Using six-bit one's and two's complement representation I am trying to solve the following problem:
12 - 7
Now, i take 12 in binary and 7 in binary first.
12 = 001100 - 6 bit
7 = 000111 - 6 bit
Then, would I then flip the bit for two's complement and add one?
12 = 110011 ones complement
+ 1
-------
001101
7 = 111000 ones complement
+ 1
---------
111001
then, add those two complement together
001101
+111001
-------
1000110 = overflow? discard the last digit? If so I get 5
Now, if I have a number like
-15 + 2
I would then add a sign magnitude on the MSB if it's a zero?
like:
-15 = 001111 6 bit
Would I add a 1 at the end here before I flip the bits?
= 101111
Using two's complement to represent negative values has the benefit that subtraction and addition are the same. In your case, you can think of 12 - 7
as 12 + (-7)
. Hence you only need to find the two's complement representation of -7 and add it to +12:
12 001100
-7 111001 -- to get this, invert all bits of 7 (000111) and add 1
----------
5 1000101
Then discard the carry (indicates overflow), and you have your result: 000101
which equals to 5 as expected.
For your example of -15 + 2
, simply follow the same procedure to get the two's complement representation of -15:
15 001111
110000 -- inverted bits
110001 -- add 1
Now do the addition as usual:
-15 110001
2 000010
-----------
res 110011
To see that res
indeed equals -13, you can see that it is negative (MSB set). For the magnitude, convert to positive (invert bits, add 1):
res 110011
001100 -- inverted bits
001101 -- add 1
Hence the magnitude is 13 as expected.