You are confusing direct assignment and computed assignment. "$b=
$b*10" creates a new PDL and puts it in the scalar named $b, dropping
the old one (and its dataflow connection to $a) on the floor. use
computed assignment (".=") instead.
(Mobile)
On Jun 26, 2010, at 8:12 AM, P Kishor <[email protected]> wrote:
> perldl> $a = sequence 5, 4, 2
> perldl> p $a
> [
> [
> [ 0 1 2 3 4]
> [ 5 6 7 8 9]
> [10 11 12 13 14]
> [15 16 17 18 19]
> ]
> [
> [20 21 22 23 24]
> [25 26 27 28 29]
> [30 31 32 33 34]
> [35 36 37 38 39]
> ]
> ]
>
> **************
> perldl> $b = $a->slice(':,1:2,:')
> perldl> $b = $b * 10
> perldl> p $a
> [
> [
> [ 0 1 2 3 4]
> [ 5 6 7 8 9]
> [10 11 12 13 14]
> [15 16 17 18 19]
> ]
> [
> [20 21 22 23 24]
> [25 26 27 28 29]
> [30 31 32 33 34]
> [35 36 37 38 39]
> ]
> ]
> perldl> p $b
> [
> [
> [ 50 60 70 80 90]
> [100 110 120 130 140]
> ]
> [
> [250 260 270 280 290]
> [300 310 320 330 340]
> ]
> ]
>
> So, what happened here? Why does multiplying the slice by 10 not
> change $a as well?
>
> Turning on dataflow at *************** above with $a->doflow doesn't
> help as well (although, by my understanding, it shouldn't be needed
>
> --
> Puneet Kishor
>
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>
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