CODEWITHSUNDEEP
perlreference
sarthak
sarthak

Reputation: 794

References in perl: Error in perl code

Here is my code:

  #!/usr/bin/perl -w
  my $name = "mark";
  my $nameRef = \$name;
  print "${$nameRef}\n";
  print "$nameRef\n";
  my $ref = $nameRef + 1;
  $$ref = "antony";
  print "$ref\n";
  print "$$ref\n";

But when I run the code I get the following error:

mark
SCALAR(0x9556cf8)
Modification of a read-only value attempted at ./stringPerl.pl line 7.

How to remove the above error?

Upvotes: 1

Views: 43

Answers (2)

Miguel Prz
Miguel Prz

Reputation: 13792

The line that causes the error is:

$$ref = "antony";

That tries to asign the value "antony" to the referenced variable by $ref, but the line:

my $ref = $nameRef + 1;

It's not a valid reference. Try to replace simply by:

my $ref = $nameRef;

So $ref is the same as $nameRef, which is a reference to $name.

If you want to change a value in an array via a reference, you may write this:

my @names = ("mark");
my $nameRef = \@names;

$nameRef->[1] = "antony";

print "$names[1]\n";
print "$nameRef->[1]\n";

Upvotes: 1

jcaron
jcaron

Reputation: 17710

Perl string references are not C string pointers. You cannot take a reference and increment it to point somewhere else.

When you do my $ref = $nameRef + 1;, it actually takes the SCALAR(0x9556cf8) string and adds 1, making it a string, not a reference.

Not sure what you are trying to do exactly.

Upvotes: 2

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