如何在ANTLR中设置规则选项计数 [英] How to set a rule option count in ANTLR

问题描述

我需要帮助来设置规则中选项匹配的计数. 此规则示例将匹配(rule2)零次或多次.

I need help setting the count for an option match in a rule. This rule example will match (rule2) zero or more times.

rule1: 'text' ( '(' rule2 ')' )*

我知道+，* 、?计数，但是如果我想让(rule2)匹配5次呢?

I know +,*,? for counts but what if I want the (rule2) to match 5 times?

+，* 、?语法支持的唯一计数修饰符，所以我需要在解析器侦听器中强制计数?

Are +,*,? the only count modifier supported for grammar so I need to enforce count in the parser listener?

推荐答案

我想让(rule2)匹配5次怎么办?

what if I want the (rule2) to match 5 times?

在语法中，有2个选项:

Inside the grammar, there are 2 options:

将其写出5次:

rule1: 'text' '(' rule2 ')' '(' rule2 ')' '(' rule2 ')' '(' rule2 ')' '(' rule2 ')'

2.

或使用语义谓词.请参阅 ANTLR4 Wiki上的使用上下文相关谓词.谓词

2.

or use a semantic predicate. See paragraph Using Context-Dependent Predicates from the ANTLR4 wiki about predicate

+，* 、?语法支持的唯一计数修饰符

Are +,*,? the only count modifier supported for grammar

是的，没有(...){5}语法可以精确匹配5次.

Yes, there is no (...){5} syntax to match exactly 5 times.

所以我需要在解析器侦听器中强制计数吗?

so I need to enforce count in the parser listener?

这是(IMO)的最佳选择:在解析器中，只需匹配( '(' rule2 ')' )+，然后在侦听器/访问者中进行解析后进行验证.

That would be the (IMO) best option: in the parser, just match ( '(' rule2 ')' )+, and in a listener/visitor validate after parsing.

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