antlr4:如何知道在给定上下文的情况下选择了哪个选项 [英] antlr4: how to know which alternative is chosen given a context

问题描述

假设有一个关于类型"的规则.它是预定义的类型(由 IDENTIFIER 引用)或 typeDescriptor.

Assume there is a rule about 'type'. It is either a predefined type (referred by IDENTIFIER) or a typeDescriptor.

type
:   IDENTIFIER
|   typeDescriptor
;

在我的程序中，我有一个 typeContext 'ctx' 的实例.我怎么知道是选择了路径 IDENTIFIER，还是选择了 typeDescriptor.

In my program, I have got an instance of typeContext 'ctx'. How do I know if the path IDENTIFIER is chosen, or typeDescriptor is chosen.

我知道一种方法是测试 ctx.IDENTIFIER() == null 和 ctx.typeDescriptor() == null.但是当有更多选择时，它似乎不太好用.有没有办法返回一个索引来指示选择了哪个规则?谢谢.

I recognise one way which is to test ctx.IDENTIFIER() == null and ctx.typeDescriptor() == null. But it seems not working very well when there are a lot more alternatives. Is there a way to return an index to indicate which rule is chosen? Thanks.

推荐答案

不，您可以使用您描述的方法(检查项目是否为非空)，或者您可以使用# 运算符.

No, you can either use the method you described (checking if an item is non-null), or you can label the outer alternatives of the rule using the # operator.

type
  : IDENTIFIER     # someType
  | typeDescriptor # someOtherType
  ;

当您标记外部替代项时，它将为每个标签生成 ParserRuleContext 类.在上面的示例中，您将获得一个 SomeTypeContext 或一个 SomeOtherTypeContext，它们同样适用于生成的侦听器和访问者接口.

When you label the outer alternatives, it will produce ParserRuleContext classes for each of the labels. In the example above, you'll either get a SomeTypeContext or a SomeOtherTypeContext, which applies equally to the generated listener and visitor interfaces.

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