09-11-25 Lab 5

Class: CSCE-312

Notes:

How does 2's complement came about

We need a way to show the sign bit for negative binary numbers

• 0000 -> 0
• 1111 -> 15

0000 -> 0
0001
0010
0011 -> +3
0100
0101
0110
0111

• If you want to show the negative numbers, just change the first byte from 0 to 1

1000 -> 0
1001
1010
1011 -> -3
1100
1101
1110 -> -6
1111

• This system has a couple of problems
• You now have two versions of zeros, 0 and -0
• You are losing one number

Another method they came up with is called 1s complement

• If you want a positive number and you want to change it to a negative version, just invert all the bits

1111 -> 0
1110
1101
1100 -> -3
1011
1010
1001
1000

• One beautiful finding

Lets add two numbers together

  0011
+ 1100
  ----
  1111 -> -0

Another example

  1
  0101 -> +5
+ 1100 -> -3
  ----
  0001 -> -0 + 1 + 1 = 2

• The finding was that if you add one to the thing after doing its complement, then you actually get the number you want.

So we have the following form of representing it:

x(1's complement + 1) = -x

• Also called 2's complement

Getting the 2's complement of numbers

+0  0000     0000  -0
+1	0001     1111  -1
+2	0010     1110  -2
+3	0011     1101  -3
+4	0100     1100  -4
+5	0101     1011  -5
+6	0110     1010  -6
+7	0111     1001  -7
			 1000  -8

• Now in twos complement we can show one more number as well
• The problem is solved

So as range we have:

Just remember that in 2's complement the bits by value are

-8 4 2 1
 -------
 0 0 0 0
 ...
 1 0 0 1 <- -8 + 1 = -7

Full adders and 2's complement

Lets say now we have 2 four bit numbers, how many full adders do we need?

• Each full adder is capable to manage 2 bits
• We will need 4 full adders

A - B -> (A) + (-B)

This is the exact circuit we use inside an ALU

			A3  B1    A2  B2    A1  B1     A0  B0
			 |   |-    |   |-    |   | -   |   |----- (XOR) Control
			-------   -------   -------   -------   |
			|     |   |     |   |     |   |     |   |
	out----	| FA4 | - | FA3 | - | FA2 | - | FA1 | --- +1
	|   	|     | | |     |   |     |   |     |
---XOR		------- | -------   -------   -------
	  |------- | -- |    |         |         |
			  sum3      sum2      sum1      sum0

• For B, we just get our control, Xor that with B, and pass it to each adder
• Now if we do A - B = A + (B') + 1
• This is the circuit we use to perform addition and subtraction with 2's complement
• The XOR is the overflow in case of any

Overflow

• The result you are seeing is not correct, you need more bits for it to show, you do not have enough bits
• In unsigned binary, it shows overflow

In 2's complement the story is a little bit different

n = 4    -8 -> +2

Lets say that we want to add:

+5   0101   
+2   0111
	 ----
	*1100  -> -4

• How can your computer understand that this result is not correct
  • There is a good indication:
    • There are both positive numbers because of their MSB, and the result is negative! how?! that could be a good indicator

Lets do another example:

+5   0101   
+7   1001
	 ----
	*0100  -> +4

• Again an overflow!
• What is the indication here?
  • That the two numbers are negative but the result is positive! that i impossible.
• If you have a way to store that bit elsewhere operations would not overflow, but sometimes there is no access to memory in order to store that extra overflow bit

To represent it in digital logic, take the output of the full adders circuit and XOR it with the carry, that is the overflow

Let's do an example where we do not see overflow

	   1
+5   0101   
-7   1001
	 ----
	 1110  -> -2 

• If you want to see this, just convert it to a positive number
• It should be +2 after your conversion check
• This is the correct result we want, there is no overflow
  • You can check that with XOR
    • The output carry is 0, the previous carry is 0
      • 0 XOR 0 = 0