Left-factoring grammar of coffeescript expressions - antlr

Question

I'm writing an Antlr/Xtext parser for coffeescript grammar. It's at the beginning yet, I just moved a subset of the original grammar, and I am stuck with expressions. It's the dreaded "rule expression has non-LL(*) decision" error. I found some related questions here, Help with left factoring a grammar to remove left recursion and ANTLR Grammar for expressions. I also tried How to remove global backtracking from your grammar, but it just demonstrates a very simple case which I cannot use in real life. The post about ANTLR Grammar Tip: LL() and Left Factoring gave me more insights, but I still can't get a handle.
My question is how to fix the following grammar (sorry, I couldn't simplify it and still keep the error). I guess the trouble maker is the term rule, so I'd appreciate a local fix to it, rather than changing the whole thing (I'm trying to stay close to the rules of the original grammar). Pointers are also welcome to tips how to "debug" this kind of erroneous grammar in your head.
grammar CoffeeScript;
options {
output=AST;
}
tokens {
AT_SIGIL; BOOL; BOUND_FUNC_ARROW; BY; CALL_END; CALL_START; CATCH; CLASS; COLON; COLON_SLASH; COMMA; COMPARE; COMPOUND_ASSIGN; DOT; DOT_DOT; DOUBLE_COLON; ELLIPSIS; ELSE; EQUAL; EXTENDS; FINALLY; FOR; FORIN; FOROF; FUNC_ARROW; FUNC_EXIST; HERECOMMENT; IDENTIFIER; IF; INDENT; INDEX_END; INDEX_PROTO; INDEX_SOAK; INDEX_START; JS; LBRACKET; LCURLY; LEADING_WHEN; LOGIC; LOOP; LPAREN; MATH; MINUS; MINUS; MINUS_MINUS; NEW; NUMBER; OUTDENT; OWN; PARAM_END; PARAM_START; PLUS; PLUS_PLUS; POST_IF; QUESTION; QUESTION_DOT; RBRACKET; RCURLY; REGEX; RELATION; RETURN; RPAREN; SHIFT; STATEMENT; STRING; SUPER; SWITCH; TERMINATOR; THEN; THIS; THROW; TRY; UNARY; UNTIL; WHEN; WHILE;
}
COMPARE : '<' | '==' | '>';
COMPOUND_ASSIGN : '+=' | '-=';
EQUAL : '=';
LOGIC : '&&' | '||';
LPAREN : '(';
MATH : '*' | '/';
MINUS : '-';
MINUS_MINUS : '--';
NEW : 'new';
NUMBER : ('0'..'9')+;
PLUS : '+';
PLUS_PLUS : '++';
QUESTION : '?';
RELATION : 'in' | 'of' | 'instanceof';
RPAREN : ')';
SHIFT : '<<' | '>>';
STRING : '"' (('a'..'z') | ' ')* '"';
TERMINATOR : '\n';
UNARY : '!' | '~' | NEW;
// Put it at the end, so keywords will be matched earlier
IDENTIFIER : ('a'..'z' | 'A'..'Z')+;
WS : (' ')+ {skip();} ;
root
: body
;
body
: line
;
line
: expression
;
assign
: assignable EQUAL expression
;
expression
: value
| assign
| operation
;
identifier
: IDENTIFIER
;
simpleAssignable
: identifier
;
assignable
: simpleAssignable
;
value
: assignable
| literal
| parenthetical
;
literal
: alphaNumeric
;
alphaNumeric
: NUMBER
| STRING;
parenthetical
: LPAREN body RPAREN
;
// term should be the same as expression except operation to avoid left-recursion
term
: value
| assign
;
questionOp
: term QUESTION?
;
mathOp
: questionOp (MATH questionOp)*
;
additiveOp
: mathOp ((PLUS | MINUS) mathOp)*
;
shiftOp
: additiveOp (SHIFT additiveOp)*
;
relationOp
: shiftOp (RELATION shiftOp)*
;
compareOp
: relationOp (COMPARE relationOp)*
;
logicOp
: compareOp (LOGIC compareOp)*
;
operation
: UNARY expression
| MINUS expression
| PLUS expression
| MINUS_MINUS simpleAssignable
| PLUS_PLUS simpleAssignable
| simpleAssignable PLUS_PLUS
| simpleAssignable MINUS_MINUS
| simpleAssignable COMPOUND_ASSIGN expression
| logicOp
;
UPDATE:
The final solution will use Xtext with an external lexer to avoid to intricacies of handling significant whitespace. Here is a snippet from my Xtext version:
CompareOp returns Operation:
AdditiveOp ({CompareOp.left=current} operator=COMPARE right=AdditiveOp)*;
My strategy is to make a working Antlr parser first without a usable AST. (Well, it would deserve a separates question if this is a feasible approach.) So I don't care about tokens at the moment, they are included to make development easier.
I am aware that the original grammar is LR. I don't know how close I can stay to it when transforming to LL.
UPDATE2 and SOLUTION:
I could simplify my problem with the insights gained from Bart's answer. Here is a working toy grammar to handle simple expressions with function calls to illustrate it. The comment before expression shows my insight.
grammar FunExp;
ID: ('a'..'z'|'A'..'Z'|'_') ('a'..'z'|'A'..'Z'|'0'..'9'|'_')*;
NUMBER: '0'..'9'+;
WS: (' ')+ {skip();};
root
: expression
;
// atom and functionCall would go here,
// but they are reachable via operation -> term
// so they are omitted here
expression
: operation
;
atom
: NUMBER
| ID
;
functionCall
: ID '(' expression (',' expression)* ')'
;
operation
: multiOp
;
multiOp
: additiveOp (('*' | '/') additiveOp)*
;
additiveOp
: term (('+' | '-') term)*
;
term
: atom
| functionCall
| '(' expression ')'
;

