# A-level Computing/AQA/Fundamentals of data representation/ASCII and unicode

ASCII normally uses 8 bits (1 byte) to store each character. However, the 8th bit is used as a check digit, meaning that only 7 bits are available to store each character. This gives ASCII the ability to store a total of
ASCII values can take many forms:
* Numbers
* Letters (capitals and lower case are separate)
* Punctuation (?/|\£$ etc.)
* non-printing commands (enter, escape, F1)
Take a look at your keyboard and see how many different keys you have. The number should be 104 for a windows keyboard, or 101 for traditional keyboard. With the shift function valus (a, A; b, B etc.) and recognising that some keys have repeated functionality (two shift keys, the num pad). We roughly have 128 functions that a keyboard can perform.
| Binary | Dec | Hex | Abbr |
| 000 0000 | 0 | 00 | NUL |
| 000 0001 | 1 | 01 | SOH |
| 000 0010 | 2 | 02 | STX |
| 000 0011 | 3 | 03 | ETX |
| 000 0100 | 4 | 04 | EOT |
| 000 0101 | 5 | 05 | ENQ |
| 000 0110 | 6 | 06 | ACK |
| 000 0111 | 7 | 07 | BEL |
| 000 1000 | 8 | 08 | BS |
| 000 1001 | 9 | 09 | HT |
| 000 1010 | 10 | 0A | LF |
| 000 1011 | 11 | 0B | VT |
| 000 1100 | 12 | 0C | FF |
| 000 1101 | 13 | 0D | CR |
| 000 1110 | 14 | 0E | SO |
| 000 1111 | 15 | 0F | SI |
| 001 0000 | 16 | 10 | DLE |
| 001 0001 | 17 | 11 | DC1 |
| 001 0010 | 18 | 12 | DC2 |
| 001 0011 | 19 | 13 | DC3 |
| 001 0100 | 20 | 14 | DC4 |
| 001 0101 | 21 | 15 | NAK |
| 001 0110 | 22 | 16 | SYN |
| 001 0111 | 23 | 17 | ETB |
| 001 1000 | 24 | 18 | CAN |
| 001 1001 | 25 | 19 | EM |
| 001 1010 | 26 | 1A | SUB |
| 001 1011 | 27 | 1B | ESC |
| 001 1100 | 28 | 1C | FS |
| 001 1101 | 29 | 1D | GS |
| 001 1110 | 30 | 1E | RS |
| 001 1111 | 31 | 1F | US |
| 111 1111 | 127 | 7F | DEL |
| Binary | Dec | Hex | Glyph |
| 010 0000 | 32 | 20 | ? |
| 010 0001 | 33 | 21 | ! |
| 010 0010 | 34 | 22 | " |
| 010 0011 | 35 | 23 | # |
| 010 0100 | 36 | 24 | $ |
| 010 0101 | 37 | 25 | % |
| 010 0110 | 38 | 26 | & |
| 010 0111 | 39 | 27 | ' |
| 010 1000 | 40 | 28 | ( |
| 010 1001 | 41 | 29 | ) |
| 010 1010 | 42 | 2A | * |
| 010 1011 | 43 | 2B | + |
| 010 1100 | 44 | 2C | , |
| 010 1101 | 45 | 2D | - |
| 010 1110 | 46 | 2E | . |
| 010 1111 | 47 | 2F | / |
| 011 0000 | 48 | 30 | 0 |
| 011 0001 | 49 | 31 | 1 |
| 011 0010 | 50 | 32 | 2 |
| 011 0011 | 51 | 33 | 3 |
| 011 0100 | 52 | 34 | 4 |
| 011 0101 | 53 | 35 | 5 |
| 011 0110 | 54 | 36 | 6 |
| 011 0111 | 55 | 37 | 7 |
| 011 1000 | 56 | 38 | 8 |
| 011 1001 | 57 | 39 | 9 |
| 011 1010 | 58 | 3A | : |
| 011 1011 | 59 | 3B | ; |
| 011 1100 | 60 | 3C | < |
| 011 1101 | 61 | 3D | = |
| 011 1110 | 62 | 3E | > |
| 011 1111 | 63 | 3F | ? |
| Binary | Dec | Hex | Glyph |
| 100 0000 | 64 | 40 | @ |
| 100 0001 | 65 | 41 | A |
| 100 0010 | 66 | 42 | B |
| 100 0011 | 67 | 43 | C |
| 100 0100 | 68 | 44 | D |
| 100 0101 | 69 | 45 | E |
| 100 0110 | 70 | 46 | F |
| 100 0111 | 71 | 47 | G |
| 100 1000 | 72 | 48 | H |
| 100 1001 | 73 | 49 | I |
| 100 1010 | 74 | 4A | J |
| 100 1011 | 75 | 4B | K |
| 100 1100 | 76 | 4C | L |
| 100 1101 | 77 | 4D | M |
| 100 1110 | 78 | 4E | N |
| 100 1111 | 79 | 4F | O |
| 101 0000 | 80 | 50 | P |
| 101 0001 | 81 | 51 | Q |
| 101 0010 | 82 | 52 | R |
| 101 0011 | 83 | 53 | S |
| 101 0100 | 84 | 54 | T |
| 101 0101 | 85 | 55 | U |
| 101 0110 | 86 | 56 | V |
| 101 0111 | 87 | 57 | W |
| 101 1000 | 88 | 58 | X |
| 101 1001 | 89 | 59 | Y |
| 101 1010 | 90 | 5A | Z |
| 101 1011 | 91 | 5B | [ |
| 101 1100 | 92 | 5C | \ |
