如何跳过类似于Haskell模式匹配的Perl列表赋值中的元素？

Question

In Haskell (and various other functional programming languages), pattern matching can be used to assign specific elements of a list while discarding others:

Prelude> let [x, _, z] = "abc"
Prelude> x
'a'
Prelude> [z, x]
"ca"

Note that ' _ ' is not a variable and hasn't been assigned anything:

Prelude> _

<interactive>:5:1: Pattern syntax in expression context: _

For an Irssi script, written in Perl, I want to do a similar thing and discard the 2 ^nd element of ' @_ ' (ie not assign it to anything):

my ($message, _, $windowItem) = @_;

This fails with the error message: “ Can't declare constant item in "my" at [...]overlength_filter.pl line 17, near ") =" ”

So what is the Perl equivalent of this underscore wildcard?

Answer 1

只需将其分配给undef 。

my ($message, undef, $windowItem) = @_;

Answer 2

你也可以拿一个数组:)

my( $message , $winItem ) = @_[ 0, 2];