How to make a bitwise NOR gate

I'm trying to understand the code from an answer I received earlier today:

a=0b01100001 b=0b01100010 bin((a ^ 0b11111111) & (b ^ 0b11111111)) 

This is my understanding:

• bin means that the result will be in binary form.
• a is the processes going through the gate
• 0b means base 2 form

Could someone explain the rest? I am confused about 11111111. & is the and gate (confused why this separates the two). And how would you change this to work for any other gate, e.g. XOR, NAND, or...?

3 Answers

a ^ 0b11111111 #exclusive or's each bit in a with 1, inverting each bit >>> a=0b01100001 >>> bin(a ^ 0b11111111) '0b10011110' >>> bin((a ^ 0b11111111) & (b ^ 0b11111111)) '0b10011100' 

This is different than using the ~ operator since ~ returns a negative binary result.

>>> bin(~a & ~b) '-0b1100100 

The reason is the ~ operator inverts all bits used in representing the number, including the leading 0's that are not typically displayed, resulting in a 2's complement negative result. By using ^ and the 8 bit binary mask, only the first 8 bits are inverted.

Starting with the original answer, which explains how a NOR gate can be implemented using AND and NOT:

You are asking for a NOR bitwise operation:

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) 

The poster than translates that pseudo-code into the Python you posted. For some reason he used ^ 0b11111111 to do a binary NOT, rather than simply ~, which is what I would have chosen. If we switch (a ^ 0b11111111) to the simpler ~ then we get:

bin(~a & ~b) 

That expression is how you write "(not a) and (not b)" in Python. ~ means NOT and & means AND.

A binary NOT flips all of the bits in a number. 0 becomes 1 and 1 becomes 0. The direct way to do that is with ~. An indirect way to flip all the bits in a number is to XOR it with all 1 bits. That has the same effect, it's just longer to write.

Or actually, to be more precise, it has almost the same effect. ^ 0b11111111 flips the first eight bits of the number because there are eight 1's. Whereas ~ flips all of the bits. If you're interested in only the first 8 bits then you can add & 0b11111111, which truncates the results to 8 bits:

>>> bin((~a & ~b) & 0b11111111) '0b10011100' 

In my opinion this is better than the mysterious ^ 0b11111111.

The ^ is the XOR operator. XOR means Exclusive OR. One of the operand to ^ is a sequence of ones. This essentially means every bit in the other operand (viz., either a or b) will be flipped. Once the two individual XOR operations are done, their results are OR-ed.

Looking outside of bits, and bitwise operations, if you see it from the realm of logic operations, the code is in essence doing (! A ) ^ (! B) which per DeMorgan's law is identically the same as ! (A v B) which is the NOR operation.