问题描述
限时送ChatGPT账号..我需要帮助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 +,*,?对于计数,但如果我希望 (rule2) 匹配 5 次怎么办?
I know +,*,? for counts but what if I want the (rule2) to match 5 times?
是 +,*,?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
如果我想让 (rule2) 匹配 5 次怎么办?
what if I want the (rule2) to match 5 times?
在语法里面,有两个选项:
Inside the grammar, there are 2 options:
写5遍:
rule1: 'text' '(' rule2 ')' '(' rule2 ')' '(' rule2 ')' '(' rule2 ')' '(' rule2 ')'
2.
或使用语义谓词.请参阅 ANTLR4 wiki 中的段落Using Context-Dependent Predicates谓词
是 +,*,?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监听器中强制计数?
so I need to enforce count in the parser listener?
这将是 (IMO) 最佳选择:在解析器中,只需匹配 ( '(' rule2 ')' )+
,并在解析后在侦听器/访问者中进行验证.
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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