4.5.4.1–4.5.4.2 - Unsigned binary arithmetic

Unsigned values and ranges

An unsigned integer uses every bit as a non-negative power-of-two contribution.

Unsigned binary represents non-negative integers. Signed binary uses a coding scheme that also represents negative integers, so the same bits may mean a different value. For example, $1111$ is 15 as a four-bit unsigned integer; it does not have the same meaning in the signed two's-complement scheme taught next.

Convert unsigned binary by adding its positive place values: $10110_2=16+4+2=22_{10}$. To encode 22 in eight bits, pad with leading zeros: $00010110_2$. The number of available bits is part of the representation, even though padding does not change an unsigned value.

The minimum is the all-zero pattern. The maximum is the all-one pattern because every positive place value contributes. Five bits have values 0–31, not 0–32: counting both endpoints gives 32 values. Eight bits have values 0–255. A desired value of 256 needs a ninth bit because it is $100000000_2$.

Binary addition and carries

Binary addition follows place-value addition, but a total of two in one column becomes zero with a carry to the next column.

The column rules are $0+0=0$, $0+1=1$, $1+0=1$ and $1+1=10_2$. With an incoming carry, $1+1+1=11_2$: write 1 and carry 1. Align the rightmost bits and work from right to left.

Consider $1011_2+0110_2$. The rightmost column is $1+0=1$. The next is $1+1=10_2$, so write 0 and carry 1. The next is $0+1+1=10_2$, so write 0 and carry 1. The leftmost original column is $1+0+1=10_2$, so write 0 and carry a new leading 1. The full sum is $10001_2$. The decimal check is $11+6=17$.

If only four result bits are available, 17 exceeds the maximum of 15. This is unsigned overflow: the correct mathematical result needs more bits than the field provides. Retaining only the low four bits would give $0001_2$, or 1, which is not the sum. State the required width and whether overflow occurs; a carry beyond the available most-significant bit detects unsigned addition overflow.

A leading carry does not automatically mean an incorrect calculation: it can be the correct extra bit of a wider answer. For example, a five-bit destination can store the full result 10001 without overflow.

Binary multiplication and result width

Multiplication builds a sum of shifted copies of the multiplicand, one for each 1 bit in the multiplier.

Multiplying by 0 produces zero; multiplying by 1 preserves the value. Multiplying a non-negative integer by $10_2$, or two, moves its digits one position left and appends zero. This is a place-value multiplication only when the result width is sufficient: dropping the leading bit can change the stored answer.

For $101_2\times11_2$, the multiplier contains a 1 in its units position and a 1 in its twos position. The partial products are $00101_2$ and $01010_2$. Adding them gives $01111_2=15_{10}$, matching $5\times3$. A zero multiplier bit contributes a zero row, not a shifted copy with value added.

For $110_2\times101_2$, the partial products are $000110_2$ for the units bit, zero for the twos bit, and $011000_2$ for the fours bit. Their sum is $011110_2=30_{10}$, matching $6\times5$. The full product fits five unsigned bits, since the maximum is 31, but not four bits, whose maximum is 15.

Multiplying an $a$-bit unsigned value by a $b$-bit unsigned value may require up to $a+b$ bits for the full product. The largest product is $(2^a-1)(2^b-1)<2^{a+b}$. A particular smaller product may need fewer bits. If the destination is fixed, compare the actual full product with its maximum rather than assuming the operand width will be sufficient.