How to fix extraneous input ' ' expecting, in antlr4 - antlr

Hello when running antlr4 with the following input i get the following error
image showing problem
[
I have been trying to fix it by doing some changes here and there but it seems it only works if I write every component of whileLoop in a new line.
Could you please tell me what i am missing here and why the problem persits?
grammar AM;
COMMENTS :
'{'~[\n|\r]*'}' -> skip
;
body : ('BODY' ' '*) anything | 'BODY' 'BEGIN' anything* 'END' ;
anything : whileLoop | write ;
write : 'WRITE' '(' '"' sentance '"' ')' ;
read : 'READ' '(' '"' sentance '"' ')' ;
whileLoop : 'WHILE' expression 'DO' ;
block : 'BODY' anything 'END';
expression : 'TRUE'|'FALSE' ;
test : ID? {System.out.println("Done");};
logicalOperators : '<' | '>' | '<>' | '<=' | '>=' | '=' ;
numberExpressionS : (NUMBER numberExpression)* ;
numberExpression : ('-' | '/' | '*' | '+' | '%') NUMBER ;
sentance : (ID)* {System.out.println("Sentance");};
WS : [ \t\r\n]+ -> skip ;
NUMBER : [0-9]+ ;
ID : [a-zA-Z0-9]* ;
**`strong text`**

Your lexer rules produce conflicts:
body : ('BODY' ' '*) anything | 'BODY' 'BEGIN' anything* 'END' ;
vs
WS : [ \t\r\n]+ -> skip ;
The critical section is the ' '*. This defines an implicit lexer token. It matches spaces and it is defined above of WS. So any sequence of spaces is not handled as WS but as implicit token.
If I am right putting tabs between the components of whileloop will work, also putting more than one space between them should work. You should simply remove ' '*, since whitespace is to be skipped anyway.

Related

ANTLR4: wrong lexer rule matches

I'm at a very beginning of learning ANTLR4 lexer rules. My goal is to create a simple grammar for Java properties files. Here is what I have so far:
lexer grammar PropertiesLexer;
LineComment
: ( LineCommentHash
| LineCommentExcl
)
-> skip
;
fragment LineCommentHash
: '#' ~[\r\n]*
;
fragment LineCommentExcl
: '!' ~[\r\n]*
;
fragment WrappedLine
: '\\'
( '\r' '\n'?
| '\n'
)
;
Newline
: ( '\r' '\n'?
| '\n'
)
-> skip
;
Key
: KeyLetterStart
( KeyLetter
| Escaped
)*
;
fragment KeyLetterStart
: ~[ \t\r\n:=]
;
fragment KeyLetter
: ~[\t\r\n:=]
;
fragment Escaped
: '\\' .?
;
Equal
: ( '\\'? ':'
| '\\'? '='
)
;
Value
: ValueLetterBegin
( ValueLetter
| Escaped
| WrappedLine
)*
;
fragment ValueLetterBegin
: ~[ \t\r\n]
;
fragment ValueLetter
: ~ [\r\n]+
;
Whitespace
: [ \t]+
-> skip
;
My test file is this one:
# comment 1
# comment 2
#
.key1= value1
key2\:sub=value2
key3 \= value3
key4=value41\
value42
# comment3
#comment4
key=value
When I run grun, I'm getting following output:
[#0,30:42='.key1= value1',<Value>,4:0]
[#1,45:60='key2\:sub=value2',<Value>,5:0]
[#2,63:76='key3 \= value3',<Value>,6:0]
[#3,81:102='key4=value41\\r\nvalue42',<Value>,8:0]
[#4,130:138='key=value',<Value>,13:0]
[#5,141:140='<EOF>',<EOF>,14:0]
I don't understand why the Value definition is matched. When commenting out the Value definition, however, it recognizes the Key and Equal definitions:
[#0,30:34='.key1',<Key>,4:0]
[#1,35:35='=',<Equal>,4:5]
[#2,37:42='value1',<Key>,4:7]
[#3,45:49='key2\',<Key>,5:0]
[#4,50:50=':',<Equal>,5:5]
[#5,51:53='sub',<Key>,5:6]
[#6,54:54='=',<Equal>,5:9]
[#7,55:60='value2',<Key>,5:10]
[#8,63:68='key3 \',<Key>,6:0]
[#9,69:69='=',<Equal>,6:6]
[#10,71:76='value3',<Key>,6:8]
[#11,81:84='key4',<Key>,8:0]
[#12,85:85='=',<Equal>,8:4]
[#13,86:93='value41\',<Key>,8:5]
[#14,96:102='value42',<Key>,9:0]
[#15,130:132='key',<Key>,13:0]
[#16,133:133='=',<Equal>,13:3]
[#17,134:138='value',<Key>,13:4]
[#18,141:140='<EOF>',<EOF>,14:0]
but how to let it recognize the Key, Equal and Value definitons?
ANTLR's lexer rules match as much characters as possible, that is why you're seeing all these Value tokens being created (they match the most characters).
Lexical modes seem like a good fit to use here. Something like this:
lexer grammar PropertiesLexer;
COMMENT
: [!#] ~[\r\n]* -> skip
;
KEY
: ( '\\' ~[\r\n] | ~[\r\n\\=:] )+
;
EQUAL
: [=:] -> pushMode(VALUE_MODE)
;
NL
: [\r\n]+ -> skip
;
mode VALUE_MODE;
VALUE
: ( ~[\\\r\n] | '\\' . )+
;
END_VALUE
: [\r\n]+ -> skip, popMode
;

