steve abrams wrote:
Lee,
Please reply to the list, if you want our personal consultation we can arrange payment of a consulting fee :)
First off, thanks for your advice! Took what you said and passed back a hash within {}, making it 'anonymous', which I learned about today after checking out your code.
It was just a shotrcut to pass a hashref instead of creating a hash and doin a refernce to it.
Now, what we just did, vs. what I was doing in the example without the Method call (if you don't mind, of course). In my example below, I used online examples and set the scalar to be equal to the reference of the subroutine (&hashit()). From the subroutine I passed a reference of the hash (\%hash). Why does this work in the non-class case? How does
Because you were calling it wrong as I pointed out in the original reply. If you do ti right the class method works great as the example I gave you illustrated.
it work to set a scalar to be a reference to a subroutine passing a reference, and end up with the referenced return value? (That sentence might take more than one read.)
You where not doing a code reference, you where doing an object.
#!/usr/bin/perl
use strict; use warnings;
my $coderef = sub { return "Perl Rocks!\n"; }print $coderef->();
A second question: I'm having trouble getting a clear picture of what
perldoc perlref
anonymous data, or more specifically, a anonymous hash, really is. Is this a reference in itself? If so, how is it different from preceding a hash variable with "\"? If not, how does passing an anon. hash from a method called with:
$var = $obj->hashit();
change the example to:
sub hashit {
my %stuff = (a => 1, b => 2);
return \%stuff;
}It will work the same.
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