Answer 1

When you generate a lexer and parser from your grammar, you see the following error printed to your console:
error(211): CoffeeScript.g:52:3: [fatal] rule expression has non-LL(*) decision due to recursive rule invocations reachable from alts 1,3. Resolve by left-factoring or using syntactic predicates or using backtrack=true option.
warning(200): CoffeeScript.g:52:3: Decision can match input such as "{NUMBER, STRING}" using multiple alternatives: 1, 3
As a result, alternative(s) 3 were disabled for that input
(I've emphasized the important bits)
This is only the first error, but you start with the first and with a bit of luck, the errors below that first one will also disappear when you fix the first one.
The error posted above means that when you're trying to parse either a NUMBER or a STRING with the parser generated from your grammar, the parser can go two ways when it ends up in the expression rule:
expression
: value // choice 1
| assign // choice 2
| operation // choice 3
;
Namely, choice 1 and choice 3 both can parse a NUMBER or a STRING, as you can see by the "paths" the parser can follow to match these 2 choices:
choice 1:
expression
value
literal
alphaNumeric : {NUMBER, STRING}
choice 3:
expression
operation
logicOp
relationOp
shiftOp
additiveOp
mathOp
questionOp
term
value
literal
alphaNumeric : {NUMBER, STRING}
In the last part of the warning, ANTLR informs you that it ignores choice 3 whenever either a NUMBER or a STRING will be parsed, causing choice 1 to match such input (since it is defined before choice 3).
So, either the CoffeeScript grammar is ambiguous in this respect (and handles this ambiguity somehow), or your implementation of it is wrong (I'm guessing the latter :)). You need to fix this ambiguity in your grammar: i.e. don't let the expression's choices 1 and 3 both match the same input.
I noticed 3 other things in your grammar:
1
Take the following lexer rules:
NEW : 'new';
...
UNARY : '!' | '~' | NEW;
Be aware that the token UNARY can never match the text 'new' since the token NEW is defined before it. If you want to let UNARY macth this, remove the NEW rule and do:
UNARY : '!' | '~' | 'new';
2
In may occasions, you're collecting multiple types of tokens in a single one, like LOGIC:
LOGIC : '&&' | '||';
and then you use that token in a parser rules like this:
logicOp
: compareOp (LOGIC compareOp)*
;
But if you're going to evaluate such an expression at a later stage, you don't know what this LOGIC token matched ('&&' or '||') and you'll have to inspect the token's inner text to find that out. You'd better do something like this (at least, if you're doing some sort of evaluating at a later stage):
AND : '&&';
OR : '||';
...
logicOp
: compareOp ( AND compareOp // easier to evaluate, you know it's an AND expression
| OR compareOp // easier to evaluate, you know it's an OR expression
)*
;
3
You're skipping white spaces (and no tabs?) with:
WS : (' ')+ {skip();} ;
but doesn't CoffeeScript indent it's code block with spaces (and tabs) just like Python? But perhaps you're going to do that in a later stage?
I just saw that the grammar you're looking at is a jison grammar (which is more or less a bison implementation in JavaScript). But bison, and therefor jison, generates LR parsers while ANTLR generates LL parsers. So trying to stay close to the rules of the original grammar will only result in more problems.