[antlr-interest] Tree Evaluation with Logical Operators (AND and OR)

[Priolo, Scott]

This is what I have so far.  ID is converted to an INT. Variable are
pulled out of a memory map that is preloaded prior to processing.

I will try | ^(AND|OR expr expr) but it seems like there should be more
of a recursive nature to "and, or"  operators.

expr returns [int value]
    :   ^(GT a=expr b=expr)   {
    			       $value = (a>b)?1:0;
    			       le.isLowScore((a>b));
    			       System.out.println("(GT " + a + " " + b +
") bool:" + (a>b));
    			       }
    |   ^(GTEQ a=expr b=expr) {
    				$value = (a>=b)?1:0;
    				le.isLowScore((a>=b));
    				System.out.println("(GTEQ " + a + " " +
b + ") bool:" + (a>=b));
    			      }   
    |   ^(LT a=expr b=expr)   {
    				$value = (a<b)?1:0;
    				le.isLowScore((a<b));
    				System.out.println("(LT " + a + " " + b
+ ") bool:" + (a<b));
    			      }
    |   ^(LTEQ a=expr b=expr) {
    				$value = (a<=b)?1:0;
    				le.isLowScore((a<=b));
    				System.out.println("(LTEQ " + a + " " +
b + ") bool:" + (a<=b));
    			      }
    |   ^(EQ a=expr b=expr)   {
    				$value = (a==b)?1:0;
    				le.isLowScore((a==b));
    				System.out.println("(EQ " + a + " " + b
+ ") bool:" + (a==b));
    			      }
    |   ^(NEQ a=expr b=expr)  {
    				$value = (a!=b)?1:0;
    				le.isLowScore((a!=b));
    				System.out.println("(NEQ " + a + " " + b
+ ") bool:" + (a!=b));
    			      }
    |   ^(OR a=expr b=expr)   {
    				// WHAT TO DO HERE
    				//System.out.println("(OR " + a + " " +
b + ") bool:" + (a|b));
    			      }
    |   ^(AND a=expr b=expr)  {
    				// WHAT TO DO HERE
    				//System.out.println("(AND " + a + " " +
b + ") bool:" + (a&b));
    			      }
    |   ID 
        {
        try {
        	//pull value out of map if this is an id "Q22_B"
        	PgmQuestionsEntry q =
(PgmQuestionsEntry)variableMap.get(($ID.text).toLowerCase());
	        if ( q!=null ) {
	        	int idvalue = (int) q.getResponseNumber();
	        	$value = idvalue;
	        	System.out.println("=> Parsing ID = " + $ID.text
+ " value = " + idvalue);	
        	} else System.err.println("undefined variable
"+$ID.text);
	} catch (Exception e)
        {
        	e.printStackTrace();
        }
        }
    |   INT  
    	// when token matches an INT pattern convert it to int.
    	{$value = Integer.valueOf($INT.text);}
    ;

-----Original Message-----
From: shmuel siegel [mailto:antlr at shmuelhome.mine.nu] 
Sent: Thursday, January 22, 2009 7:39 AM
To: Oliver Zeigermann
Cc: Priolo, Scott; antlr-interest at antlr.org
Subject: Re: [antlr-interest] Tree Evaluation with Logical Operators
(AND and OR)

Oliver Zeigermann wrote:
> OK, so you need a tree grammar that can parse complex boolean
> expressions, right?
>
> expression
>   : ^((AND|OR) expression expression)
>   | ^(('<'|'>'|'='|'!=') expression expression)
>   | ^(('+'|'-'|'*'|'/') expression expression)
>   | ID
>   | literal
>   ;
>
> Would that work?
>
> -Oliver
>
>   
This grammar is not syntactically restrictive since it allows ^(AND 3 
4). This will cause problems if you want to evaluate the tree since you 
won't know the type of the return value of expression. It is better to 
separate out arithmetic and logical expressions. Care must be taken to 
establish if ID is arithmetic or logical.