How to convert bytearray into integer?

[Jacky]

Hi Mark,

Thanks for your reply.  Agree and I'll use your suggestions.  Thanks!

-Jacky

On Aug 17, 3:36 am, Mark Dickinson <dicki... at gmail.com> wrote:
> On Aug 16, 8:08 pm, Jacky <jacky.chao.w... at gmail.com> wrote:
>
> > Hi Thomas,
>
> > Thanks for your comments!  Please check mine inline.
>
> > On Aug 17, 1:50 am, Thomas Jollans <tho... at jollybox.de> wrote:
>
> > > On Monday 16 August 2010, it occurred to Jacky to exclaim:
>
> > > > Hi there,
>
> > > > Recently I'm facing a problem to convert 4 bytes on an bytearray into
> > > > an 32-bit integer.  So far as I can see, there're 3 ways:
> > > > a) using struct module,
>
> > > Yes, that's what it's for, and that's what you should be using.
>
> > My concern is that struct may need to parse the format string,
> > construct the list, and de-reference index=0 for this generated list
> > to get the int out.
>
> > There should be some way more efficient?
>
> Well, you can improve on the struct solution by using the
> struct.Struct class to avoid parsing the format string repeatedly:
>
> >>> import struct
> >>> S = struct.Struct('<I')
> >>> S.unpack_from(buffer(bytearray([1,2,3,4,5])))
>
> (67305985,)
>
> This doesn't make a huge difference on my machine (OS X 10.6.4, 64-bit
> build of Python 2.6) though;  it's probably more effective for long
> format strings. Adding:
>
> def test_struct2(buf, offset, S=struct.Struct('<I')):
>     return S.unpack_from(buf, offset)[0]
>
> to your test code, I see a speedup of around 8% over your test_struct.
>
> By the way, you may want to consider using an explicit byte-order/size
> marker in your format string;  i.e., use '<I' instead of 'I'.  This
> forces a 4-byte little-endian interpretation, regardless of the
> platform you're running Python on.
>
> --
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>
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