Why is 128 in one and two's complement using 8 bits overflow?

Question

Suppose I want to represent 128 in one and two's complement using 8 bits, no sign bit

Wouldn't that be:

One's complement: 0111 1111

Two's complement: 0111 1110

No overflow

But the correct answer is:

One's complement: 0111 1111

Two's complement: 0111 1111

Overflow

Additional Question:

How come 1 in one's and two's complement is 0000 0001 and 0000 0001 respectively. How come you don't flip the bits like we did with 128?

128 is 10000000, which in 8-bit twos complement is -128. 01111111 is 127. — Lee Daniel Crocker, Nov 07 '17 at 18:46
128 is not representable in an 8-bit one's or two's compliment format. — Christian Gibbons, Nov 07 '17 at 18:46
@ChristianGibbons but, if my understanding is correct, -128 IS representable, right? — Daniel, May 27 '20 at 08:09
yes, right, because the 1's complement has a -0, which 2's complement doesn't have. — Daniel, May 27 '20 at 13:04

score 4 · Accepted Answer · answered Nov 07 '17 at 19:27

One's and Two's Complements are both ways to represent signed integers.

For One's Complement Representation:

Positive Numbers: Represented with its regular binary representation
- For example: decimal value 1 will be represented in 8 bit One's Complement as 0000 0001
Negative Numbers: Represented by complementing the binary representation of its magnitude
- For example: decimal value of -127 will be represented in 8 bit One's Complement as 1000 0000 because the binary representation of 127 is 0111 1111 when complemented that will be 1000 0000

For Two's Complement Representation:

Positive Numbers: Represented with its regular binary representation
- For example: decimal value 1 will be represented in 8 bit One's Complement as 0000 0001
Negative Numbers: Represented by complementing the binary representation of its magnitude then adding 1 to the value
- For example: decimal value of -127 will be represented in 8 bit One's Complement as 1000 0001 because the binary representation of 127 is 0111 1111 when complemented that will be 1000 0000 then add 0000 0001 to get 1000 0001

Therefore, 128 overflows in both instances because the binary representation of 128 is 1000 0000 which in ones complement represents -127 and in twos complement represents -128. In order to be able to represent 128 in both ones and twos complement you would need 9 bits and it would be represented as 0 1000 0000.

Wait so how come in one's compliment 128 is 0111 1111? With the logic above shouldn't it be 1000 0000? — James Mitchell, Nov 07 '17 at 19:36
@JamesMitchell In one's complement, 0111 1111 is 127, ***not*** 128. 1000 0000 in 8-bit one's complement is -127. — 8bittree, Nov 07 '17 at 19:48
@JamesMitchell Since 127 is the highest magnitude that can be represented, the correct answer you have is just showing that 0111 1111 is the highest magnitude you can represent with 8 bit ones/twos complement and that in order to represent 128 it would overflow. — cass-m-b, Nov 07 '17 at 20:01

score 3 · Answer 2 · edited Nov 07 '17 at 20:54

In 8-bit unsigned, 128 is 1000 0000. In 8-bit two's complement, that binary sequence is interpreted as -128. There is no representation for 128 in 8-bit two's complement.

0111 1110 is 126.

As mentioned in a comment, 0111 1111 is 127.

See https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html.

Both two's complement and one's complement are ways to represent negative numbers. Positive numbers are simply binary numbers; there is no complementing involved.

I worked on one computer with one's complement arithmetic (LINC). I vastly prefer two's complement, as there is only one representation for zero. The disadvantage to two's complement is that there is one value (-128, for 8-bit numbers) that can't be negated -- causing the overflow you're asking about. One's complement doesn't have that issue.

Why is 128 in one and two's complement using 8 bits overflow?

2 Answers2