Convert hex-string to integer with python

Question

Note that the problem is not hex to decimal but a string of hex values to integer.

Say I've got a sting from a hexdump (eg. '6c 02 00 00') so i need to convert that into actual hex first, and then get the integer it represents... (this particular one would be 620 as an int16 and int32)

I tried a lot of things but confused myself more. Is there a quick way to do such a conversion in python (preferably 3.x)?

Answer 1

Not only that is a string, but it is in little endian order - meanng that just removing the spaces, and using int(xx, 16) call will work. Neither does it have the actual byte values as 4 arbitrary 0-255 numbers (in which case struct.unpack would work).

I think a nice approach is to swap the components back into "human readable" order, and use the int call - thus:

number = int("".join("6c 02 00 00".split()[::-1]), 16)

What happens there: the first part of th expession is the split - it breaks the string at the spaces, and provides a list with four strings, two digits in each. The [::-1] special slice goes next - it means roughly "provide me a subset of elements from the former sequence, starting at the edges, and going back 1 element at a time" - which is a common Python idiom to reverse any sequence.

This reversed sequence is used in the call to "".join(...) - which basically uses the empty string as a concatenator to every element on the sequence - the result of the this call is "0000026c". With this value, we just call Python's int class which accepts a secondary optional paramter denoting the base that should be used to interpret the number denoted in the first argument.

>>> int("".join("6c 02 00 00".split()[::-1]), 16)
620

Another option, is to cummulatively add the conversion of each 2 digits, properly shifted to their weight according to their position - this can also be done in a single expression using reduce, though a 4 line Python for loop would be more readable:

>>> from functools import reduce #not needed in Python2.x
>>> reduce(lambda x, y: x  + (int(y[1], 16)<<(8 * y[0]) ), enumerate("6c 02 00 00".split()), 0)
620

update The OP just said he does not actually have the "spaces" in the string - in that case, one can use just abotu the same methods, but taking each two digits instead of the split() call:

reduce(lambda x, y: x  + (int(y[1], 16)<<(8 * y[0]//2) ), ((i, a[i:i+2]) for i in range(0, len(a), 2)) , 0)

(where a is the variable with your digits, of course) - Or, convert it to an actual 4 byte number in memory, usign the hex codec, and unpack the number with struct - this may be more semantic correct for your code:

import codecs
import struct
struct.unpack("<I", codecs.decode("6c020000", "hex") )[0]

So the approach here is to pass each 2 digits to an actual byte in memory in a bytes object returned by the codecs.decode call, and struct to read the 4 bytes in the buffer as a single 32bit integer.

Answer 2

You can use unhexlify() to convert the hex string to its binary form, and then use struct.unpack() to decode the little endian value into an int:

>>> from struct import unpack
>>> from binascii import unhexlify
>>> n = unpack('<i', unhexlify('6c 02 00 00'.replace(' ','')))[0]
>>> n

The format string '<i' means little endian signed integer. You can substitute with '<I' or '<L' for unsigned int or long (both 4 bytes).

If the data does not contain spaces this simplifies to

>>> n = unpack('<i', unhexlify('6c020000'))[0]