Why is this grammar giving me a "non LL(*) decision" error? - antlr

I am trying to add support for expressions in my grammar. I am following the example given by Scott Stanchfield's Antlr Tutorial. For some reason the add rule is causing an error. It is causing a non-LL(*) error saying, "Decision can match input such as "'+'..'-' IDENT" using multiple alternatives"
Simple input like:
a.b.c + 4
causes the error. I am using the AntlrWorks Interpreter to test my grammar as I go. There seems to be a problem with how the tree is built for the unary +/- and the add rule. I don't understand why there are two possible parses.
Here's the grammar:
path : (IDENT)('.'IDENT)* //(NAME | LCSTNAME)('.'(NAME | LCSTNAME))*
;
term : path
| '(' expression ')'
| NUMBER
;
negation
: '!'* term
;
unary : ('+' | '-')* negation
;
mult : unary (('*' | '/' | '%') unary)*
;
add : mult (( '+' | '-' ) mult)*
;
relation
: add (('==' | '!=' | '<' | '>' | '>=' | '<=') add)*
;
expression
: relation (('&&' | '||') relation)*
;
multiFunc
: IDENT expression+
;
NUMBER : DIGIT+ ('.'DIGIT+)?
;
IDENT : (LCLETTER|UCLETTER)(LCLETTER|UCLETTER|DIGIT|'_')*
;
COMMENT
: '//' ~('\n'|'\r')* '\r'? '\n' {$channel=HIDDEN;}
| '/*' ( options {greedy=false;} : . )* '*/' {$channel=HIDDEN;}
;
WS : (' ' | '\t' | '\r' | '\n' | '\f')+ {$channel = HIDDEN;}
;
fragment
LCLETTER
: 'a'..'z'
;
fragment
UCLETTER: 'A'..'Z'
;
fragment
DIGIT : '0'..'9'
;
I need an extra set of eyes. What am I missing?

The fact that you let one or more expressions match in:
multiFunc
: IDENT expression+
;
makes your grammar ambiguous. Let's say you're trying to match "a 1 - - 2" using the multiFunc rule. The parser now has 2 possible ways to parse this: a is matched by IDENT, but the 2 minus signs 1 - - 2 cause trouble for expression+. The following 2 parses are possible:
parse 1
parse 2

Your grammar in rule multiFunc has a list of expressions. An expression can begin with + or - on behalf of unary, thus due to the list, it can also be followed by the same tokens. This is in conflict with the add rule: there is a problem deciding between continuation and termination.

Related

ANTLR4 - How to close "longest-match-wins" and use first match rule?

Orignial question:
My code to parse:
N100G1M4
What I expcted: N100 G1 M4
But ANTLR can not idetify this because ANTLR always match longest substring?
How to handle the case?
Update
What I am going to do:
I am trying to parse CNC G-Code txt and get keywords from a file stream, which is usually used to control a machine and drive motors to move.
The G-Code rule is :
// Define a grammar called Hello
grammar GCode;
script : blocks+ EOF;
blocks:
assign_stat
| ncblock
| NEWLINE
;
ncblock :
ncelements NEWLINE //
;
ncelements :
ncelement+
;
ncelement
:
LINENUMEXPR // linenumber N100
| GCODEEXPR // G10 G54.1
| MCODEEXPR // M30
| coordexpr // X100 Y100 Z[A+b*c]
| FeedExpr // F10.12
| AccExpr // E2.0
// | callSubroutine
;
assign_stat:
VARNAME '=' expression NEWLINE
;
expression:
multiplyingExpression ('+' | '-') multiplyingExpression
;
multiplyingExpression
: powExpression (('*' | '/') powExpression)*
;
powExpression
: signedAtom ('^' signedAtom)*
;
signedAtom
: '+' signedAtom
| '-' signedAtom
| atom
;
atom
: scientific
| variable
| '(' expression ')'
;
LINENUMEXPR: 'N' Digit+ ;
GCODEEXPR : 'G' GPOSTFIX;
MCODEEXPR : 'M' INT;
coordexpr:
CoordExpr
| ParameterKeyword getValueExpr
;
getValueExpr:
'[' expression ']'
;
CoordExpr
:
ParameterKeyword SCIENTIFIC_NUMBER
;
ParameterKeyword: [XYZABCUVWIJKR];
FeedExpr: 'F' SCIENTIFIC_NUMBER;
AccExpr: 'E' SCIENTIFIC_NUMBER;
fragment
GPOSTFIX
: Digit+ ('.' Digit+)*
;
variable
: VARNAME
;
scientific
: SCIENTIFIC_NUMBER
;
SCIENTIFIC_NUMBER
: SIGN? NUMBER (('E' | 'e') SIGN? NUMBER)?
;
fragment NUMBER
: ('0' .. '9') + ('.' ('0' .. '9') +)?
;
HEX_INTEGER
: '0' [xX] HEX_DIGIT+
;
fragment HEX_DIGIT
: [0-9a-fA-F]
;
INT : Digit+;
fragment
Digit : [0-9];
fragment
SIGN
: ('+' | '-')
;
VARNAME
: [a-zA-Z_][a-zA-Z_0-9]*
;
NEWLINE
: '\r'? '\n'
;
WS : [ \t]+ -> skip ; // skip spaces, tabs, newlines
Sample program(it works well except the last line):
N200 G54.1
a = 100
b = 10
c = a + b
Z[a + b*c]
N002 G2 X30.1 Y20.1 I20.1 J0.1 K0.2 R20
N100 G1X100.5Z[VAR1+100]M3H3 // it works well except the last line
I want to parse N100G1X100.5YE5Z[VAR1+100]M3H3 to
-> N100 G1 X100 Z[VAR1+100]
-> or it will be better to split the node X100 to two subnode X 100:
I am trying to use ANTLR, but ANTLR always take the rule "longest match wins". N100G1X100 is identified to a word.
Append question:
What's the best tool to finish the task?
ANTLR has a strict separation between pasrer and lexer, and therefor the lexer operates in a predictable way (longest match wins). So if you have some sort of identifier rule that matches N100G1M4 but sometimes want to match N100, G1 and M4 separately, you're out of luck.
How to handle the case?
The only answer one can give (with the amount of details given) is: remove the rule that matches N100G1M4 as 1 token. If that is something you cannot do, then don't use ANTLR, but use a "scannerless" parser.
Scannerless Parser Generators

