Perl modifying hash reference in subroutine

Question

I am having trouble understanding the hash references and changing the hash in place, instead of returning it. I want to write a sub routine which will return a value from hash and also modify the hash. I was facing some issues while coding for it. So, I wrote the following basic code to understand modifying the hash in place. ``` #!/usr/local/bin/perl #Check hash and array references #Author: Sidartha Karna use warnings; use strict; use Data::Dumper; sub checkHashRef{ my ($hashRef, $arrVal) = @_; my %hashDeref = %{$hashRef}; $hashDeref{'check'} = 2; push(@{$arrVal}, 3); print "There:" ; print Dumper $hashRef; print Dumper %hashDeref; print Dumper $arrVal } my %hashVal = ('check', 1); my @arrVal = (1, 2); checkHashRef(\%hashVal, \@arrVal); print "here
"; print Dumper %hashVal; print Dumper @arrVal; ``` The output observed is: ``` There:$VAR1 = { 'check' => 1 }; $VAR1 = 'check'; $VAR2 = 2; $VAR1 = [ 1, 2, 3 ]; here $VAR1 = 'check'; $VAR2 = 1; $VAR1 = 1; $VAR2 = 2; $VAR3 = 3; ``` From…

Accepted Answer

This is making a (shallow) copy of `%hashVal`:

```
my %hashDeref = %{$hashRef};
```

The hash-ref `$hashRef` still points to `%hashVal` but `%hashDeref` doesn't, it is just a copy. If you want to modify the passed hash-ref in-place, then work with the passed hash-ref:

```
sub checkHashRef{
   my ($hashRef, $arrVal) = @_;
   $hashRef->{'check'} = 2;
   #...
```

That will leave your changes in `%hashVal`. In the array case, you never make a copy, you just dereference it in-place:

```
push(@{$arrVal}, 3);
```

and the change to `$arrVal` shows up in `@arrVal`.

Perl modifying hash reference in subroutine

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