Reputation: 7301
How is operator precedence implemented in ANTLR?
I'm using the XText/Antlr package at the moment.
Edit:
I did what sepp2k suggested, and operator precedence works now, but stuff like 3 +* also work now. The operators are basically "falling through" the tree.
Also, I tried the C grammar on ANTLR's website and the same thing happened in ANTLRworks.
Anyone know what the issue is?
BinaryExpression:
'or'? AndOp; //or op
AndOp:
'and'? ComparisonOp;
ComparisonOp:
('>'|'<'|'>='|'<='|'=='|'~=')? ConcatOp;
ConcatOp:
'..'? AddSubOp;
AddSubOp:
('+' | '-')? MultDivOp;
MultDivOp:
('*' | '/')? ExpOp;
ExpOp:
'^'? expr=Expression;
Upvotes: 11
Views: 9273
Reputation: 370172
With Xtext / ANTLR 3 you encode the precedence in the grammar rules like this:
Expr: mult ('+' mult)* ;
Mult: atom ('*' atom)* ;
Atom: INT | '(' expr ')' ;
This would parse "1 + 2 * 3 + ( 4 * 5 + 6)" as "(1 + (2 * 3)) + ((4 * 5) + 6)"
Upvotes: 15
Reputation: 6729
Since you use Xtext, I'd recommend to use the action concept of Xtext. That is, a simple expression grammar would typically look similar to this one:
Sum: Product ({Sum.left=current} operator=('+'|'-') right=Product)*;
Product: Atom ({Product.left=current} operator=('+'|'-') right=Atom)*;
Atom: Number | Paren;
Paren: '(' Sum ')';
Number: value=INT;
Please have a look at the docs for details.
Upvotes: 3