"The following sets of rules are mutually left-recursive"

I have tried to write a grammar to recognize expressions like:
(A + MAX(B) ) / ( C - AVERAGE(A) )
IF( A > AVERAGE(A), 0, 1 )
X / (MAX(X)
Unfortunately antlr3 fails with these errors:
error(210): The following sets of rules are mutually left-recursive [unaryExpression, additiveExpression, primaryExpression, formula, multiplicativeExpression]
error(211): DerivedKeywords.g:110:13: [fatal] rule booleanTerm has non-LL(*) decision due to recursive rule invocations reachable from alts 1,2. Resolve by left-factoring or using syntactic predicates or using backtrack=true option.
error(206): DerivedKeywords.g:110:13: Alternative 1: after matching input such as decision cannot predict what comes next due to recursion overflow to additiveExpression from formula
I have spent some hours trying to fix these, it would be great if anyone could at least help me fix the first problem. Thanks
Code:
grammar DerivedKeywords;
options {
output=AST;
//backtrack=true;
}
WS : ( ' ' | '\t' | '\n' | '\r' )
{ skip(); }
;
//for numbers
DIGIT
: '0'..'9'
;
//for both integer and real number
NUMBER
: (DIGIT)+ ( '.' (DIGIT)+ )?( ('E'|'e')('+'|'-')?(DIGIT)+ )?
;
// Boolean operatos
AND : 'AND';
OR : 'OR';
NOT : 'NOT';
EQ : '=';
NEQ : '!=';
GT : '>';
LT : '<';
GTE : '>=';
LTE : '<=';
COMMA : ',';
// Token for Functions
IF : 'IF';
MAX : 'MAX';
MIN : 'MIN';
AVERAGE : 'AVERAGE';
VARIABLE : 'A'..'Z' ('A'..'Z' | '0'..'9')*
;
// OPERATORS
LPAREN : '(' ;
RPAREN : ')' ;
DIV : '/' ;
PLUS : '+' ;
MINUS : '-' ;
STAR : '*' ;
expression : formula;
formula
: functionExpression
| additiveExpression
| LPAREN! a=formula RPAREN! // First Problem
;
additiveExpression
: a=multiplicativeExpression ( (MINUS^ | PLUS^ ) b=multiplicativeExpression )*
;
multiplicativeExpression
: a=unaryExpression ( (STAR^ | DIV^ ) b=unaryExpression )*
;
unaryExpression
: MINUS^ u=unaryExpression
| primaryExpression
;
functionExpression
: f=functionOperator LPAREN e=formula RPAREN
| IF LPAREN b=booleanExpression COMMA p=formula COMMA s=formula RPAREN
;
functionOperator :
MAX | MIN | AVERAGE;
primaryExpression
: NUMBER
// Used for scientific numbers
| DIGIT
| VARIABLE
| formula
;
// Boolean stuff
booleanExpression
: orExpression;
orExpression : a=andExpression (OR^ b=andExpression )*
;
andExpression
: a=notExpression (AND^ b=notExpression )*
;
notExpression
: NOT^ t=booleanTerm
| booleanTerm
;
booleanOperator :
GT | LT | EQ | GTE | LTE | NEQ;
booleanTerm : a=formula op=booleanOperator b=formula
| LPAREN! booleanTerm RPAREN! // Second problem
;
error(210): The following sets of rules are mutually left-recursive [unaryExpression, additiveExpression, primaryExpression, formula, multiplicativeExpression]
- this means that if the parser enters unaryExpression rule, it has the possibility to match additiveExpression, primaryExpression, formula, multiplicativeExpression and unaryExpression again without ever consuming a single token from input - so it cannot decide whether to use those rules or not, because even if it uses the rules, the input will be the same.
You're probably trying to allow subexpressions in expressions by this sequence of rules - you need to make sure that path will consume the left parenthesis of the subexpression. Probably the formula alternative in primaryExpression should be changed to LPAREN formula RPAREN, and the rest of grammar be adjusted accordingly.

