CODEWITHSUNDEEP
binary
user3232476
user3232476

Reputation: 41

Problems with subtracting in binary aritmetics

I wonder how I implement the 2-complements method for binary-aritmetics when the number that is subtracted is less than the other number. For instance - If I want to subtract 13 from 12 ... i.e 12 -13 = -1.

I have tried the following steps:

1) 12 to binary using 8 bits (0000 1100)

2) -13 to binary using 2-complements:

a) 13 into binary using 8 bits: 0000 1101

b) invert: 1111 0010

c) add 1: 1111 0010 + 1 = 1111 0011

So if I calculated correctly the result (1111 0011) represents -13. So lets add this negative representation with 12:

  00001100 + 11110011 = 11111111

So I do not understand the result - only only ones:s (1111...)

This happens when the result is negative. Obviously I missed something.

Grateful for help!!!!

Upvotes: 2

Views: 228

Answers (1)

OnoSendai
OnoSendai

Reputation: 3970

Congratulations, you've just discovered the difference between signed/unsigned integer representations!

Let's take, for example, the Uint16 format (unsigned 16 bit). It can represent any integer value from 0 (binary 0000000000000000) to 65535 (binary (1111111111111111).

So, what happens when you add one to 65535? You'll need another Uint16 to express that value, and the first will be all zeroes again:

0000000000000001 0000000000000000

Now, the signed version of Uint16, Int16, may range from −32,768 to 32,767. Why? Because it may use its most significant bit to determine the signal.

Like this, for Int8 (signed integer, 8 bits), which happens to be exactly the format you're using on your example:

00000000 -> 0
00000001 -> 1
00000010 -> 2
[...]
01111110 -> 126
01111111 -> 127
10000000 -> -128 (MSB set - number is negative)
10000001 -> -127
10000010 -> -126
[...]
11111110 -> -2
11111111 -> -1 <- (This is your result!)

Notice that you're already taking that into consideration when you do step B - by inverting you're setting the MSB to 1, thus indicating that the value is negative.

What's happening to your calculations is that you'll reach a state of arithmetic overflow if you don't take into consideration the signal bit.

Here's some links that may help with your issue:

Bitwise Operator on positive and negative numbers

Signed number representations - Wikipedia

Integer (computer science) - Wikipedia

Upvotes: 2

Related Questions