Hi , 



     # The $x is a hashreference and here :

      my $ones = {};

      $ones{a}={b=>1};

      print "[$ones{a}]\n";
      print "[$ones->{a}]\n";
      print "[$ones{a}->{b}]\n";
      print "[$ones{a}{b}]\n";


    __END__

    [HASH(0xe01d8)]
    []
    [1]
    [1]

[download]


$x->{a} is not the same as $x{a}

( $x->{a} will be a lookup of the key of a hashrefernce (the declaration in the above my $ones = {} kind of decalaration available ),

$x{a} will be a lookup of the key of %x)

$x{a}->{$b} is the same as $x{a}{$b}

Please read the perldoc perldsc for all kinds of data structure decalaration definition and access

In reply to Re: dereference syntax by OM_Zen
in thread dereference syntax by mifflin

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