| 101 1101 | 93 | 5D | ] |
| 101 1110 | 94 | 5E | ^ |
| 101 1111 | 95 | 5F | _ |
| Binary | Dec | Hex | Glyph |
| 110 0000 | 96 | 60 | ` |
| 110 0001 | 97 | 61 | a |
| 110 0010 | 98 | 62 | b |
| 110 0011 | 99 | 63 | c |
| 110 0100 | 100 | 64 | d |
| 110 0101 | 101 | 65 | e |
| 110 0110 | 102 | 66 | f |
| 110 0111 | 103 | 67 | g |
| 110 1000 | 104 | 68 | h |
| 110 1001 | 105 | 69 | i |
| 110 1010 | 106 | 6A | j |
| 110 1011 | 107 | 6B | k |
| 110 1100 | 108 | 6C | l |
| 110 1101 | 109 | 6D | m |
| 110 1110 | 110 | 6E | n |
| 110 1111 | 111 | 6F | o |
| 111 0000 | 112 | 70 | p |
| 111 0001 | 113 | 71 | q |
| 111 0010 | 114 | 72 | r |
| 111 0011 | 115 | 73 | s |
| 111 0100 | 116 | 74 | t |
| 111 0101 | 117 | 75 | u |
| 111 0110 | 118 | 76 | v |
| 111 0111 | 119 | 77 | w |
| 111 1000 | 120 | 78 | x |
| 111 1001 | 121 | 79 | y |
| 111 1010 | 122 | 7A | z |
| 111 1011 | 123 | 7B | { |
| 111 1100 | 124 | 7C | | |
| 111 1101 | 125 | 7D | } |
| 111 1110 | 126 | 7E | ~ |
If you look carefully at the ASCII representation of each character you might notice some patterns. For example:
| Binary | Dec | Hex | Glyph |
| 110 0001 | 97 | 61 | a |
| 110 0010 | 98 | 62 | b |
| 110 0011 | 99 | 63 | c |
As you can see, a = 97, b = 98, c = 99. This means that if we are told what value a character is we can easily work out the value of subsequent or prior characters.
* Example: ASCII characters
* Without looking at the ASCII table above! If we are told that the ASCII value for the character '5' is 011 0101, what is the ASCII value for '8'.
* We know that '8' is three characters after '5', as 5,6,7,8. This means that the ASCII value of '8' will be three bigger than that for '5':
* 011 0101 ASCII '5'
* + 011
* 011 1000 ASCII '8'
* Checking above this is the correct value.
* If you are worried about making mistakes with binary addition, you can deal with the decimal numbers instead. Take the example where you are given the ASCII value of 'g', 110 0111, what is 'e'?
* We know that 'e' is two characters before 'g', as e, f, g. This means that the ASCII value of 'e' will be two smaller than that for 'g'.
* 64 32 16 8 4 2 1
* 1 1 0 0 1 1 1 = 103$_{10}$ = ASCII value of 'g'
* 103 - 2 = 101$_{10}$
* 64 32 16 8 4 2 1
* 1 1 0 0 1 0 1 = 101$_{10}$ = ASCII value of 'e'
* Exercise: ASCII
* Without using the crib table (you won't get it in the exam!) answer the following questions:
* The ASCII code for the letter 'Z' is 90(base10), what is the letter 'X' stored as
* Answer:
* 88 - as it is 2 characters down in the alphabet
* How many ASCII 'characters' does the following piece of text use:
* Hello Pete,
* ASCII rocks!
* Answer:
* 27 or 26. If you said 23 you'd be wrong because you must include the non-printing characters at the end of each line. Each end of line needs a EOL command, and a new line needs a carriage return (CR), making the text like so:
* Hello Pete,[EOL][CR]
* ASCII rocks![EOL]
For the Latin alphabet ASCII is generally fine, but what if you wanted to write something in Mandarin, or Hindi? We need another coding scheme!
* Extension: Coding ASCII
* You might have to use ASCII codes when reading from text files. To see what each ASCII code means we can use the folliwing function ChrW(x) which returns the ASCII code with denary value x. Try out the following code to see the first 128 characters. What is special about character 10?
* For x = 0 To 127
* Console.WriteLine("ASCII for " & x & " = " & ChrW(x))
* Next
* Console.ReadLine