Need elegant way to cast four bytes into a long

[sismex01 at hebmex.com]

> From: Skip Montanaro [mailto:skip at pobox.com] 
> Sent: Viernes, 08 de Agosto de 2003 10:55 a.m.
> 
>     wsh> I have been getting what I want in this sort of manner :
> 
>     wsh>        l  = 0L
>     wsh>        l  = a[0]
>     wsh>        l += a[1] <<  8
>     wsh>        l += a[2] << 16
>     wsh>        l += a[3] << 24
> 
>     wsh> but I think that's too wordy.  Is there a more intrinsic and
>     wsh> elegant way to do this?
> 
> You mean like this:
> 
>     l = long(a[0] + a[1] << 8 + a[2] << 16 + a[3] << 24)
> 
> You can use struct.unpack as well, though I'd be hard-pressed 
> to get the details correct.  You'd be better off with "pydoc struct".
> 
> Skip
>

Kinda like this?

Python 2.2.3 (#42, May 30 2003, 18:12:08) [MSC 32 bit (Intel)] on win32
Type "copyright", "credits" or "license" for more information.
IDLE 0.8 -- press F1 for help
>>> a = [ 0x11, 0x33, 0x66, 0x99 ]
>>> import struct
>>> I = struct.unpack("=i","".join([chr(i) for i in a]))[0]
>>> print I, hex(I)
-1721355503 0x99663311
>>> 

As you can see, this is subject to platform byte-ordering
gimmics, but you can force it by changing the "=" in the
format string to "<" to force little-endian and ">" for
bigger-endian.

HTH

-gustavo

pd: one little, two little, three little endians, ...


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