On Aug 7, 7:47 pm, [EMAIL PROTECTED] (Chris Cosner) wrote:
> I've re-read perlref and have been trying to tease an answer out of the
> Perl Cookbook.
>
> If you put a hash reference into an array
> push @array, \%hash;
> you do not store any actual hash data in the array.
> So if you change the hash, then later pull the hash reference from the
> array and access it, you get changed data.
> If you do this in a loop, you store an array of references to the exact
> same hash, rather than lots of different hashes.
> At least, this seems to be what's happening to me in the pseudo code below:
>
> Question: Is there an efficient way (resembling push @array, \%hash) to
> do this that will work? Or do I need to 'unpack' the hash into key =>
> value notation to truly add the hash as an element in an array?
>
> while ( condition ) {
> (add data to %address)
> push @addresses, \%address;
> # print statement here shows the hash is different each time
> through the loop
>
> }
Shawn already gave you one answer. Allow me to point out that if your
comment "(add data to %address)" actually means to reassign the entire
hash, rather than to just add or delete elements from the hash, then
the far more correct solution is to just scope your variables
correctly:
my @addresses;
while (keep_going()) {
my %address = ( ... => ..., ... => .... );
push @addresses, \%address;
}
instead of:
my @addresses;
my %address;
while (keep_going()) {
%address = ( ... => ..., ... => .... );
push @addresses, \%address;
}
This way, as soon as one iteration of your loop ends, the %address
that was declared in that iteration goes away, and you get a brand new
hash the next time around.
If, however, you're only modifying an existing hash, then Shawn's
solution is the only correct one:
my @addresses;
my %address;
while (keep_going() ) {
$address{...} = ....;
push @addresses, { %address };
}
Paul Lalli
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