[Numpy-discussion] Native byteorder representation
Francesc Altet
faltet at carabos.com
Fri Feb 2 13:42:24 EST 2007
El dv 02 de 02 del 2007 a les 19:11 +0100, en/na Francesc Altet va
escriure:
> El dv 02 de 02 del 2007 a les 10:22 -0700, en/na Travis Oliphant va
> escriure:
> > Francesc Altet wrote:
> >
> > >Hi,
> > >
> > >We have been bitten by a small glitch related with the representation of
> > >native byteorders. Here is an example exposing the problem:
> > >
> > >
> > >
> > >>>>numpy.dtype('<i4').byteorder
> > >>>>
> > >>>>
> > >'='
> > >
> > >
> > >>>>numpy.dtype('>i4').newbyteorder('little').byteorder
> > >>>>
> > >>>>
> > >'<'
> > >
> > >
> > >
> > This is somewhat inconsistent. But, I'm not sure it's worth changing.
> >
> > In the second case, you request a "little" byteorder data-type. Keeping
> > this as '<' seems O.K.
> >
> > One could instead ask why the first example did not report a byte-order
> > of "<" when that's what was explicitly asked for.
>
> Well, just because of the same reason that
>
> numpy.dtype('<i4').byteorder
>
> returns a '=' instead of a '<' (the latter being explicitely set in the
> constructor).
Ops. I was confused about what you was saying here, sorry. Forget this.
>
> I think that returning a '=' whenever the byteorder is the same than the
> underlying machine is desirable because the user can quickly see whether
> her data is in native order or not.
I think that this is the only reason I can argue.
--
Francesc Altet | Be careful about using the following code --
Carabos Coop. V. | I've only proven that it works,
www.carabos.com | I haven't tested it. -- Donald Knuth
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