Peter Fleck wrote:
>
> I swear that some day I will understand references. In the meantime...
>
> I wrote some perl that used a hash where each value was a reference
> to an array. No problem. Here is the segment from within a foreach
> loop that populates the hash.
>
> @storedates = ($sdates, $idates);
> $datestore{$oldgrantid} = [EMAIL PROTECTED];
>
> I thought this would work. But it doesn't. I end up with the same
> array reference for each key and the same values, the last ones of
> the series.
>
> The fix was to change it to:
>
> $datestore{$oldgrantid} = [$sdates, $idates];
>
> Using the same array reference name (@storedata) does not work even
> though the values are different.
>
> I'm interested in a bit more explanation about why this didn't work.
> I thought I was 'getting' it until this latest struggle.
It depends on the scope of the variable. You do have warnings and
strict enabled don't you?
my @array1; # file scope
foreach ( @something ) {
my @array2; # block scope
# In the loop both @array1 and @array2 are visible
# The address of @array1 stays the same for each
# iteration of the loop.
# The address of @array2 is different for each
# iteration because my() creates a new variable
}
If you are not using my() to declare the variable inside the loop then
the variable will always have the same address.
John
--
use Perl;
program
fulfillment
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