You said:
What I wish to do is a + b = 10011100
My solution:
>>> a=0b01100001
>>> b=0b01100010
>>> bin((a | b) ^ 0b11111111)
'0b10011100'
And now, for the explanation:
You are asking for a NOR bitwise operation (http://en.wikipedia.org/wiki/NOR_gate if it's not obvious):
r = not (a or b)
Also, you can use De Morgan's law, that says that it's equivalent to:
r = (not a) and (not b)
In Python:
>>> bin((a ^ 0b11111111) & (b ^ 0b11111111))
'0b10011100'
You may also wonder what's that ^ 0b11111111
. Well, not a
is equivalent to a xor 1
and xor
is written ^
in python. I'd suggest you write down the logic table if you are not 100% convinced. So basically, ^ 0b11111111
changes the 0 to 1 and the 1 to 0.
The bin
function gives the binary representation of the number given as a parameter. The 0b
at the beginning of a number means that the number is given in base 2 (otherwise it's base 10).
Edit:
Initially, my first thought for this problem was:
bin(~(a|b))
But the result is '-0b1100100'
. This is because in Python the number are signed. But it is also possible to get the good result by only keeping the first byte:
>>> bin(~(a|b) & 0xff)
'0b10011100'
Edit 2:
I've just found that OP asked another question in order to better understand my answer. So, if you wonder why I used a XOR to do the NOT, see a good explanation here: https://stackoverflow.com/a/19203069/1787973
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