antlr4：运算符优先级更改

如何解决antlr4：运算符优先级更改

我对antlr4及其令牌的优先级有疑问。我有以下语法：

grammar TestGrammar;

@header {
package some.package;
}

fragment A : ('A'|'a') ;
fragment E : ('E'|'e') ;
fragment F : ('F'|'f') ;
fragment L : ('L'|'l') ;
fragment R : ('R'|'r') ;
fragment S : ('S'|'s') ;
fragment T : ('T'|'t') ;
fragment U : ('U'|'u') ;
BOOL : (T R U E | F A L S E) ;

AND : '&' ;
OR : '|' ;
IMPLIES : '=>' ;

AS : 'als' ;

ID : [a-zA-Z_][a-zA-Z0-9_]+ ;

value_assignment : AS name=ID ;

formula  :
  BOOL /*(variable=value_assignment)?*/  #ExpressionBoolean
  | identifier=ID /*(variable=value_assignment)?*/  #ExpressionIdentifier
  | leftFormula=formula operator=(AND | OR) rightFormula=formula /*(variable=value_assignment)?*/  #ExpressionBinaryAndOr
  | leftFormula=formula operator=IMPLIES rightFormula=formula /*(variable=value_assignment)?*/  #ExpressionBinaryImplies
;

它用于执行一些命题逻辑。我希望它首先评估和或或，然后说明含义。如果我使用的是建议的语法，则可以按预期工作。请注意， value_assignment 规则已被注释掉。我有一些测试用例可以使用该功能：

public class TestGrammarTest {

    private static ParserRuleContext parse(final String input) {
        final TestGrammarLexer lexer = new TestGrammarLexer(CharStreams.fromString(input));
        final CommonTokenStream tokens = new CommonTokenStream(lexer);
        return new TestGrammarParser(tokens).formula();
    }

    private static Set<Object> states() {
        final Set<Object> states = new HashSet<Object>();

        states.add(0);
        states.add(1);
        states.add(2);

        return states;
    }

    @DataProvider (name = "testEvaluationData")
    public Object[][] testEvaluationData() {
        return new Object [][] {
            {"true & false => true",states(),states()},{"false & true => true",};
    }

    @Test (dataProvider = "testEvaluationData")
    public void testEvaluation(final String input,final Set<Object> states,final Set<Object> expectedResult) {
        System.out.println("test evaluation of <" + input + ">");
        Assert.assertEquals(
                new TestGrammarVisitor(states).visit(parse(input)),expectedResult
            );
    }

}

我认为我还需要向访问者展示（缩小版本），以使行为更清晰。如您所愿，实现很简单：

public class TestGrammarVisitor extends TestGrammarBaseVisitor<Set<Object>> {

    final Set<Object> states;

    public TestGrammarVisitor(final Set<Object> theStates) {
        states = Collections.unmodifiableSet(theStates);
    }

    @Override
    public Set<Object> visitExpressionBoolean(final ExpressionBooleanContext ctx) {
        System.out.println("\nvisitExpressionBoolean called ...\n");
        final TerminalNode node = ctx.BOOL();
        final Set<Object> result;
        if (node.getText().equalsIgnoreCase("true")) {
            result = new HashSet<>(states);
            return result;
        }
        result = Collections.emptySet();
        return result;
    }

    @Override
    public Set<Object> visitExpressionBinaryAndOr(final ExpressionBinaryAndOrContext ctx) {
        System.out.println("\nvisitExpressionBinaryAndOr called ...\n");
        final Set<Object> result = new HashSet<>(super.visit(ctx.leftFormula));
        switch (ctx.operator.getText()) {
        case "&":
            result.retainAll(super.visit(ctx.rightFormula));
            return result;
        case "|":
            result.addAll(super.visit(ctx.rightFormula));
            return result;
        default:
            throw new UnsupportedOperationException();
        }
    }

    @Override
    public Set<Object> visitExpressionBinaryImplies(final ExpressionBinaryImpliesContext ctx) {
        System.out.println("\nvisitExpressionBinaryImplies called ...\n");
        final Set<Object> result = new HashSet<>(states);
        result.removeAll(super.visit(ctx.leftFormula));
        result.addAll(super.visit(ctx.rightFormula));
        return result;
    }

    @Override
    protected Set<Object> aggregateResult(Set<Object> aggregate,Set<Object> nextResult) {
        if (aggregate == null) {
            return nextResult;
        }
        if (nextResult == null) {
            return aggregate;
        }
        Set<Object> clone = new HashSet<>(aggregate);
        clone.addAll(nextResult);
        return clone;
    }

}

我使用println语句来查看何时调用不同的规则。如果我测试显示的语法，

(variable=value_assignment)?

被注释掉，输出如预期：

test evaluation of <true & false => true>
visitExpressionBinaryImplies called ...
visitExpressionBinaryAndOr called ...
visitExpressionBoolean called ...
visitExpressionBoolean called ...
visitExpressionBoolean called ...

test evaluation of <false & true => true>
visitExpressionBinaryImplies called ...
visitExpressionBinaryAndOr called ...
visitExpressionBoolean called ...
visitExpressionBoolean called ...
visitExpressionBoolean called ...

PASSED: testEvaluation("true & false => true",[0,1,2],2])
PASSED: testEvaluation("false & true => true",2])

但是，当我包含这些语句时，优先级会发生变化：

test evaluation of <true & false => true>
visitExpressionBinaryAndOr called ...
visitExpressionBoolean called ...
visitExpressionBinaryImplies called ...
visitExpressionBoolean called ...
visitExpressionBoolean called ...

test evaluation of <false & true => true>
visitExpressionBinaryAndOr called ...
visitExpressionBoolean called ...
visitExpressionBinaryImplies called ...
visitExpressionBoolean called ...
visitExpressionBoolean called ...

PASSED: testEvaluation("true & false => true",2])
Failed: testEvaluation("false & true => true",2])
java.lang.AssertionError: Sets differ: expected [0,2] but got []

如您所见，该蕴涵将在合取之后被称为，这不是我想要的。同样，第一个测试用例意外通过了测试，因为预期的操作员优先级将无法得到满足。谁能向我解释为什么使用 value_assignment -规则（我只是删除了周围的注释符号）而导致运算符优先级发生变化？

非常感谢您的帮助！

解决方法

经过一些尝试，我设法解决了以下问题：

grammar TestGrammar;

@header {
package some.package;
}

fragment A : ('A'|'a') ;
fragment E : ('E'|'e') ;
fragment F : ('F'|'f') ;
fragment L : ('L'|'l') ;
fragment R : ('R'|'r') ;
fragment S : ('S'|'s') ;
fragment T : ('T'|'t') ;
fragment U : ('U'|'u') ;
BOOL : (T R U E | F A L S E) ;

AND : '&' ;
OR : '|' ;
IMPLIES : '=>' ;

AS : 'als' ;

ID : [a-zA-Z_][a-zA-Z0-9_]+ ;

formula  :
  BOOL #ExpressionBoolean
  | leftFormula=formula operator=(AND | OR) rightFormula=formula #ExpressionBinaryAndOr
  | leftFormula=formula operator=IMPLIES rightFormula=formula #ExpressionBinaryImplies
  | innerFormula=formula AS storageName=ID  #ExpressionAssignment
  | identifier=ID #ExpressionIdentifier
;

因此，我将存储能力作为一个单独的公式进行处理。这并不是我想要做的（这迫使我向每个子公式提供存储选项，并且如果需要特定子公式是否需要存储行为，则必须在访问者中管理它）。但是，我可以忍受这种解决方法。