97 lines
2.9 KiB
Typst
97 lines
2.9 KiB
Typst
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#import "@local/handout:0.1.0": *
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= Integers
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#definition()
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A _bit string_ is a string of binary digits. \
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In this handout, we'll denote bit strings with the prefix `0b`. \
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That is, $1010 =$ "one thousand and one," while $#text([`0b1001`]) = 2^3 + 2^0 = 9$
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#v(2mm)
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We will separate long bit strings with underscores for readability. \
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Underscores have no meaning: $#text([`0b1111_0000`]) = #text([`0b11110000`])$.
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#problem()
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What is the value of the following bit strings, if we interpret them as integers in base 2?
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- `0b0001_1010`
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- `0b0110_0001`
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#solution([
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- $#text([`0b0001_1010`]) = 2 + 8 + 16 = 26$
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- $#text([`0b0110_0001`]) = 1 + 32 + 64 = 95$
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])
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#v(1fr)
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#definition()
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We can interpret a bit string in any number of ways. \
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One such interpretation is the _unsigned integer_, or `uint` for short. \
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`uint`s allow us to represent positive (hence "unsigned") integers using 32-bit strings.
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#v(2mm)
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The value of a `uint` is simply its value as a binary number:
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- $#text([`0b00000000_00000000_00000000_00000000`]) = 0$
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- $#text([`0b00000000_00000000_00000000_00000011`]) = 3$
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- $#text([`0b00000000_00000000_00000000_00100000`]) = 32$
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- $#text([`0b00000000_00000000_00000000_10000010`]) = 130$
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#problem()
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What is the largest number we can represent with a 32-bit `uint`?
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#solution([
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$#text([`0b01111111_11111111_11111111_11111111`]) = 2^(31)$
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])
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#v(1fr)
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#pagebreak()
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#problem()
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Find the value of each of the following 32-bit unsigned integers:
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- `0b00000000_00000000_00000101_00111001`
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- `0b00000000_00000000_00000001_00101100`
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- `0b00000000_00000000_00000100_10110000`
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#hint([The third conversion is easy---look carefully at the second.])
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#solution([
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- $#text([`0b00000000_00000000_00000101_00111001`]) = 1337$
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- $#text([`0b00000000_00000000_00000001_00101100`]) = 300$
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- $#text([`0b00000000_00000000_00000010_01011000`]) = 1200$
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Notice that the third int is the second shifted left twice (i.e, multiplied by 4)
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])
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#v(1fr)
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#definition()
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In general, division of `uints` is nontrivial#footnote([One may use repeated subtraction, but that isn't efficient.]). \
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Division by powers of two, however, is incredibly easy: \
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To divide by two, all we need to do is shift the bits of our integer right.
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#v(2mm)
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For example, consider $#text[`0b0000_0110`] = 6$. \
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If we insert a zero at the left end of this bit string and delete the digit at the right \
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(thus "shifting" each bit right), we get `0b0000_0011`, which is 3. \
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#v(2mm)
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Of course, we loose the remainder when we left-shift an odd number: \
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$9 div 2 = 4$, since `0b0000_1001` shifted right is `0b0000_0100`.
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#problem()
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Right shifts are denoted by the `>>` symbol: \
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$#text[`00110`] #text[`>>`] n$ means "shift `0b0110` right $n$ times." \
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Find the value of the following:
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- $12 #text[`>>`] 1$
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- $27 #text[`>>`] 3$
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- $16 #text[`>>`] 8$
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#solution[
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- $12 #text[`>>`] 1 = 6$
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- $27 #text[`>>`] 3 = 3$
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- $16 #text[`>>`] 8 = 0$
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]
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#v(1fr)
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