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Home/ Questions/Q 7445019
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T11:43:18+00:00 2026-05-29T11:43:18+00:00

With respect to the following code segment, I would like to know whether my

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With respect to the following code segment, I would like to know whether my understanding on several issues are correct?

1) In the structure of $model->{in1}->{tra1}->{data}} , “in1”, “tra1”, and “data” all represent specific keys at different levels of hash structures.

2) Does $#{$model->{in1}->{tra1}->{data}}represent an array?

3) What does my @cus = sort keys %cus; aim to do? Are the “cus” at the right side and the “cus” at the left side the same thing?

my %cus = ();
for my $i ( 0 .. $#{$model->{in1}->{tra1}->{data}})
{
  foreach  my $cu (keys %{$model->{in1}->{tra1}->{data}->[$i]->{concept}}
   {
       $cus{$cu} = 1;
   }
 }
my  @cus = sort keys %cus;
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T11:43:19+00:00Added an answer on May 29, 2026 at 11:43 am

    1)

    They are keys to different hashes, yes.

    • in1 is used as the key to the hash referenced by $model.
    • tra1 is used as the key to the hash referenced by $model->{in1}.
    • data is used as the key to the hash referenced by $model->{in1}->{tra1}.

    2)

    • $#a returns the last index of array @a.

    so

    • $#{ $ref } (or $#$ref for short) returns the last index of @{ $ref } (or @$ref for short), the array referenced by $ref.

    so

    • $#{ $model->{in1}->{tra1}->{data} } returns the last index of @{ $model->{in1}->{tra1}->{data} }, the array referenced by $model->{in1}->{tra1}->{data}.

    3)

    The statement sorts the keys of the hash %cus and places them in array @cus. No, %cus and @cus aren’t the same variable.

    “4”)

    The code can be simplified to:

    my %cus;
    my $data = $model->{in1}->{tra1}->{data};
    for my $i (0 .. $#$data) {
       for my $cu (keys %{ $data->[$i]->{concept} }) {
          ++$cus{$cu};
       }
    }
    
    my @cus = sort keys %cus;
    

    Or even:

    my %cus;
    for my $data_item (@{ $model->{in1}->{tra1}->{data} }) {
       for my $cu (keys %{ $data_item->{concept} }) {
          ++$cus{$cu};
       }
    }
    
    my @cus = sort keys %cus;
    
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