Need elegant way to cast four bytes into a long

Anton Vredegoor anton at
Mon Aug 11 10:30:23 CEST 2003

"William S. Huizinga" <wiiliam.huizinga at> wrote:

>I've got an array.array of unsigned char and would like to make a slice
>of that array (e.g. a[0:4]) become one long like I would in "C" :
>	l = ((unsigned long *) (&a[0]))[0];
>I have been getting what I want in this sort of manner :
>	l  = 0L
>	l  = a[0]
>	l += a[1] <<  8
>	l += a[2] << 16
>	l += a[3] << 24
>but I think that's too wordy.  Is there a more intrinsic and elegant way 
>to do this?

Just for completeness, it's also possible to go back to basic style


from string  import hexdigits

def hexchr(i):  return hexdigits[i/16]+hexdigits[i%16] 
asc = dict([(chr(i), hexchr(i)) for i in range(256)])
def tolong(s): return long(''.join(map(asc.get,s[::-1])),16)

def test():
    a = '\x01\x02\x03\x04'
    print tolong(a)
if __name__=='__main__':

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