139 lines
4.2 KiB
Markdown
139 lines
4.2 KiB
Markdown
---
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tags: [binary, binary-encoding]
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---
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# Signed and unsigned numbers
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## Summary
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- To represent negative integers in binary we use signed numbers._Signed binary_
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includes negative integers, _unsigned binary_ does not.
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- The principal methnod for encoding signed binary numbers is called **two's
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complement**.
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## Related notes
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In order to represent negative integers alonside positive integers a natural
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approach is to divide the available
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[encoding space / word length](Binary_encoding.md) into two subsets: one for
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representing non-negative integers and one for representing negative integers.
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The primary method for doing this is to use _two's complement_. This method
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makes signed numbers possible in a way that complicates the hardware
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implementation of the binary arithmetic operations as little as possible.
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## Two's complement
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Signed numbers can be implemented in binary in a number of ways. The differences
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come down to how you choose to encode the negative integers. A common method is
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to use "two's complement".
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> The two's complement of a given binary number is its negative equivalent
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For example the two's complement of $0101$ (binary 5) is $1011$. There is a
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simple algorithm at work to generate the complement for 4-bit number:
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1. Take the unsigned number, and flip the bits. In other words: invert the
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values, so $0$ becomes $1$ and $1$ becomes $0$.
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2. Add one
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To translate a signed number to an unsigned number you flip them back and still
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add one:
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### Formal expresssion: $2^n - x$
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The heuristic account of deriving two's complement above can be formalised as
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follows:
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> in a binary system that uses a word size of $n$ bits, the two's complement
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> binary code that represents negative $x$ is taken to be the code that
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> represents $2^n - x$
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Let's demonstrate this, again using -5 as the value we wish to encode.
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Applying the formula to a 4-bit system, negative 5 can be derived as follows:
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$$
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2^4 -5
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$$
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$$
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16 -5 = 11
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$$
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$$
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11 = 1011
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$$
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So basically we carry out the subtraction against the word length and then
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convert the decimal result to binary.
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We can also confirm the output by noting that when the binary form of the number
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and its negative are added the sum will be `0000` if we ignore the overflow bit:
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$$
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1011 + 0101 = 0000
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$$
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(This makes sense if we recall that earlier we derived the complement by
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inverting the bits.)
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### Advantages
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The chief advantage of the two's complement technique of signing numbers is that
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its circuit implementation is no different from the adding of two unsigned
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numbers. Once the signing algorithm is applied the addition can be passed
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through an [adder](Half_adder_and_full_adder.md) component without any special
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handling or additional hardware.
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Let's demonstrate this with the following addition:
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$$
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7 + -3 = 4
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$$
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First we convert $7$ to binary: $0111$.
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Then we convert $-3$ to unsigned binary, by converting $3$ to its two's
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complement
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$$
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0011 \rArr 1100 \rArr 1101
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$$
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Then we simply add the binary numbers regardless of whether they happen to be
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positive or negative integers in decimal:
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$$
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0111 \\
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1101 \\
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====\\
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0100
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$$
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Which is 4. This means the calculation above would be identical whether we were
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calculating $7 + -3$ or $7 + 13$.
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The ease by which we conduct signed arithmetic with standard hardware contrasts
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with alternative approaches to signing numbers. An example of another approach
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is **signed magnitude representation**. A basic implemetation of this would be
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to say that for a given bit-length (6, 16, 32...) if the
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[most significant bit](Half_adder_and_full_adder.md#binary-arithmetic) is a 0
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then the number is positive. If it is 1 then it is negative.
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This works but it requires extra complexity to in a system's design to account
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for the bit that has a special meaning. Adder components would need to be
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modified to account for it.
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## Considerations
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A limitation of signed numbers via two's complement is that it reduces the total
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informational capacity of a 4-bit number. Instead 16 permutations of bits giving
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you sixteen integers you instead have 8 integers and 8 of their negative
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equivalents. So if you wanted to represent integers greater than decimal 8 you
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would need to increase the bit length.
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