Question

I have written this grammar:

expr        : multExpr ( ('+' | '-') multExpr )*;
multExpr    : atom ( ('*' | '/') atom )*;
atom    : INT | FLOAT | ID | '(' expr ')';
condition   : cond ('or' cond)*;
cond    : c1 ('and' c1)*;
c1      : ('not')? c2;
c2      : '(' condition ')' | boolean;
boolean : expr (relop expr | ²) | 'true' | 'false';
relop   : '<' | '<=' | '>' | '>=' | '==' | '!=';

I have omitted the lexer rules for INT,FLOAT,ID as it is obvious.

The problem is c2 rule, it is ambiguous because of ‘(‘, I could not find the solution, can you offer me a solution?

Answer 1

Why not simply do:

expr      : orExpr; 
orExpr    : andExpr ('or' andExpr)*;
andExpr   : relExpr ('and' relExpr)*;
relExpr   : addExpr (relop addExpr)?;
relop     : '<' | '<=' | '>' | '>=' | '==' | '!=';
addExpr   : multExpr (('+' | '-') multExpr)*;
multExpr  : unaryExpr (('*' | '/') unaryExpr)*;
unaryExpr : 'not'? atom;
atom      : INT | FLOAT | ID | 'true' | 'false' | '(' expr ')';

The unary not usually has a higher precedence than you’re trying to do now.

This will allow for expressions like 42 > true, but checking such semantics can come when you’re walking the AST/tree.

EDIT

The input "not(a+b >= 2 * foo/3.14159) == false" would now be parsed like this (ignoring spaces):

And if you set the output to AST and mix in some tree rewrite operators (^ and !):

options {
  output=AST;
}

// ...

expr      : orExpr; 
orExpr    : andExpr ('or'^ andExpr)*;
andExpr   : relExpr ('and'^ relExpr)*;
relExpr   : addExpr (relop^ addExpr)?;
relop     : '<' | '<=' | '>' | '>=' | '==' | '!=';
addExpr   : multExpr (('+' | '-')^ multExpr)*;
multExpr  : unaryExpr (('*' | '/')^ unaryExpr)*;
unaryExpr : 'not'^ atom | atom;
atom      : INT | FLOAT | ID | 'true' | 'false' | '('! expr ')'!;

you’d get: