No, it will not. Perl automatic memory handling takes care of these things.

You can also use the [] operator as ...

$g_User{user1} = [@l_UserDetails];
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if i delare array with my / local , they will get destroyed when the control comes out of the function in which the addition happens

This is wrong, Perl5 counts the references to a variable/data structure and only destroys the container if no more references are held. Since you are keeping a reference to your array in the the hash, the array will only be destroyed when your hash goes away (i.e. goes out of scope with no more references held to it) and is destroyed itself.

How did you check that the array gets destroyed when the control leaves your function?

use strict;

my %g_User;
sub setup_user_details {
  my ($username) = @_;
  my @l_UserDetails = ("John","Doe");
  $g_user{$username} = \@l_UserDetails;
};

setup_user_details('user1');
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Your naming convention is a bit unwieldly, but I attribute that to your company coding guidelines. In Perl it's conventional to use lower_caps_names instead of CamelCase, and prefixing of variables with their scope is uncommon because the strict pragma enforces proper scoping.

Use an anonymous array-ref "in situ":

$g_User{'user1'} = [];
$g_User{'user1'}->[0] = 'Hans';
$g_User{'user1'}->[1] = 'Wurst';
print join ' ', @{$g_User{'user1'}};
__END__
Hans Wurst
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To give it a life of its own instead, a copy of the array first has to be declared with dynamic scope:

my %g_User;
my @l_UserDetails = ('hallo','goodbye');
InsertArray( \@l_UserDetails, \%g_User, 'user1' );
sub InsertArray {
    my $aref = shift;
    my $href = shift;
    my $elt = shift;
    my @dynamic = @$aref;
    $href -> { $elt } = \@dynamic;
}
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$g_User{'user1'} = \@l_UserDetails;

To answer your question:

$g_User{'user1'} = [@l_UserDetails];
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But then chances are you may avoid @l_UserDetails altogether, if the code that fills it is simple enough, that is.