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perl6/rakudo: dereferencing-question

开发者 https://www.devze.com 2023-02-10 22:00 出处:网络
#!perl6 use v6; my $list = \'a\' .. \'f\'; sub my_function( $list ) { for ^$list.elems -> $e { $li开发者_如何学Gost[$e].say;
#!perl6
use v6;

my $list = 'a' .. 'f';

sub my_function( $list ) {
    for ^$list.elems -> $e {
        $li开发者_如何学Gost[$e].say;
    }
}

my_function( $list );

First I tried this in perl5-style, but it didn't work:

for @$list -> $e {
    $e.say;
}
# Non-declarative sigil is missing its name at line ..., near "@$list -> "

How could I do this in perl6?


You don't dereference variables like this in Perl 6. Just use for $list

But that proably won't do what you want to do. 'a'..'f' doesn't construct a list in Perl 6, but rather a built-in data type called Range. You can check that with say $list.WHAT. To turn it into a list and iterate over each element, you'd use for $list.list


These should work:

.say for @( $list );
.say for $list.list;
.say for $list.flat;

Since $listis a scalar, for $list will just iterate over a single item.


Now, Rakudo 2015.02 works it ok.

You'd better use @ as twigil of variable name as array.

Perl 6 is context sensitive language, so if you want array act as 'true array', you'd better give it a suitable name.

#!perl6
use v6;

my @list = 'a' .. 'f';

for @list -> $e { $e.say };
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