Antlr3具有良好AST的表达式的非递归/左因子分解语法

Question

我为表达式定义了以下生成规则。语法不允许有回溯，并且k更好或等于3。目前的版本有一些歧义，但我不知道在哪里。我在这里删除了AST规则，但语法应该创建一个很好的AST，其中根据优先级显示操作，并显示操作的左关联性。

Antlr 3.2.1与Antlerworks 1.5.1

disjunctionExpresion
:   (conjunctionExpresion Disjunction conjunctionExpresion disjunctionExpresionDash) | conjunctionExpresion;
disjunctionExpresionDash
:   (Disjunction conjunctionExpresion disjunctionExpresionDash) |;
conjunctionExpresion
:   (relationalExpresion Conjunction relationalExpresion conjunctionExpresionDash) | relationalExpresion;
conjunctionExpresionDash
:   (Conjunction relationalExpresion conjunctionExpresionDash)|;

relationalExpresion
:   (addExpresion RelationalOperator addExpresion relationalExpresionDash) | addExpresion;
relationalExpresionDash
:   (RelationalOperator addExpresion relationalExpresionDash)|;
addExpresion
:   (multiExpresion addOperator multiExpresion addExpresionDash)| multiExpresion;

addExpresionDash
:   (addOperator multiExpresion addExpresionDash)|;
multiExpresion
:   (unaryExpresion MultiOperator unaryExpresion multiExpresionDash) | unaryExpresion;
multiExpresionDash
:   (MultiOperator unaryExpresion multiExpresionDash) | ;   
unaryExpresion 
:   (unaryOperator basicExpr)->^(unaryOperator basicExpr) | basicExpr -> basicExpr;
basicExpr
:   Number | var basicExprDash  | ('(' expr ')')->expr;
basicExprDash
:   'touches' var | ;

Answer 1

有了@kaby76关于使用EBNF和查找示例语法的提示，我最终得到了类似于这个C++Antlr3示例的东西。Antrl3树构造文档也非常有帮助。

"^"快速操作员允许我创建所需的AST。

expr    :   disjunctionExpression;
disjunctionExpression
:   conjunctionExpression (Disjunction^ conjunctionExpression)*;
/*equal to:
disjunctionExpression
:   (a=conjunctionExpression->$a) (o=Disjunction b=conjunctionExpression -> ^($o $disjunctionExpression $b) )*;

*/
conjunctionExpression
:   relationalExpression (Conjunction^ relationalExpression)*;
relationalExpression
:   additiveExpression (relationalOperator^ additiveExpression)*;
additiveExpression
:   multiExpression (addOperator^ multiExpression)*;
multiExpression
:   unaryExpression (multiOperator^ unaryExpression)*;
unaryExpression
:   (unaryOperator^)? basicExpr;
basicExpr
:   Number | var ('touches'^ var)? | '(' expr ')' -> expr;

这实际上是我以前的精简版：

expr    :   disjunctionExpresion;
disjunctionExpresion
:   conjunctionExpresion disjunctionExpresionDash;
disjunctionExpresionDash
:   (Disjunction conjunctionExpresion disjunctionExpresionDash) |;