Re: [Numpy-discussion] integer power in scalarmath: how to handle overflow?

[David D. <dav...@lo...>]

On Mon, Jun 05, 2006 at 05:10:23PM -0400, David M. Cooke wrote:
> I just ran into the fact that the power function for integer types
> isn't handled in scalarmath yet. I'm going to add it, but I'm wondering
> what people want when power overflows the integer type?
>=20
> Taking the concrete example of a =3D uint8(3), b =3D uint8(10), then shou=
ld
> a ** b return
>=20
> 1) the maximum integer for the type (255 here)
> 2) 0
> 3) upcast to the largest type that will hold it (but what if it's
>    larger than our largest type? Return a Python long?)
> 4) do the power using "long" like Python does, then downcast it to the
>    type (that would return 169 for the above example)
> 5) something else?
>=20
> I'm leaning towards #3; if you do a ** 10, you get the right
> answer (59049 as an int64scalar), because 'a' is upcasted to
> int64scalar, since '10', a Python int, is converted to that type.
> Otherwise, I would choose #1.

I agree, #1 seems the better solution for me.

BTW, I'm quite new on this list, and I don't know is this has already
been discussed (I guess I has): why does uint_n arithmetics are done
in the  Z/(2**n)Z field (not sure about the maths correctness here)? I mean:
>>> a =3D numpy.uint8(10)
>>> a*a
100
>>> a*a*a # I'd like to have 255 here
232
>>> 1000%256
232

It would be really a nice feature to be able (by the mean of a numpy flag o=
r so)
to have bound limited uint operations (especially when doing image processi=
ng...).


David


--=20
David Douard                             LOGILAB, Paris (France)
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