Unpacking a hex value

[Bengt Richter]

On Sat, 18 May 2002 02:36:51 GMT, Matthew Diephouse <fokke_wulf at hotmail.com> wrote:

>The code is very much like the Perl code I submitted. It is within a 
>subroutine.
>
>def hex2bin(input)
>      output = hex( input )
>      output = pack("!l", output)
>      ...
>
> From this, I get the error "required argument is not an integer", as 
>previously stated. There's no other info in the Traceback that's 
>important. I've tried wrapping output with a call to the int() function, 
>but that doesn't work either.
>
>You did get me on the "!l" instead of "l!", but it didn't specify order 
>in the struct docs, so I was guessing.
>
 Notice that hex in perl and python effectively are inverses of each other:

 [ 1:47] C:\pywk>python -c "print hex(65539)"
 0x10003

 [ 1:47] C:\pywk>perl -e "print hex('0x10003')"
 65539

So if I understand right I would guess you need something like

 >>> from struct import pack
 >>> def hex2bin(input):
 ...     output = int( input, 16)       # makes 32-bit native int from hex string
 ...     output = pack("!l", output)    # takes native int and packs its bytes into 'network' order string
 ...     return output
 ...
 >>> hex2bin('12345678')
 '\x124Vx'
 >>> ['%02X' % ord(x) for x in hex2bin('12345678')]  # hex values of char seq generated
 ['12', '34', '56', '78']

Which you can see is the hex values of the characters of the above '\x124Vx' in order,
appropriate for stuffing a big-endian long to get the same numeric value:

 >>> [chr(int(x,16),) for x in ['12', '34', '56', '78']]
 ['\x12', '4', 'V', 'x']

Note that this is on NT/Pentium, so if we leave out the network '!' we get little endianness:

 >>> def hex2bin(input):
 ...     output = int( input, 16)
 ...     output = pack("l", output)
 ...     return output
 ...
 >>> hex2bin('12345678')
 'xV4\x12'
 >>> ['%02X' % ord(x) for x in hex2bin('12345678')]
 ['78', '56', '34', '12']
 >>> [chr(int(x,16),) for x in ['78', '56', '34', '12']]
 ['x', 'V', '4', '\x12']

Regards,
Bengt Richter