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CS253 Bases

Bases

• We’re used to base 10, because we have that many toes. Really.
• There’s nothing intrinsically good or bad about base 10.
• Vulcans use base 12. Well, in my head-canon, they do.
• In base N, the digits are 0…N−1.
  • Base 8 has no digit 8, just as base 10 has no digit ⑩.
• Digits after 9 are usually letters: A…Z or a…z.

Writing integers (mathematics)

• When writing numbers in mathematics, they are by default in base 10.
• A subscript indicates an explicit base: 27₈ + 43₆ = 32₁₆

Writing integers (C++ source code)

0b… or 0B…

It’s binary, base 2.

0x… or 0x…

It’s hexadecimal (“hex”), base 16. “Hexa” is Greek, “decimal” is Latin. 😼

0digits

It’s octal (base 8). Plain 0 is octal, not decimal!

Otherwise

It’s decimal (base 10). “Decimal” ≠ “digits after the decimal point” here.

Write for readability: sizes in decimal, bitmasks in binary, octal, or hex.

cerr << 0b11 << ' ' << 0XfF << ' ' << 012 << ' ' << 34;

3 255 10 34

Place notation

Consider an ordinary, three-digit base 10 number. It has a hundreds place, a tens place, and a ones place. You know this so well that you’ve forgotten it.

So, 789₁₀ means:

• 7⋅10² + 8⋅10¹ + 9⋅10⁰
• 7⋅100 + 8⋅10 + 9⋅1
• 700 + 80 + 9
• 789

Place notation

Now, consider a base 8 number, such as 375₈. This means:

• 3⋅8² + 7⋅8¹ + 5⋅8⁰
• 3⋅64 + 7⋅8 + 5⋅1
• 192 + 56 + 5
• 253

Place notation

A binary number, such as 1 0110₂, means:

• 1⋅2⁴ + 0⋅2³ + 1⋅2² + 1⋅2¹ + 0⋅2⁰
• 1⋅16 + 0⋅8 + 1⋅4 + 1⋅2 + 0⋅1
• 16 + 0 + 4 + 2 + 0
• 22

Place notation

Of course, in practice, we wouldn’t bother writing down the terms with zero coefficients, since they contribute only zeroes to the sum.

A binary number, such as 1 0110₂, means:

• 1⋅2⁴ + 1⋅2² + 1⋅2¹
• 1⋅16 + 1⋅4 + 1⋅2
• 16 + 4 + 2
• 22

I’m thinking of a number…

• First we need a number for examples.
• How about 0644?
• The 0 in the beginning just means it is in octal.
• What is it though?
  • In C++, 0644 is an int, just as 42 is.

0644 in octal… okayyyyyy…

• What is it in different bases?
• So, first what does 644 mean in octal?
  • 6 in the 64’s place (64 = 8²)
  • 4 in the 8’s place (8 = 8¹)
  • 4 in the 1’s place (1 = 8⁰)

What is 0644 in decimal?

• Octal to decimal
• Not hard, we just need to do the math.
• It’s just arithmetic.
  • 644₈ = 6⋅8² + 4⋅8¹ + 4⋅8⁰
  • 644₈ = 6⋅64 + 4⋅8 + 4⋅1
  • 644₈ = 384 + 32 + 4
  • 644₈ = 420

What is 0644 in binary?

Hardest way is convert to decimal, then convert to binary.

• Hard way: Translate to decimal, then to octal.
  • 6⋅8² = 384
  • 4⋅8¹ = 32
  • 4⋅8⁰ = 4
  • Total = 420, or 256+128+32+4, or 1 1010 0100₂
• Easy Way: Break it into pieces of 3 each.
  • Why 3? 8¹ = 2³, that’s why!
  • One digit in octal is three digits in binary!
  • 6₈ = 110₂ and 4₈ = 100₂, so 0644₈ = 110 100 100₂

What is 0644 in hexadecimal?

Decimal	Hex
0	0
1	1
2	2
3	3
4	4
5	5
6	6
7	7
8	8
9	9
10	a
11	b
12	c
13	d
14	e
15	f

• Hard way: Translate to decimal (420), then to hex.
  • Hexadecimal is base 16, so the places are 4096, 256, 16, 1
  • (16³, 16², 16¹, 16⁰)
• How many of each?
  • 420 = 1⋅256 + 10⋅16 + 4⋅1
• So, it’s 1a4₁₆
  • or 0x1a4 in C++

What is 0644 in hexadecimal?

• Easy way:
  • We already know it is 1 1010 0100 in binary.
  • 16¹ = 2⁴
    • Each hex digit is 4 binary digits.
  • 1 1010 0100
  • 1₂ is easy: 1
  • 1010₂ is 8+2 = 10 = a
  • 0100₂ is 4 … so 4
  • 1a4₁₆

Why doesn’t the easy way always work?

It’s all about the ratios:

From	To	Ratio	Hard?
Octal	Binary	8 = 23	Easy
Hexadecimal	Binary	16 = 24	Easy
Hexadecimal	Octal	16 = 8log₈16 = 84/3	OK
Hexadecimal	Decimal	16 = 10log₁₀16 = 101.20412	Hard
Decimal	Octal	10 = 8log₈10 = 81.1073	Hard
Decimal	Binary	10 = 2log₂10 = 23.3219	Hard

Everything’s simple, unless stupid decimal (🖐🖑) is involved.

So, our number is:

1 1010 0100₂ or 644₈ or 420₁₀ or 1a4₁₆

• What’s the difference?!
  • To a human? It’s the difference between a number and a headache.
  • To the computer? Nothing!

The computer doesn’t care!

cout << 0b10000 << ' ' << 020 << ' ' << 16 << ' ' << 0x10 << '\n';

16 16 16 16

• It stores it all in binary.
  • They all look the same in binary.
  • They’re all just ints.
• In fact, even if it DID care, it can’t tell the difference anyway!
  • If a memory cell contains 0001 1011 0000 1011, was it binary, octal, decimal, or hex to begin with?

How does it know?

It doesn’t know—you have to tell it!

int x = 0b10101;
cout << x << '\n'
     << oct << x << '\n'
     << hex << x << '\n'
     << dec << x << '\n';

21
25
15
21

• Alas, there is no bin.
• I/O manipulators such as hex and oct are “sticky”; they persist until changed.