Art Davis wrote:
> perldl> $a=sequence(5)
> perldl> p$a
> [0 1 2 3 4]
> perldl> $b=pdl(1)
> perldl> $c=$a->shiftleft($b,0);
> perldl> p$c
> [0 2 4 6 8]
>
> Looks like the syntax is OK and it's being interpreted by perldl. Pete
> said that the shift command works bitwise and I haven't invested the
> mental energy to comprehend what that means yet.
>
It just means that shiftleft works like the C << operator (or >> - I mix
them up!).
Anyway, it is easy to see if you print out in binary, e.g.:
perldl> $a=pdl([0,1,2,4,8])
perldl> wcols "%2d %10b", $a, $a
0 0
1 1
2 10
4 100
8 1000
perldl> $b=pdl(1)
perldl> $c = $a->shiftleft($b, 0)
perldl> wcols "%2d %2d %10b", $a, $c, $c
0 0 0
1 2 10
2 4 100
4 8 1000
8 16 10000
(the last column is the binary representation in case %b is unfamiliar).
So you see that what happens is that you shift the bit-pattern leftwise
- adding zeros at the end. shiftright does the opposite and when it
reaches zero it remains zero.
Hope that clears it up!
Cheers,
Jarle.
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