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Is there an easy way to produce a one's complement in python?

For instance, if you take the hex value 0x9E, I need to convert it to 0x61.

I need to swap the binary 1's for 0's and 0's for 1's. It feels like this should be simple.

2 Answers 2

29

Just use the XOR operator ^ against 0xFF:

>>> hex(0x9E ^ 0xFF) '0x61' 

If you need to work with values larger than a byte, you could create the mask from the int.bit_length() method on your value:

>>> value = 0x9E >>> mask = (1 << value.bit_length()) - 1 >>> hex(value ^ mask) '0x61' >>> value = 0x9E9E >>> mask = (1 << value.bit_length()) - 1 >>> hex(value ^ mask) '0x6161' 
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4 Comments

OK I'm confused, why is ~x wrong? It's supposed to be the binary negation operator after all. I can't wrap my mind over the lack of signed/unsigned distinction when binary ops are concerned
@Kos: ~ literally returns -(x+1), which is great when dealing with signed values. So ~0x9E is -159, or -0x9f hex. For unsigned bitwise work, not so great..
OK I got this, Python's "conceptually infinite string of 1's in front of negative numbers" approach has messed with my head. Analogies to C bitwise ops won't work here :(
@aka.nice: Sure, that works (I used | bitwise or accidentally). You still need to create the mask though.
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Hah. just found out that python bin() return a string!

so we can have some fun at this!

for x in numbers: # numbers is a list of int b = bin(x) #print b # e.g. String 0b1011100101 b = b.replace('0', 'x') b = b.replace('1', '0') b = b.replace('x', '1') b = b.replace('1b', '0b') #print b # String 0b0100011010 print int(b, 2) # the decimal representation 

1 Comment

This just made it to the top five of my most-pragmatic-code list :)

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