Inline comments and empty line in antlr4 grammar

please can anyone explain me, what i need to change i this grammar to support inline comments (such as // some text) and empty line (which contains any number of space characters). I write following grammar, but this doesn't work.
program: line* EOF ;
line: (expression | assignment) (NEWLINE | EOF);
assignment : VARIABLE '=' expression ;
expression : '(' expression ')' #parenthesisExpression
| '-' expression #unaryExpression
| left=expression OP1 right=expression #firstPriorityExpression
| left=expression OP2 right=expression #secondPriorityExpression
| number=NUMBER #numericExpression
| variable=VARIABLE #variableExpression
;
NUMBER : [0-9]+ ;
VARIABLE : [a-zA-Z][a-zA-Z0-9]* ;
OP1 : '*' | '/' ;
OP2 : '+' | '-' ;
NEWLINE : '\r'? '\n' ;
WHITESPACE : [ \t\r]+ -> skip ;
COMMENT : '//' ~[\n\r]* -> skip ;
The fact you added - in a parser rule as a literal token, and also made OP2 match this character causes OP2 to never match a -. You need to have a lexer rule that matches only the single minus sign (as I showed earlier):
op1
: MUL
| DIV
;
op2
: ADD
| MIN
;
...
MUL : '*' ;
DIV : '/' ;
ADD : '+' ;
MIN : '-' ;
and then use MIN in your unary alternative:
...
| MIN expression #unaryExpression
...
When you have a separate MIN : '-' ; rule, you could do this:
...
| '-' expression #unaryExpression
...
because now ANTLR "knows" you mean the rule that matches a single -, but ANTLR does not "know" this when you have a lexer rule that matches a either a - or + like your OP2 rule:
OP2 : '+' | '-' ;

How to allow an identifer which can start with a digit without causing MismatchedTokenException

I want to match the following input:
statement span=1m 0_dur=12
with the following grammar:
options {
language = Java;
output=AST;
ASTLabelType=CommonTree;
}
statement :'statement' 'span' '=' INTEGER 'm' ident '=' INTEGER;
INTEGER
: DIGIT+
;
ident : IDENT | 'AVG' | 'COUNT';
IDENT
: (LETTER | DIGIT | '_')+ ;
WHITESPACE
: ( ' '
| '\t'
| '\r'
| '\n'
) {$channel=HIDDEN;}
;
fragment
LETTER : ('a'..'z' | 'A'..'Z') ;
fragment
DIGIT : '0'..'9';
but it cause an error:
MismatchedTokenException : line 1:15 mismatched input '1m' expecting '\u0004'
Does anyone has any idea how to solve this?
THanks
Charles
I think your grammar is context sensitive, even at the lexical analyser(Tokenizer) level. The string "1m" is recognized as IDENT, not INTEGER followed by 'm'. You either redefine your syntax, or use predicated parsing, or embed Java code in your grammar to detect the context (e.g. If the number is presented after "span" followed by "=", then parse it as INTEGER).

How to create a rule to match 2^3 to create a power operator?

