John W. Krahn wrote:
> 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.
Just for the sake of persnicketiness, the address of @array2 will
only be difference each time around the loop if its reference
count is non-zero. If a reference to it isn't saved then it will
just be garbage collected and reallocated in the same place.
:)
Rob
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