如何将参数传递给Perl 6语法? [英] How can I pass arguments to a Perl 6 grammar?

问题描述

在编辑距离:忽略开始/结束中，我为模糊模糊匹配问题提供了Perl 6解决方案.我有一个这样的语法(尽管也许我在Edit#3之后已经改进了它):

In Edit distance: Ignore start/end, I offered a Perl 6 solution to a fuzzy fuzzy matching problem. I had a grammar like this (although maybe I've improved it after Edit #3):

grammar NString {
    regex n-chars { [<.ignore>* \w]**4 }
    regex ignore  { \s }
    }

文字4本身就是示例中目标字符串的长度.但是下一个问题可能是其他长度.那么我怎样才能告诉语法我要匹配多长时间呢?

The literal 4 itself was the length of the target string in the example. But the next problem might be some other length. So how can I tell the grammar how long I want that match to be?

推荐答案

尽管文档未显示示例或使用$args参数，但我在

Although the docs don't show an example or using the $args parameter, I found one in S05-grammar/example.t in roast.

在:args中指定参数，并为正则表达式赋予适当的签名.在正则表达式中，访问代码块中的参数:

Specify the arguments in :args and give the regex an appropriate signature. Inside the regex, access the arguments in a code block:

grammar NString {
    regex n-chars ($length) { [<.ignore>* \w]**{ $length } }
    regex ignore { \s }
    }

class NString::Actions {
    method n-chars ($/) {
        put "Found $/";
        }
    }

my $string = 'The quick, brown butterfly';

loop {
    state $from = 0;
    my $match = NString.subparse(
        $string,
        :rule('n-chars'),
        :actions(NString::Actions),
        :c($from++),
        :args( \(5) )
        );

    last unless ?$match;
    }

我仍然不确定传递参数的规则.这不起作用:

I'm still not sure about the rules for passing the arguments though. This doesn't work:

        :args( 5 )

我得到:

通过的位置太少；预期有2个参数，但有1个

Too few positionals passed; expected 2 arguments but got 1

这有效:

        :args( 5, )

但这足以让我们考虑一个晚上.

But that's enough thinking about this for one night.

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