perl6是否可以在匹配中使用结点? [英] perl6 Is using junctions in matching possible?

问题描述

是否可以使用结点匹配结点中的任何值?我想匹配数组中的任何值.正确的方法是什么?

Is it possible to use junction to match any of the values in a junction? I want to match any of the values in an array. What is the proper way to do it?

lisprog$ perl6
To exit type 'exit' or '^D'
> my @a=<a b c>
[a b c]
> any(@a)
any(a, b, c)
> my $x=any(@a)
any(a, b, c)
> my $y = "a 1"
a 1
> say $y ~~ m/ $x /
False
> say $y ~~ m/ "$x" /
False
> my $x = any(@a).Str
any("a", "b", "c")
> say $y ~~ m/ $x /
False
> say $y ~~ m/ || $x /
False
> say $y ~~ m/ || @a /
｢a｣
> 

谢谢!!

推荐答案

结点不能插入到正则表达式中.它们应在正常的Perl 6表达式中使用，尤其是在比较运算符(例如eq)中:

Junctions are not meant to be interpolated into regexes. They're meant to be used in normal Perl 6 expressions, particularly with comparison operators (such as eq):

my @a = <x y z>;
say    "y" eq any(@a);  # any(False, True, False)
say so "y" eq any(@a);  # True

要匹配正则表达式中数组的任何值，只需在正则表达式中写入数组变量的名称(以@开头).默认情况下，这被解释为|替代(最长匹配")，但是您也可以将其指定为||替代(第一匹配"):

To match any of the values of an array in a regex, simply write the name of the array variable (starting with @) in the regex. By default, this is interpreted as an | alternation ("longest match"), but you can also specify it to be a || alternation ("first match"):

my @a = <foo bar barkeep>;
say "barkeeper" ~~ / @a /;     # ｢barkeep｣
say "barkeeper" ~~ / || @a /;  # ｢bar｣

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