perl6/rakudo: dereferencing-question
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14-11-2019 - |
質問
#!perl6
use v6;
my $list = 'a' .. 'f';
sub my_function( $list ) {
for ^$list.elems -> $e {
$list[$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?
解決
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
他のヒント
These should work:
.say for @( $list );
.say for $list.list;
.say for $list.flat;
Since $list
is 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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