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Re: unpack('>f', b'\x00\x01\x00\x00')

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On Wed, Nov 30, 2011 at 3:24 PM, kuaile xu <kuaile.xu@gmail.com> wrote:
> Hi:
>
> I am working on a python script that parses mp4 video header. Once of
> the field is a 32-bit fixed-point number.
>
> I know that the four bytes are: 00, 01, 00, 00. I have a third party
> mp4 parsing program which displays this field's value is:1.0.
>
> However, the struct.unpack gets a value of 0.0.
>
> Python 3.2 (r32:88445, Feb 20 2011, 21:30:00) [MSC v.1500 64 bit
> (AMD64)] on win32
> Type "help", "copyright", "credits" or "license" for more information.
>>>> from struct import *
>>>> unpack('>f', b'\x00\x01\x00\x00')
> (9.183549615799121e-41,)
>>>>

Like Chris said, you can't unpack it as a float.  Assuming that's Q16
fixed-point, what you can do is unpack it as an int and then multiply
by 2.0 ** -16 to get the corresponding float value.  There should be
no loss of precision converting to a float since floats store 52 bits
of precision, but be aware that you could have overflow or loss of
precision when converting in the opposite direction.

> In addition, there is another field which is a 16-bit fixed point
> number. How do I unpack two bytes into a float?

Same as above, just change the multiplier to match the scale.

Cheers,
Ian

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Thread

unpack('>f', b'\x00\x01\x00\x00') kuaile xu <kuaile.xu@gmail.com> - 2011-11-30 14:24 -0800
  Re: unpack('>f', b'\x00\x01\x00\x00') Chris Rebert <clp2@rebertia.com> - 2011-11-30 15:02 -0800
    Re: unpack('>f', b'\x00\x01\x00\x00') kuaile xu <kuaile.xu@gmail.com> - 2011-11-30 15:25 -0800
    Re: unpack('>f', b'\x00\x01\x00\x00') Hrvoje Niksic <hniksic@xemacs.org> - 2011-12-01 11:21 +0100
      Re: unpack('>f', b'\x00\x01\x00\x00') Mark Dickinson <mdickinson@enthought.com> - 2011-12-02 08:25 -0800
  Re: unpack('>f', b'\x00\x01\x00\x00') Ian Kelly <ian.g.kelly@gmail.com> - 2011-11-30 16:49 -0700

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