Guide or approval for ANTLR example

I have an AlgebraRelacional.g4 file with this. I need to read a file with a syntax like a CSV file, put the content in some memory tables and then resolve relational algebra operations with that. Can you tell me if I am doing it right?
Example data file to read:
cod_buy(char);name_suc(char);Import(int);date_buy(date)
“P-11”;”DC Med”;900;01/03/14
“P-14”;”Center”;1500;02/05/14
Current ANTLR grammar:
grammar AlgebraRelacional;
SEL : '\u03C3'
;
PRO : '\u220F'
;
UNI : '\u222A'
;
DIF : '\u002D'
;
PROC : '\u0058'
;
INT : '\u2229'
;
AND : 'AND'
;
OR : 'OR'
;
NOT : 'NOT'
;
EQ : '='
;
DIFERENTE : '!='
;
MAYOR : '>'
;
MENOR : '<'
;
SUMA : '+'
;
MULTI : '*'
;
IPAREN : '('
;
DPAREN : ')'
;
COMA : ','
;
PCOMA : ';'
;
Comillas: '"'
;
file : hdr row+ ;
hdr : row ;
row : field (',' field)* '\r'? '\n' ;
field : TEXT | STRING | ;
TEXT : ~[,\n\r"]+ ;
STRING : '"' ('""'|~'"')* '"' ;
I suggest you that read this document (http://is.muni.cz/th/208197/fi_b/bc_thesis.pdf), It contains usefull information about how to write a parser for relational algebra. That is not ANTLR, but you only has to translate the grammar in BNF to EBNF.

Intellij Antlr4 Plugin Left direct recursion doesn't work

I'm trying to make parser using Antlr4 for the sql select statement, in which contains the following part
expr: '1' | expr('*'|'/'|'+'|'-'|'||') expr; // As the re-factored form of expression: compound expression;
WS :[ \t\r\n]+ -> skip ;
I suppose this rule will allow the following sets of result:
1
1+1
1+1-1
....
But in the graph it shows that it cannot be parsed
Does anyone get the idea why it cannot be parsed like what i expected?
This slightly adjusted grammar works for me. Tested on input 1+1-1||1*1-1/1. Tested in ANTLRWorks2.1
grammar myGrammar;
top : expr EOF ;
expr : '1'
| expr '+' expr
| expr '*' expr
| expr '/' expr
| expr '+' expr
| expr '-' expr
| expr '||' expr
;
WS :[ \t\r\n]+ -> skip ;
One : '1' ;
Times : '*' ;
Div : '/' ;
Plus : '+' ;
Minus : '-' ;
Or : '||' ;
EDIT
I was able to get this to work, too, when matching the rule top:
grammar newEmptyCombinedGrammar;
top : expr EOF ;
expr: one
| expr op=(Times|Div|Plus|Minus|Or) expr
;
one : One ;
One : '1' ;
Times : '*' ;
Div : '/' ;
Plus : '+' ;
Minus : '-' ;
Or : '||' ;
WS :[ \t\r\n]+ -> skip ;

