"When I removed the "\" it works just fine. I can not exaclty (sic) understand the error or the reason...."
The backslash "\" you removed was escaping the "@" sigil, identifying your variable. That prevented Perl from understanding what you wanted.
Without getting into referencing and dereferencing, observe what happens with a simple array:
#!/usr/bin/perl
use 5.016;
use strict;
use warnings;
my @foo = ("bar", "baz", "bat");
say "\@foo with the backslash is: \@foo";
say "\@foo without the slash is @foo";
=head
C:\>thanos.pl
@foo with the backslash is: @foo # i.e., the escape makes it print as
+ a literal "@" here
@foo without the slash is bar baz bat # but here, Perl gives you the
+values in @foo!
=cut
In other words, be alert to the different uses of the "\" and the "/".
For your precise case (and you seem to have done most of your homework here), see:
- perldoc perlref read down to referencing and dereferencing.
- References quick reference and look for the slash
Questions containing the words "doesn't work" (or their moral equivalent) will usually get a downvote from me unless accompanied by:
- code
- verbatim error and/or warning messages
- a coherent explanation of what "doesn't work actually means.
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