How to do a bitwise NOR Gate in Python (editing python maths to work for me)

Question

Say I was to write this:

a=01100001 
b=01100010 
c=01100011 
d=01100100 
e=01100101 

each letter resembles the given numbers now how would

Answer 1

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