ANTLR grammar error

I'm trying to built C-- compiler using ANTLR 3.4.
Full set of the grammar listed here,
program : (vardeclaration | fundeclaration)* ;
vardeclaration : INT ID (OPENSQ NUM CLOSESQ)? SEMICOL ;
fundeclaration : typespecifier ID OPENP params CLOSEP compoundstmt ;
typespecifier : INT | VOID ;
params : VOID | paramlist ;
paramlist : param (COMMA param)* ;
param : INT ID (OPENSQ CLOSESQ)? ;
compoundstmt : OPENCUR vardeclaration* statement* CLOSECUR ;
statementlist : statement* ;
statement : expressionstmt | compoundstmt | selectionstmt | iterationstmt | returnstmt;
expressionstmt : (expression)? SEMICOL;
selectionstmt : IF OPENP expression CLOSEP statement (options {greedy=true;}: ELSE statement)?;
iterationstmt : WHILE OPENP expression CLOSEP statement;
returnstmt : RETURN (expression)? SEMICOL;
expression : (var EQUAL expression) | sampleexpression;
var : ID ( OPENSQ expression CLOSESQ )? ;
sampleexpression: addexpr ( ( LOREQ | LESS | GRTR | GOREQ | EQUAL | NTEQL) addexpr)?;
addexpr : mulexpr ( ( PLUS | MINUS ) mulexpr)*;
mulexpr : factor ( ( MULTI | DIV ) factor )*;
factor : ( OPENP expression CLOSEP ) | var | call | NUM;
call : ID OPENP arglist? CLOSEP;
arglist : expression ( COMMA expression)*;
Used lexer rules as following,
ELSE : 'else' ;
IF : 'if' ;
INT : 'int' ;
RETURN : 'return' ;
VOID : 'void' ;
WHILE : 'while' ;
PLUS : '+' ;
MINUS : '-' ;
MULTI : '*' ;
DIV : '/' ;
LESS : '<' ;
LOREQ : '<=' ;
GRTR : '>' ;
GOREQ : '>=' ;
EQUAL : '==' ;
NTEQL : '!=' ;
ASSIGN : '=' ;
SEMICOL : ';' ;
COMMA : ',' ;
OPENP : '(' ;
CLOSEP : ')' ;
OPENSQ : '[' ;
CLOSESQ : ']' ;
OPENCUR : '{' ;
CLOSECUR: '}' ;
SCOMMENT: '/*' ;
ECOMMENT: '*/' ;
ID : ('a'..'z' | 'A'..'Z')+/*(' ')*/ ;
NUM : ('0'..'9')+ ;
WS : (' ' | '\t' | '\n' | '\r')+ {$channel = HIDDEN;};
COMMENT: '/*' .* '*/' {$channel = HIDDEN;};
But I try to save this it give me the error,
error(211): /CMinusMinus/src/CMinusMinus/CMinusMinus.g:33:13: [fatal] rule expression has non-LL(*) decision due to recursive rule invocations reachable from alts 1,2. Resolve by left-factoring or using syntactic predicates or using backtrack=true option.
|---> expression : (var EQUAL expression) | sampleexpression;
1 error
How can I resolve this problem?
As already mentioned: your grammar rule expression is ambiguous: both alternatives in that rule start, or can be, a var.
You need to "help" your parser a bit. If the parse can see a var followed by an EQUAL, it should choose alternative 1, else alternative 2. This can be done by using a syntactic predicate (the (var EQUAL)=> part in the rule below).
expression
: (var EQUAL)=> var EQUAL expression
| sampleexpression
;
More about predicates in this Q&A: What is a 'semantic predicate' in ANTLR?
The problem is this:
expression : (var EQUAL expression) | sampleexpression;
where you either start with var or sampleexpression. But sampleexpression can be reduced to var as well by doing sampleexpression->addExpr->MultExpr->Factor->var
So there is no way to find a k-length predicate for the compiler.
You can as suggested by the error message set backtrack=true to see whether this solves your problem, but it might lead not to the AST - parsetrees you would expect and might also be slow on special input conditions.
You could also try to refactor your grammar to avoid such recursions.