Evaluating Multidimensional Arrays and References

mr_dont has asked for the wisdom of the Perl Monks concerning the following question:

Hi PerlMonks!

Last night I was writing a script and I was stumped by its behaviour... Basically, it contains a data object which is a hash containing a multidimensional array (a 2n one)...

So I might have data like this: @{$self->{_mydata}[2][3]} = "John";

In one of my methods, when I print this out, it gives my the correct answer:

print "My name is @{$self->{_mydata}[2][3]}\n";
Output: My name is John

However, when I try to evaluate it, It doesn't work:

if ((@{$self->{_mydata}[2][3]}) eq "John")
  {
  print "My name is @{$self->{_mydata}[2][3]}\n";
  }
[download]

The above code won't print anything... Why isn't this working? Anyone know what I am doing wrong?
Thanks Fellow Monks..!

Comment on Evaluating Multidimensional Arrays and References Select or Download Code

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Re: Evaluating Multidimensional Arrays and References by c-era (Curate) on Aug 10, 2001 at 19:46 UTC
You are evaluating it in array context, try: `if ($self->{_mydata}[2][3] eq "John") or if ($$self{_mydata}[2][3] eq "John")` [download] and your print statment should be: `print $self->{_mydata}[2][3]; or print $$self{_mydata}[2][3];` [download]	[reply] [d/l] [select]
Re: Evaluating Multidimensional Arrays and References by bikeNomad (Priest) on Aug 10, 2001 at 19:48 UTC
The problem is that the @{...} returns an array. When you compare this with "John", it gets converted into a scalar, which is the size of the array (probably 1 in this case). So you're saying 'if 1 eq "John"'... The reason that it prints is that its contents are being interpolated into the string (separated by $, but that doesn't matter if you have one element). What you probably want is to say: `if ($self->{_mydata}[2][3][0] eq "John") { print "My name is $self->{_mydata}[2][3][0]\n"; }` [download]	[reply] [d/l]


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