oops! I'm sorry. I thought i was posting this in coders corner! my bad!

i'm not sure but i think -3(101)

we use this system because we can use the exact same set of logic gates to add two normal binary integers, as we can to add two two's comlement binary integers. without any overhead such as 'if first bit is 1, then do this instead because it is negative' it can simply be treat as any other positive integer, and due to overflow's ends up giving the correct answer

example: 4 bit signed integers:

6 = 0110. -6 = !6+1 = 1001+1 = 1010

now if we add them together we get

0110+1010 = (1)0000. since the 1 is lost in overflow we get 0. the correct result

another example:

5 = 0100. -7 = 1001

add them together you get 1101 which is -2. the correct result.

Last edited by luca-deltodesco (2008-01-18 02:11:29)

also note that this doesn't just work for binary numbers. if we define the negative integer as k-x where k is the number 1 followed by 'n' 0's where n is how many bits the number is. i.e. -0101 = 10000-0101 = 1011.

then lets do it in decimal.

-0485 = 10000-0485 = 9515.
0785+9515 = (1)0300 = 300. the correct result for 785-485.

Wow! look at the response this threads getting!

Ricky wrote:

After a great amount of confusion i've hypothesized that they use the regular binary system to store positive values, and two's complement numbers for negative values.

Two's complement is "regular binary" for positive values.  I'm not certain if that's what you were trying to say.

by 'regular binary' i meant place value, as in 1010 = 1*2^3 + 0*2^2 + 1*2^1 + 0*^0 = 12d

What i really think my book is doing is using the ordinary binary system for positive numbers (as in, 1010 = 12d) and using two's complement for negative numbers. I was actually able to prove that there is no overlap so there is no ambiguity between the two.

we are using an online text for this course, and its accesible from anywhere:
http://webster.cs.ucr.edu/AoA/DOS/pdf/0_AoAPDF.html

if you click on chapter 1 and go to the pages 13 of the pdf (not of the actual page numbers) you can find the part on signed numbers and on the next page they do positive to negative conversions.

The reason i think they're using 'place value binary' for positive numbers and 'two's complement' for negatives is because supposedly, you can calculate a numbers negative by using NOT and adding 1. They begin with 5 written as 0101 and then write -5 as 1011. 1011 = 11d which by my understanding, is how we write positive 5 in two's complement form.

also, that pdf never actually told me what two's complement meant and i'm currently under the impression that if x is the value you want to represent (who's binary form has n bits) then
x + complement = 2^n
and we write x in two's complement form by writing 'complement' in binary form.


Thats my understanding of the subject. Care to point out where i'm going wrong, if anywhere?

Last edited by mikau (2008-01-18 12:38:05)