Ones' complement
| Bits | Unsigned value | Ones' complement value |
| 0111 1111 | 127  | 127  |
| 0111 1110 | 126  | 126  |
| 0000 0010 | 2  | 2  |
| 0000 0001 | 1  | 1  |
| 0000 0000 | 0  | 0  |
| 1111 1111 | 255  | −0  |
| 1111 1110 | 254  | −1  |
| 1111 1101 | 253  | −2  |
| 1000 0001 | 129  | −126  |
| 1000 0000 | 128  | −127  |
The ones' complement of a binary number is defined as the value obtained by inverting all the bits in the binary representation of the number. The ones' complement of the number then behaves like the negative of the original number in some arithmetic operations. To within a constant, the ones' complement behaves like the negative of the original number with binary addition. However, unlike two's complement, these numbers have not seen widespread use because of issues such as the offset of −1, that negating zero results in a distinct negative zero bit pattern, less simplicity with arithmetic borrowing, etc.
A ones' complement system or ones' complement arithmetic is a system in which negative numbers are represented by the inverse of the binary representations of their corresponding positive numbers. In such a system, a number is negated by computing its ones' complement. An N-bit ones' complement numeral system can only represent integers in the range − to 2N−1−1 while two's complement can express −2N−1 to 2N−1−1.
The ones' complement binary numeral system is characterized by the bit complement of any integer value being the arithmetic negative of the value. That is, inverting all of the bits of a number produces the same result as subtracting the value from 0.
Many early computers, including the CDC 6600, the LINC, the PDP-1, and the UNIVAC 1107, used ones' complement notation. Successors of the CDC 6600 continued to use ones' complement until the late 1980s, and the descendants of the UNIVAC 1107 still do, but the majority of modern computers use two's complement.
Number representation
Positive numbers are the same simple, binary system used by two's complement and sign-magnitude. Negative values are the bit complement of the corresponding positive value. The largest positive value is characterized by the sign bit being off and all other bits being on. The lowest negative value is characterized by the sign bit being 1, and all other bits being 0. The table below shows all possible values in a 4-bit system, from −7 to +7.+ −
0 0000 1111 — Note that both +0 and −0 return TRUE when tested for zero
1 0001 1110 — and FALSE when tested for non-zero.
2 0010 1101
3 0011 1100
4 0100 1011
5 0101 1010
6 0110 1001
7 0111 1000
Basics
Adding two values is straightforward. Simply align the values on the least significant bit and add, propagating any carry to the bit one position left. If the carry extends past the end of the word it is said to have "wrapped around", a condition called an "end-around carry". When this occurs, the bit must be added back in at the right-most bit. This phenomenon does not occur in two's complement arithmetic.0001 0110 22
+ 0000 0011 3
= =
0001 1001 25
Subtraction is similar, except that borrows, rather than carries, are propagated to the left. If the borrow extends past the end of the word it is said to have "wrapped around", a condition called an "end-around borrow". When this occurs, the bit must be subtracted from the right-most bit. This phenomenon does not occur in two's complement arithmetic.
0000 0110 6
− 0001 0011 19
= =
1 1111 0011 −12 —An end-around borrow is produced, and the sign bit of the intermediate result is 1.
− 0000 0001 1 —Subtract the end-around borrow from the result.
= =
1111 0010 −13 —The correct result
It is easy to demonstrate that the bit complement of a positive value is the negative magnitude of the positive value. The computation of 19 + 3 produces the same result as 19 − .
Add 3 to 19.
0001 0011 19
+ 0000 0011 3
= =
0001 0110 22
Subtract −3 from 19.
0001 0011 19
− 1111 1100 −3
= =
1 0001 0111 23 —An end-around borrow is produced.
− 0000 0001 1 —Subtract the end-around borrow from the result.
= =
0001 0110 22 —The correct result.Negative zero
Negative zero is the condition where all bits in a signed word are 1. This follows the ones' complement rules that a value is negative when the left-most bit is 1, and that a negative number is the bit complement of the number's magnitude. The value also behaves as zero when computing. Adding or subtracting negative zero to/from another value produces the original value.
Adding negative zero:
0001 0110 22
+ 1111 1111 −0
= =
1 0001 0101 21 An end-around carry is produced.
+ 0000 0001 1
= =
0001 0110 22 The correct result
Subtracting negative zero:
0001 0110 22
− 1111 1111 −0
= =
1 0001 0111 23 An end-around borrow is produced.
− 0000 0001 1
= =
0001 0110 22 The correct result
Negative zero is easily produced in a 1's complement adder. Simply add the positive and negative of the same magnitude.
0001 0110 22
+ 1110 1001 −22
= =
1111 1111 −0 Negative zero.
