNORED_ESCAPE
: '\\' . -> skip
;
which can be used in the parser as follows:
parser grammar DemoParser;
options {
tokenVocab=DemoLexer;
}
parse
: ( string | OTHER )* EOF
;
string
: STRING_OPEN ( ESCAPE | STRING_PART | string )* STRING_CLOSE
;
If you now parse the string FU(abc(def)\#\))BAR, you will get the following parse tree:
As you can see, the \) is left in the tree, but \# is omitted.

ANTLR4 Grammar picks up 'and' and 'or' in variable names

Please help me with my ANTLR4 Grammar.
Sample "formel":
(Arbejde.ArbejderIKommuneNr=860) and (Arbejde.ErIArbejde = 'J') &
(Arbejde.ArbejdsTimerPrUge = 40)
(Ansogeren.BorIKommunen = 'J') and (BeregnDato(Ansogeren.Fodselsdato;
'+62Å') < DagsDato)
(Arb.BorI=860)
My problem is that Arb.BorI=860 is not handled correct. I get this error:
Error: no viable alternative at input '(Arb.Bor' at linenr/position: 1/6 \r\nException: Der blev udløst en undtagelse af typen 'Antlr4.Runtime.NoViableAltException
Please notis that Arb.BorI contains the word 'or'.
I think my problem is that my 'booleanOps' in the grammar override 'datakildefelt'
So... My problem is how do I get my grammar correct - I am stuck, so any help will be appreciated.
My Grammar:
grammar UnikFormel;
formel : boolExpression # BooleanExpr
| expression # Expr
| '(' formel ')' # Parentes;
boolExpression : ( '(' expression ')' ) ( booleanOps '(' expression ')' )+;
expression : element compareOps element # Compare;
element : datakildefelt # DatakildeId
| function # Funktion
| int # Integer
| decimal # Real
| string # Text;
datakildefelt : datakilde '.' felt;
datakilde : identifyer;
felt : identifyer;
function : funktionsnavn ('(' funcParameters? ')')?;
funktionsnavn : identifyer;
funcParameters : funcParameter (';' funcParameter)*;
funcParameter : element;
identifyer : LETTER+;
int : DIGIT+;
decimal : DIGIT+ '.' DIGIT+ | '.' DIGIT+;
string : QUOTE .*? QUOTE;
booleanOps : (AND | OR);
compareOps : (LT | GT | EQ | GTEQ | LTEQ);
QUOTE : '\'';
OPERATOR: '+';
DIGIT: [0-9];
LETTER: [a-åA-Å];
MUL : '*';
DIV : '/';
ADD : '+';
SUB : '-';
GT : '>';
LT : '<';
EQ : '=';
GTEQ : '>=';
LTEQ : '<=';
AND : '&' | 'and';
OR : '?' | 'or';
WS : ' '+ -> skip;
Rules that come first always have precedence. In your case you need to move AND and OR before LETTER. Also there is the same problem with GTEQ and LTEQ, maybe somewhere else too.
EDIT
Additionally, you should make identifyer a lexer rule, i.e. start with capital letter (IDENTIFIER or Identifier). The same goes for int, decimal and string. Input is initially a stream of characters and is first processed into a stream of tokens, using only lexer rules. At this point parser rules (those starting with lowercase letter) do not come to play yet. So, to make "BorI" parse as single entity (token), you need to create a lexer rule that matches identifiers. Currently it would be parsed as 3 tokens: LETTER (B) OR (or) LETTER (I).
Thanks for your help. There were multiple problems. Reading the ANTLR4 book and using "TestRig -gui" got me on the right track. The working grammar is:
grammar UnikFormel;
formel : '(' formel ')' # Parentes
| expression # Expr
| boolExpression # BooleanExpr
;
boolExpression : '(' expression ')' ( booleanOps '(' expression ')' )+
| '(' formel ')' ( booleanOps '(' formel ')' )+;
expression : element compareOps element # Compare;
datakildefelt : ID '.' ID;
function : ID ('(' funcParameters? ')')?;
funcParameters : funcParameter (';' funcParameter)*;
funcParameter : element;
element : datakildefelt # DatakildeId
| function # Funktion
| INT # Integer
| DECIMAL # Real
| STRING # Text;
booleanOps : (AND | OR);
compareOps : ( GTEQ | LTEQ | LT | GT | EQ |);
AND : '&' | 'and';
OR : '?' | 'or';
GTEQ : '>=';
LTEQ : '<=';
GT : '>';
LT : '<';
EQ : '=';
ID : LETTER ( LETTER | DIGIT)*;
INT : DIGIT+;
DECIMAL : DIGIT+ '.' DIGIT+ | '.' DIGIT+;
STRING : QUOTE .*? QUOTE;
fragment QUOTE : '\'';
fragment DIGIT: [0-9];
fragment LETTER: [a-åA-Å];
WS : [ \t\r\n]+ -> skip;

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