I have generated a random 16 byte string. It looks like this:
b'\xb68 \xe9L\xbd\x97\xe0\xd6Q\x91c\t\xc3z\\'
I want to convert this to a (positive) integer. What's the best way to do this in Python?
I appreciate the help.
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What's its intended endianness?kindall– kindall2016-10-23 21:02:26 +00:00Commented Oct 23, 2016 at 21:02
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Possibly related: stackoverflow.com/questions/444591/…Logan– Logan2016-10-23 21:04:57 +00:00Commented Oct 23, 2016 at 21:04
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@Logan I saw that post too, but I don't understand why to use struct? That can't be the solution.awaelchli– awaelchli2016-10-23 21:06:00 +00:00Commented Oct 23, 2016 at 21:06
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@kindall Little Endianawaelchli– awaelchli2016-10-23 21:10:30 +00:00Commented Oct 23, 2016 at 21:10
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2 Answers
In Python 3.2+, you can use int.from_bytes():
>>> int.from_bytes(b'\xb68 \xe9L\xbd\x97\xe0\xd6Q\x91c\t\xc3z\\', byteorder='little')
122926391642694380673571917327050487990
You can also use 'big' byteorder:
>>> int.from_bytes(b'\xb68 \xe9L\xbd\x97\xe0\xd6Q\x91c\t\xc3z\\', byteorder='big')
242210931377951886843917078789492013660
You can also specify if you want to use two's complement representation. For more info: https://docs.python.org/3/library/stdtypes.html
4 Comments
awaelchli
Oh I think that's exactly what I need. If my data is random, I think it does not matter if I use little or big endian, it will just shuffle the number, right?
Logan
I guess it would depend on what exactly you're planning on doing with the integer, but yes, the number will be different if you change the byteorder.
awaelchli
What is the command to go back to the original string from that number?
Logan
Looks like you could use int.to_bytes(): docs.python.org/3/library/stdtypes.html#int.to_bytes
A solution compatible both with Python 2 and Python 3 is to use struct.unpack:
import struct
n = b'\xb68 \xe9L\xbd\x97\xe0\xd6Q\x91c\t\xc3z\\'
m = struct.unpack("<QQ", n)
res = (m[0] << 64) | m[1]
print(res)
Result: 298534947350364316483256053893818307030L
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