Although the math always produces the correct results, a side effect of negative zero is that software must test for negative zero.Avoiding negative zero
The generation of negative zero becomes a non-issue if addition is achieved with a complementing subtractor. The first operand is passed to the subtract unmodified, the second operand is complemented, and the subtraction generates the correct result, avoiding negative zero. The previous example added 22 and −22 and produced −0.
0001 0110 22 0001 0110 22 1110 1001 −22 1110 1001 −22
+ 1110 1001 −22 − 0001 0110 22 + 0001 0110 22 − 1110 1001 −22
= = but = = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
1 1111 0011 −12 —An end-around borrow is produced, and the sign bit of the intermediate result is 1.
− 0000 0001 1 —Subtract the end-around borrow from the result.
= =
1111 0010 −13 —The correct result
It is easy to demonstrate that the bit complement of a positive value is the negative magnitude of the positive value. The computation of 19 + 3 produces the same result as 19 − .
Add 3 to 19.
0001 0011 19
+ 0000 0011 3
= =
0001 0110 22
Subtract −3 from 19.
0001 0011 19
− 1111 1100 −3
= =
1 0001 0111 23 —An end-around borrow is produced.
− 0000 0001 1 —Subtract the end-around borrow from the result.
= =
0001 0110 22 —The correct result.Negative zero
Negative zero is the condition where all bits in a signed word are 1. This follows the ones' complement rules that a value is negative when the left-most bit is 1, and that a negative number is the bit complement of the number's magnitude. The value also behaves as zero when computing. Adding or subtracting negative zero to/from another value produces the original value.
Adding negative zero:
0001 0110 22
+ 1111 1111 −0
= =
1 0001 0101 21 An end-around carry is produced.
+ 0000 0001 1
= =
0001 0110 22 The correct result
Subtracting negative zero:
0001 0110 22
− 1111 1111 −0
= =
1 0001 0111 23 An end-around borrow is produced.
− 0000 0001 1
= =
0001 0110 22 The correct result
Negative zero is easily produced in a 1's complement adder. Simply add the positive and negative of the same magnitude.
0001 0110 22
+ 1110 1001 −22
= =
1111 1111 −0 Negative zero.
Although the math always produces the correct results, a side effect of negative zero is that software must test for negative zero.Avoiding negative zero
The generation of negative zero becomes a non-issue if addition is achieved with a complementing subtractor. The first operand is passed to the subtract unmodified, the second operand is complemented, and the subtraction generates the correct result, avoiding negative zero. The previous example added 22 and −22 and produced −0.
0001 0110 22 0001 0110 22 1110 1001 −22 1110 1001 −22
+ 1110 1001 −22 − 0001 0110 22 + 0001 0110 22 − 1110 1001 −22
= = but = = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
0001 0110 22
Subtract −3 from 19.
0001 0011 19
− 1111 1100 −3
= =
1 0001 0111 23 —An end-around borrow is produced.
− 0000 0001 1 —Subtract the end-around borrow from the result.
= =
0001 0110 22 —The correct result.Negative zero
Negative zero is the condition where all bits in a signed word are 1. This follows the ones' complement rules that a value is negative when the left-most bit is 1, and that a negative number is the bit complement of the number's magnitude. The value also behaves as zero when computing. Adding or subtracting negative zero to/from another value produces the original value.
Adding negative zero:
0001 0110 22
+ 1111 1111 −0
= =
1 0001 0101 21 An end-around carry is produced.
+ 0000 0001 1
= =
0001 0110 22 The correct result
Subtracting negative zero:
0001 0110 22
− 1111 1111 −0
= =
1 0001 0111 23 An end-around borrow is produced.
− 0000 0001 1
= =
0001 0110 22 The correct result
Negative zero is easily produced in a 1's complement adder. Simply add the positive and negative of the same magnitude.
0001 0110 22
+ 1110 1001 −22
= =
1111 1111 −0 Negative zero.
Although the math always produces the correct results, a side effect of negative zero is that software must test for negative zero.Avoiding negative zero
The generation of negative zero becomes a non-issue if addition is achieved with a complementing subtractor. The first operand is passed to the subtract unmodified, the second operand is complemented, and the subtraction generates the correct result, avoiding negative zero. The previous example added 22 and −22 and produced −0.
0001 0110 22 0001 0110 22 1110 1001 −22 1110 1001 −22
+ 1110 1001 −22 − 0001 0110 22 + 0001 0110 22 − 1110 1001 −22
= = but = = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
0001 0110 22 —The correct result.
Negative zero
Negative zero is the condition where all bits in a signed word are 1. This follows the ones' complement rules that a value is negative when the left-most bit is 1, and that a negative number is the bit complement of the number's magnitude. The value also behaves as zero when computing. Adding or subtracting negative zero to/from another value produces the original value.Adding negative zero:
0001 0110 22
+ 1111 1111 −0
= =
1 0001 0101 21 An end-around carry is produced.
