Perl 6 中的递归正则表达式? [英] Recursive regular expression in Perl 6?

问题描述

我一直在试图弄清楚如何在 Perl 6 中执行递归正则表达式.例如，一个平衡括号匹配器，它将匹配 ((())())(((((())()).

I've been trying to figure out how to do a recursive regular expression in Perl 6. For a toy example, a balanced parentheses matcher, which would match ((())()) inside (((((())()).

• PCRE 示例:/\((?R)?\)/

Onigmo 示例:(?<paren>\(\g<括号>*\))

我认为这样做可以:

my regex paren {
  '(' ~ ')' <paren>*
}

或者更简单的

my regex paren {
  '(' <paren>* ')'
}

但是失败了

No such method 'paren' for invocant of type 'Match'
in regex paren at ...

推荐答案

你需要明确你正在调用一个 my 范围的正则表达式:

You need to make explicit that you're calling a my-scoped regex:

my regex paren {
    '(' ~ ')' <&paren>*
}

注意已添加的 &.与此:

Notice the & that has been added. With that:

say "(()())" ~~ /^<&paren>$/    # ｢(()())｣
say "(()()" ~~ /^<&paren>$/     # Nil

虽然有时确实可以在不明确编写 & 的情况下逃脱，但在使用它时确实可以:

While it's true that you can sometimes get away without explicitly writing the &, and indeed could when using it:

say "(()())" ~~ /^<paren>$/    # ｢(()())｣
say "(()()" ~~ /^<paren>$/     # Nil

这只是因为编译器发现在词法作用域中定义了一个名为 paren 的正则表达式，所以将 语法编译成那个.对于递归情况，在解析正则表达式之后才安装声明，因此需要明确.

This only works because the compiler spots there is a regex defined in the lexical scope with the name paren so compiles the <paren> syntax into that. With the recursive case, the declaration isn't installed until after the regex is parsed, so one needs to be explicit.

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