RE: copying structures
That helps a lot, thanks. I don't often use references, especially to
anonymous things. I'm used to it being a reference to a "real" hash, like
my $hashref = \%hash;
Thanks for the clarification!
> -Original Message-
> From: Thomas, Mark - BLS CTR [mailto:[EMAIL PROTECTED]
> Sent: Thursday, March 17, 2005 12:52 PM
> To: 'Peter Eisengrein'; perl-win32-users@listserv.ActiveState.com
> Subject: RE: copying structures
>
>
> Peter Eisengrein wrote:
>
> > OK, I sort of get it. But what hash is it a reference to? If
> > you wanted to
> > access or modify the hash directly where is it?
>
> You can modify it through either reference. Maybe the
> following code will
> help you. Other references are 'perldoc perldata', 'perldoc perldsc',
> Learning Perl, etc.
>
> #---
>
> my $hashref1 = { foo => 'bar' };
>
> my $hashref2 = $hashref1;
>
> # Only the reference is copied
> print $hashref1->{foo}; # prints "bar"
> print $hashref2->{foo}; # prints "bar"
>
> # You can use either reference to modify the hash
> $hashref2->{foo} = 'baz';
>
> print $hashref1->{foo}; # prints "baz"
> print $hashref2->{foo}; # prints "baz"
>
> my %newhash = %$hashref1; #dereferenced, therefore values are copied
>
> $hashref1->{foo} = 'qux';
>
> print $hashref1->{foo}; # prints "qux"
> print $hashref2->{foo}; # prints "qux"
> print $newhash{foo}; # still prints "baz"
>
> #---
>
>
>
>
> __
> This message was scanned by ATX
> 12:52:33 PM ET - 3/17/2005
>
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RE: copying structures
It's nowhere. It's anonymous, u can only access it through the reference in ur data structure. At 12:33 PM 3/17/05 -0500, Peter Eisengrein wrote: >OK, I sort of get it. But what hash is it a reference to? If you wanted to >access or modify the hash directly where is it? -- REMEMBER THE WORLD TRADE CENTER ---=< WTC 911 >=-- "...ne cede males" 0100 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: copying structures
Peter Eisengrein wrote:
> OK, I sort of get it. But what hash is it a reference to? If
> you wanted to
> access or modify the hash directly where is it?
You can modify it through either reference. Maybe the following code will
help you. Other references are 'perldoc perldata', 'perldoc perldsc',
Learning Perl, etc.
#---
my $hashref1 = { foo => 'bar' };
my $hashref2 = $hashref1;
# Only the reference is copied
print $hashref1->{foo}; # prints "bar"
print $hashref2->{foo}; # prints "bar"
# You can use either reference to modify the hash
$hashref2->{foo} = 'baz';
print $hashref1->{foo}; # prints "baz"
print $hashref2->{foo}; # prints "baz"
my %newhash = %$hashref1; #dereferenced, therefore values are copied
$hashref1->{foo} = 'qux';
print $hashref1->{foo}; # prints "qux"
print $hashref2->{foo}; # prints "qux"
print $newhash{foo}; # still prints "baz"
#---
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RE: copying structures
> > > Its great that it's already done but the bigger question, to > > me, is why does it copy a reference in the second example. I > > don't get it. > > In the first example, it's a list of scalars. The scalars are > copied. In the > second example, it's a list of anonymous hash references. The > references are > copied. And the scalars they refer to are of course one and the same. OK, I sort of get it. But what hash is it a reference to? If you wanted to access or modify the hash directly where is it? ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: copying structures
> Its great that it's already done but the bigger question, to > me, is why does it copy a reference in the second example. I > don't get it. In the first example, it's a list of scalars. The scalars are copied. In the second example, it's a list of anonymous hash references. The references are copied. And the scalars they refer to are of course one and the same. -- Mark Thomas Internet Systems Architect ___ BAE SYSTEMS Information Technology 2525 Network Place Herndon, VA 20171 USA ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: copying structures
> Roll your own? This is Perl, remember? It's already done :) Its great that it's already done but the bigger question, to me, is why does it copy a reference in the second example. I don't get it. Confused... ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: copying structures
At 15:22 2005-03-17, [EMAIL PROTECTED] wrote: Is there a module that 'really' copies the values instead of the references or do I have to do that on my own? Roll your own? This is Perl, remember? It's already done :) perldoc Storable use Storable qw(dclone); # Deep (recursive) cloning $cloneref = dclone($ref); /J -- --- -- -- -- - - -- - Johan LindströmSourcerer @ Boss Casinos johanl AT DarSerMan.com Latest bookmark: "TCP Connection Passing" http://tcpcp.sourceforge.net/ dmoz: /Computers/Programming/Languages/JavaScript/ 12 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: copying structures
One way:
use Clone;
$bref = clone ([EMAIL PROTECTED]);
or
@b = @{ clone ([EMAIL PROTECTED]) };
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copying structures
Hi all,
this is more a general Perl question:
When I do
perl -e "@a=qw(a b c); @[EMAIL PROTECTED]; $b[1] = 'v'; print $a[1]"
I get
b
as output
bus when I do
perl -e "$a[1]{var} = 1; @[EMAIL PROTECTED]; $b[1]{var} = 2; print $a[1]{var}"
I get
2
as output
So obviously, which I copy a simple array, I get a real copy
of the values, but when I copy a structure (like an array of hashes
as in the second example), I get the hash references in the array
copied instead of all the key and value pairs, right?
Is there a module that 'really' copies the values instead of the
references or do I have to do that on my own?
Thanks and regards,
Gerhard
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