+ 0000 0001 1
= =
0001 0110 22 The correct result
Subtracting negative zero:
0001 0110 22
− 1111 1111 −0
= =
1 0001 0111 23 An end-around borrow is produced.
− 0000 0001 1
= =
0001 0110 22 The correct result
Negative zero is easily produced in a 1's complement adder. Simply add the positive and negative of the same magnitude.
0001 0110 22
+ 1110 1001 −22
= =
1111 1111 −0 Negative zero.
Although the math always produces the correct results, a side effect of negative zero is that software must test for negative zero.Avoiding negative zero
The generation of negative zero becomes a non-issue if addition is achieved with a complementing subtractor. The first operand is passed to the subtract unmodified, the second operand is complemented, and the subtraction generates the correct result, avoiding negative zero. The previous example added 22 and −22 and produced −0.
0001 0110 22 0001 0110 22 1110 1001 −22 1110 1001 −22
+ 1110 1001 −22 − 0001 0110 22 + 0001 0110 22 − 1110 1001 −22
= = but = = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
0001 0110 22 The correct result
Subtracting negative zero:
0001 0110 22
− 1111 1111 −0
= =
1 0001 0111 23 An end-around borrow is produced.
− 0000 0001 1
= =
0001 0110 22 The correct result
Negative zero is easily produced in a 1's complement adder. Simply add the positive and negative of the same magnitude.
0001 0110 22
+ 1110 1001 −22
= =
1111 1111 −0 Negative zero.
Although the math always produces the correct results, a side effect of negative zero is that software must test for negative zero.Avoiding negative zero
The generation of negative zero becomes a non-issue if addition is achieved with a complementing subtractor. The first operand is passed to the subtract unmodified, the second operand is complemented, and the subtraction generates the correct result, avoiding negative zero. The previous example added 22 and −22 and produced −0.
0001 0110 22 0001 0110 22 1110 1001 −22 1110 1001 −22
+ 1110 1001 −22 − 0001 0110 22 + 0001 0110 22 − 1110 1001 −22
= = but = = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
0001 0110 22 The correct result
Negative zero is easily produced in a 1's complement adder. Simply add the positive and negative of the same magnitude.
0001 0110 22
+ 1110 1001 −22
= =
1111 1111 −0 Negative zero.
Although the math always produces the correct results, a side effect of negative zero is that software must test for negative zero.Avoiding negative zero
The generation of negative zero becomes a non-issue if addition is achieved with a complementing subtractor. The first operand is passed to the subtract unmodified, the second operand is complemented, and the subtraction generates the correct result, avoiding negative zero. The previous example added 22 and −22 and produced −0.
0001 0110 22 0001 0110 22 1110 1001 −22 1110 1001 −22
+ 1110 1001 −22 − 0001 0110 22 + 0001 0110 22 − 1110 1001 −22
= = but = = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
= = likewise, = but =
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
=
1111 1111 −0 0000 0000 0 1111 1111 −0 0000 0000 0
"Corner cases" arise when one or both operands are zero and/or negative zero.
0001 0010 18 0001 0010 18
− 0000 0000 0 − 1111 1111 −0
= = = =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
= =
0001 0010 18 1 0001 0011 19
− 0000 0001 1
= =
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
0001 0010 18
Subtracting +0 is trivial. If the second operand is negative zero it is inverted and the original value of the first operand is the result. Subtracting −0 is also trivial. The result can be only 1 of two cases. In case 1, operand 1 is −0 so the result is produced simply by subtracting 1 from 1 at every bit position. In case 2, the subtraction will generate a value that is 1 larger than operand 1 and an end-around borrow. Completing the borrow generates the same value as operand 1.
The next example shows what happens when both operands are plus or minus zero:
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
+ 0000 0000 0 + 1111 1111 −0 + 0000 0000 0 + 1111 1111 −0
= = = = = = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
= = = =
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
0000 0000 0 1111 1111 −0 1111 1111 −0 1 1111 1110 −1
+ 0000 0001 1
=
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
1111 1111 −0
0000 0000 0 0000 0000 0 1111 1111 −0 1111 1111 −0
− 1111 1111 −0 − 0000 0000 0 − 1111 1111 −0 − 0000 0000 0
= = = = = = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
= = = =
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1
= =
0000 0000 0
This example shows that of the 4 possible conditions when adding only ±0, an adder will produce −0 in three of them. A complementing subtractor will produce −0 only when both operands are −0.
1 0000 0001 1 0000 0000 0 0000 0000 0 1111 1111 −0
− 0000 0001 1