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
-- Joshua Juran Metamage Software Creations - Mac Software and Consulting http://www.metamage.com/
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