如何以非递归方式编写 Antlr 规则? [英] How to write Antlr rules in a non-recursive way?

问题描述

我有以下需要解析的表达式

I have the following the expressions I need to parse

and(true, false)
or(true, false, true)
not(or(true, false, true))
and(and(true, false), false)
or(or(true, false, true), true)

到目前为止我有以下语法

so far I have the following grammar

expr
    : orExpr
    ;

orExpr
    : OR '(' andExpr (',' andExpr)+ ')'
    | andExpr
    ;

andExpr
    : AND '(' equalityExpr (',' equalityExpr)+ ')'
    | equalityExpr
    ;

equalityExpr
    : comparison ((EQUALS | NOT_EQUALS) comparison)*
    ;

comparison
    : notExpr ((GREATER_THAN_EQUALS | LESS_THAN_EQUALS | GREATER_THAN | LESS_THAN ) notExpr)?
    ;

notExpr
    : NOT '(' expr ')'
    | primary
    ;

primary
    : '(' expr ')'
    | true
    | false
    ;

我无法用上述语法解析and(and(true, false), false)?我哪里出错了?

I am not able to parse and(and(true, false), false) with the above grammar? where am I going wrong?

请假设 AND OR NOT 之间有优先级(虽然我理解它可能看起来没有必要)

Please assume there is precedence between AND OR NOT (although I understand it may look not necessary)

推荐答案

但现在，我需要能够解析 true=and(5=5, 4=4, 3>2) 或反之亦然 and(5=5,4=4, 3>2)=true ?

在那种情况下，绝对没有必要把事情复杂化.你所要做的就是:

In that case, there is absolutely no need to complicate things. All you have to do is this:

grammar Test;

parse
 : expr EOF
 ;

expr
 : '(' expr ')'
 | OR '(' expr (',' expr)+ ')'
 | AND '(' expr (',' expr)+ ')'
 | NOT '(' expr ')'
 | expr (LT | LTE | GT | GTE) expr
 | expr (EQ | NEQ) expr
 | TRUE
 | FALSE
 | INT
 ;

LT    : '<';
LTE   : '<=';
GT    : '>';
GTE   : '>=';
NEQ   : '!=';
EQ    : '=';
NOT   : 'not';
TRUE  : 'true';
FALSE : 'false';
AND   : 'and';
OR    : 'or';

INT
 : [0-9]+
 ;

SPACES
 : [ \t\r\n] -> skip
 ;

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