in reply to Two arrays into hash

Methinx you could use rudimentary algebraic methods to arrive at your goal, consider: 
@array1=qw(A B C);
@array2=qw(1 2 3);

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is equivalent to/can be re-written as: 
@array1=('A', 'B', 'C');
@array2=('1', '2', '3');

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Also, the requirement can be re-written as: 
%hash = (
array1=>[('A','B','C')],
array2=>[('1','2','3')]
};

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So, substituting for ('A','B','C') & ('1','2','3'), into the re-written required result, gives: 
%hash = (
array1=>[@array1],
array2=>[@array2]
};

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But, [@array] can be/usually is written as \@array, hence, 
%hash = (
array1=>\@array1,
array2=>\@array2
};

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As required - also discoverable by reference to perlreftut - as moritz has previously suggested.

A user level that continues to overstate my experience :-))

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Re^2: Two arrays into hash
by davorg (Chancellor) on Jul 16, 2009 at 10:40 UTC
    [@array] can be/usually is written as \@array

    Careful. The two are not the same. \@array returns a reference to the existing array. [@array] creates a new (anonymous) copy of the original array and returns a reference to the copy. It's a subtle difference, but an important one.

    #!/usr/bin/perl

use strict;
use warnings;

my @arr = (1 .. 5);

my %hash = (
            aref => \@arr,
            copy => [ @arr ],
           );

print "@{$hash{copy}}\n";
print "@{$hash{aref}}\n";

@arr = ('a' .. 'e');

print "@{$hash{copy}}\n"; # original values
print "@{$hash{aref}}\n"; # changed values

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[reply]
[d/l]
      Good point, well made - I rather overlooked that didn't I ?

      Thanx davorg.

      A user level that continues to overstate my experience :-))
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