Using floats in range and grammar mistakes? - antlr

Question

I am trying to convert a LALR grammar to LL using ANTLR and I am running into a few problems. So far, I think converting the expressions into a Top-Down approach is straight forward to me. The problem is when I include Range (1..10) and (1.0..10.0) with floats.
I have tried to use the answer found here and somehow it is not even running correctly with my code, let alone solving a range of float, i.e. (float..float).
Float literal and range parameter in ANTLR
Attached is a sample of my grammar that just focuses on this issue.
grammar Test;
options {
language = Java;
output = AST;
}
parse: 'in' rangeExpression ';'
;
rangeExpression : expression ('..' expression)?
;
expression : addingExpression (('=='|'!='|'<='|'<'|'>='|'>') addingExpression)*
;
addingExpression : multiplyingExpression (('+'|'-') multiplyingExpression)*
;
multiplyingExpression : unaryExpression
(('*'|'/'|'div') unaryExpression)*
;
unaryExpression: ('+'|'-')* primitiveElement;
primitiveElement : literalExpression
| id ('.' id)?
| '(' expression ')'
;
literalExpression : NUMBER
| BOOLEAN_LITERAL
| 'infinity'
;
id : IDENTIFIER
;
// L E X I C A L R U L E S
Range
: '..'
;
NUMBER
: (DIGITS Range) => DIGITS {$type=DIGITS;}
| (FloatLiteral) => FloatLiteral {$type=FloatLiteral;}
| DIGITS {$type=DIGITS;}
;
// fragments
fragment FloatLiteral : Float;
fragment Float
: DIGITS ( options {greedy = true; } : '.' DIGIT* EXPONENT?)
| '.' DIGITS EXPONENT?
| DIGITS EXPONENT
;
BOOLEAN_LITERAL : 'false'
| 'true'
;
IDENTIFIER : LETTER (LETTER | DIGIT)*;
WS : ( ' '
| '\t'
| '\r'
| '\n'
) {$channel=HIDDEN;}
;
fragment LETTER : ('a'..'z' | 'A'..'Z' | '_') ;
fragment DIGITS: DIGIT+;
fragment DIGIT : '0'..'9';
fragment EXPONENT : ('e'|'E') ('+'|'-')? ('0'..'9')+ ;
Any reason why it is not even taking:
in 10;
or
in 10.0;
Thanks in advance!

Answer 1

The following things are not correct:
you're never matching a FloatLiteral in your literalExpression rule
in every alternative of NUMBER you're changing the type of the token, therefor a NUMBER token will never be created
Something like this should work for both 11..22 and 1.1..2.2:
...
literalExpression : INT
| BOOLEAN_LITERAL
| FLOAT
| 'infinity'
;
id : IDENTIFIER
;
// L E X I C A L R U L E S
Range
: '..'
;
INT
: (DIGITS Range)=> DIGITS
| DIGITS (('.' DIGITS EXPONENT? | EXPONENT) {$type=FLOAT;})?
;
BOOLEAN_LITERAL : 'false'
| 'true'
;
IDENTIFIER : LETTER (LETTER | DIGIT)*;
WS : ( ' '
| '\t'
| '\r'
| '\n'
) {$channel=HIDDEN;}
;
fragment LETTER : ('a'..'z' | 'A'..'Z' | '_') ;
fragment DIGITS: DIGIT+;
fragment DIGIT : '0'..'9';
fragment EXPONENT : ('e'|'E') ('+'|'-')? ('0'..'9')+ ;
fragment FLOAT : ;

Answer 2

To your question about handling (1.0 .. 10.0):
Notice that parser rule primitiveElement defines an alternative as '(' expression ')', but rule expression can never reach rule rangeExpression.
Consider redefining expression and rangeExpression like so:
expression : rangeExpression
;
rangeExpression : compExpression ('..' compExpression)?
;
compExpression : addingExpression (('=='|'!='|'<='|'<'|'>='|'>') addingExpression)*
;
This ensures that the expression rule sits above all forms of expressions and will work as expected in parentheses.