Given that I have the following grammar how would I add a rule to match something like 2^3 to create a power operator?
negation : '!'* term ;
unary : ('+'!|'-'^)* negation ;
mult : unary (('*' | '/' | ('%'|'mod') ) unary)* ;
add : mult (('+' | '-') mult)* ;
relation : add (('=' | '!=' | '<' | '<=' | '>=' | '>') add)* ;
expression : relation (('&&' | '||') relation)* ;
// LEXER ================================================================
HEX_NUMBER : '0x' HEX_DIGIT+;
fragment
FLOAT: ;
INTEGER : DIGIT+ ({input.LA(1)=='.' && input.LA(2)>='0' && input.LA(2)<='9'}?=> '.' DIGIT+ {$type=FLOAT;})? ;
fragment
HEX_DIGIT : (DIGIT|'a'..'f'|'A'..'F') ;
fragment
DIGIT : ('0'..'9') ;
What I have tried:
I tried something like power : ('+' | '-') unary'^' unary but that doesn't seem to work.
I also tried mult : unary (('*' | '/' | ('%'|'mod') | '^' ) unary)* ; but that doesn't work either.
To give ^ higher precedence than negation, do this:
pow : term ('^' term)* ;
negation : '!' negation | pow ;
unary : ('+'! | '-'^)* negation ;
If you want to consider the right-associativity already in the grammar, you can also use recursion:
pow : term ('^'^ pow)?
;
negation : '!'* pow;
...

Parse sentences with different word types

I'm looking for a grammar for analyzing two type of sentences, that
means words separated by white spaces:
ID1: sentences with words not beginning with numbers
ID2: sentences with words not beginning with numbers and numbers
Basically, the structure of the grammar should look like
ID1 separator ID2
ID1: Word can contain number like Var1234 but not start with a number
ID2: Same as above but 1234 is allowed
separator: e. g. '='
#Bart
I just tried to add two tokens '_' and '"' as lexer-rule Special for later use in lexer-rule Word.
Even I haven't used Special in the following grammar, I get the following error in ANTLRWorks 1.4.2:
The following token definitions can never be matched because prior tokens match the same input: Special
But when I add fragment before Special, I don't get that error. Why?
grammar Sentence1b1;
tokens
{
TCUnderscore = '_' ;
TCQuote = '"' ;
}
assignment
: id1 '=' id2
;
id1
: Word+
;
id2
: ( Word | Int )+
;
Int
: Digit+
;
// A word must start with a letter
Word
: ( 'a'..'z' | 'A'..'Z') ('a'..'z' | 'A'..'Z' | Digit )*
;
Special
: ( TCUnderscore | TCQuote )
;
Space
: ( ' ' | '\t' | '\r' | '\n' ) { $channel = HIDDEN; }
;
fragment Digit
: '0'..'9'
;
Lexer-rule Special shall then be used in lexer-rule Word:
Word
: ( 'a'..'z' | 'A'..'Z' | Special ) ('a'..'z' | 'A'..'Z' | Special | Digit )*
;
I'd go for something like this:
grammar Sentence;
assignment
: id1 '=' id2
;
id1
: Word+
;
id2
: (Word | Int)+
;
Int
: Digit+
;
// A word must start with a letter
Word
: ('a'..'z' | 'A'..'Z') ('a'..'z' | 'A'..'Z' | Digit)*
;
Space
: (' ' | '\t' | '\r' | '\n') {skip();}
;
fragment Digit
: '0'..'9'
;
which will parse the input:
Word can contain number like Var1234 but not start with a number = Same as above but 1234 is allowed
as follows:
EDIT
To keep lexer rule nicely packed together, I'd keep them all at the bottom of the grammar instead of partly in the tokens { ... } block, which I only use for defining "imaginary tokens" (used in AST creation):
// wrong!
Special : (TCUnderscore | TCQuote);
TCUnderscore : '_';
TCQuote : '"';
Now, with the rules above, TCUnderscore and TCQuote can never become a token because when the lexer stumbles upon a _ or ", a Special token is created. Or in this case:
// wrong!
TCUnderscore : '_';
TCQuote : '"';
Special : (TCUnderscore | TCQuote);
the Special token can never be created because the lexer would first create TCUnderscore and TCQuote tokens. Hence the error:
The following token definitions can never be matched because prior tokens match the same input: ...
If you make TCUnderscore and TCQuote a fragment rule, you don't have that problem because fragment rules only "serve" other lexer rules. So this works:
// good!
Special : (TCUnderscore | TCQuote);
fragment TCUnderscore : '_';
fragment TCQuote : '"';
Also, fragment rules can therefor never be "visible" in any of your parser rules (the lexer will never create a TCUnderscore or TCQuote token!).
// wrong!
parse : TCUnderscore;
Special : (TCUnderscore | TCQuote);
fragment TCUnderscore : '_';
fragment TCQuote : '"';
I'm not sure if that fits your needs but with Bart's help in my post
ANTLR - identifier with whitespace
i came to this grammar:
grammar PropertyAssignment;
assignment
: id_nodigitstart '=' id_digitstart EOF
;
id_nodigitstart
: ID_NODIGITSTART+
;
id_digitstart
: (ID_DIGITSTART|ID_NODIGITSTART)+
;
ID_NODIGITSTART
: ('a'..'z'|'A'..'Z') ('a'..'z'|'A'..'Z'|'0'..'9')*
;
ID_DIGITSTART
: ('0'..'9'|'a'..'z'|'A'..'Z')+
;
WS : (' ')+ {skip();}
;
"a name = my 4value" works while "4a name = my 4value" causes an exception.