Re: [MacPerl] passing a reference to a hash element

On Jan 24, 2005, at 9:09 AM, Louis Pouzin wrote:

> I call a subroutine with a reference to a hash element.
> In the subroutine I want to test if the element exists.

If you've taken a reference to the hash element, then it exists (after 
autovivification, if necessary).  In the cat example, you're passing a 
hash reference and a key.

> If so, I want to get the key and the value of the element.
> I can't figure out the correct syntax.
> So far I can only get the value. Could someone prompt me ?

It's been a few years, but off the top of my head...

> #!perl -w
> use strict; $\ = "\n";
> my %H;
> $H{one}{two}{three} = 'nut';
> print exists $H{one}{two}{three}; # 1
> cat(\$H{one}{two},'three');
> dog(\$H{one}{two}{three});
>
> sub cat{
> 	print $_[1]; # three
> #	exists ${$_[0]}; # not a HASH element
> #	exists ${$_[0]}{$_[1]}; # not a HASH reference
> #	exists ${$_[0]}{"$_[1]"}; # not a HASH reference
> #	exists $_[0]{"$_[1]"}; # not a HASH reference
> }1;

My recollection is that $H{one}{two} is shorthand for $H{one}->{two}.  
That is, a hash only stores references, and a nested hash is stored in 
a containing hash by its reference.  So $H{one}{two} evaluates to a 
reference to a hash with a single key 'three' and the value 'nut'.  
Therefore, \$H{one}{two} is a scalar reference, a reference to a 
reference to a hash.  The correct syntax would be $(${$_[0]}}{$_[1]}, 
or just ${$_[0]}{$_[1]} if you drop the '\'.

> sub dog{
> 	print ${$_[0]}; # nut
> #	exists $_[0]; # not a HASH element
> #	exists ${$_[0]}; # not a HASH element
> #	exists ${${$_[0]}}; # not a HASH element
> }1;

Again, you're passing a reference to a scalar.  This time, the scalar 
is the string 'nut' instead of a hash reference.  You can't test for 
existence because you've already subscripted the hash.

Josh

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Joshua Juran
Metamage Software Creations - Mac Software and Consulting
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