ENGSCI 232 Computer SystemsLecture 4: Information Representation in

Computers

The basic unit of information on modern computers is the bit.A bit may assume one of two possible values: on/off; 0/1; true/false; 0V/5V, …Because there are only two possible values, a single bit cannot represent anything more complicated than a Boolean (logical) variable.

If we combine bits, we can represent a greater variety of values. 1 bit can assume one of 2 possible values {0, 1}. 2 bits can assume one of … possible values {00, 01, 10, 11}. 3 bits can assume one of … possible values {000, 001, 010, 011, 100, 101, 110, 111}. 4 bits - termed a …………… - can assume one of …… possible values {0000, 0001, …,

1111} 8 bits - termed a ………… - can assume one of ……… possible values {0000 0000, 0000

0001, …, 11111111}. 16 bits - often termed a 16-bit………… - can assume one of ……… possible values

{00000000 00000000, 00000000 00000001, …, 11111111 11111111}.

n 2n

1 2248

1632 429496729664 18446744073709551616

128 340282366920938463463374607431768211456256 115792089237316195423570985008687907853…

…269984665640564039457584007913129639936

The computer’s memory consists of a large sequence of bytes, with data stored in successive bytes. (Each byte actually is given a numerical address to help manage the data.)

We have to choose what meaning to associate with each state a byte can represent. For example, if we see 01001011 sitting in some byte of a computer’s memory, what does it represent? It can, in fact represent many things, depending on the context.

Firstly, consider how we represent non-negative integers. These are also termed unsigned integers.

Base-k Representations of Nonnegative (Unsigned) integersRecall that the same concept may have different representations. The base (or radix) of a number system defines the range of possible values that a digit may have.

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In a base 10 (decimal) number system the possible digits are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Most humans are familiar with base 10.

In a base 2 (binary) number system the possible digits are {0, 1} In a base 8 (………..) number system the possible digits are {0, 1, …, In a base 16 (………..….………..) number system the possible digits are

{0, 1, …,

The general form of an n digit number represented in base k is

(bn-1 bn-2 … b2 b1 b0)k

where bi is the value of the digit in position i.

The value of this number is

value =

Examples:4210 = 4*101 +2*100

528 =

1010102 =

2A16 =

Notes: bn-1 is called the most significant digit, and b0 is called the least significant digit In base 2, bn-1 is called the most significant bit (msb), and b0 is called the least significant

bit (lsb). For example, 0101001010111

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Addition in Base kWe use the standard rules to add in base k, making sure we ‘carry’ whenever we get a value of k or larger.

Example:0001001012

+0011001002

= 2

01200021103

+00120021023

= 3

Converting from Decimal into Base kThe remainder method can be used to convert a base 10 integer to its equivalent base-k value.

Remainder Method:Let value = (cn-1 cn-2 … c2 c1 c0)10.

First divide value by k, the remainder is the least significant digit b0.Divide the result by k, the remainder is b1.Continue this process until the result is less than k, giving the most significant digit, bn-1.

Formally, we can put this into a VB algorithmi = 0while value <> 0

bi=value mod kvalue = value \ ki i+1

loop

Notes: In VB, x mod y gives the remainder when x is divided by y x \ y divides two numbers and returns an integer result, discarding any fractional

component

Example: What is the base 3 representation of 4210?42 /3 = ……, remainder , b0 = (least significant digit)

/3 = ……, remainder , b1 = /3 = ……, remainder , b2 = /3 = ……, remainder , b3 =

Thus 4210 =

Check:

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Example: What is the base 2 (binary) representation of 4210?Base= 2

value value \ 2 value mod 242 21 0 (lsb)  

Answer 4210=

Example: What is the base 16 (hexadecimal) representation of 4210?42 /16 = ……, remainder , b0 = (least significant digit)

/16 = ……, remainder , b1 = Thus 4210 =

Check:Common Hexadecimal & Binary ValuesAlmost all computers use base 2 at the hardware level, but programmers prefer to think in octal or, more commonly, in hexadecimal. Hexadecimal is particularly convenient as 1 hexadecimal digit allows 16 combinations, corresponding to a nibble, while 2 digits allow 256 different values, matching a byte. The 16 hexadecimal values are shown in the table.

Changing Base using ExcelThere are some useful Excel functions for converting numbers into a string representation in some other base:

BIN2DEC Converts a binary number to decimalBIN2HEX Converts a binary number to hexadecimalBIN2OCT Converts a binary number to octalDEC2BIN Converts a decimal number to binaryDEC2HEX Converts a decimal number to hexadecimalDEC2OCT Converts a decimal number to octalHEX2BIN Converts a hexadecimal number to binaryHEX2DEC Converts a hexadecimal number to decimalHEX2OCT Converts a hexadecimal number to octalOCT2BIN Converts an octal number to binaryOCT2DEC Converts an octal number to decimalOCT2HEX Converts an octal number to hexadecimal

Examples: Hex2Bin("2F")="101111", Dec2Bin(123)="1111011”

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Decimal Base 2 Base 16010 00002 016

110 00012 116

210 00102 216

310 00112 316

410 01002 416

510 01012 516

610 01102 616

710 01112 716

810 10002 816

910 10012 916

1010 10102 A16

1110 10112 B16

1210 11002 C16

1310 11012 D16

1410 11102 E16

1510 11112 F16

Counting from 0 to 15 in 3 bases

Non-Negative Integers, Hexadecimal and Overflow in Visual BasicOn computers nonnegative (unsigned) integers are represented by a fixed number of bits (typically 8, 16, 32, and/or 64). Thus, there are only a finite set of numbers that can be represented: With 8 bits, 0…255 (0016…FF16) can be represented; With 16 bits, 0…65535 (000016…FFFF16) can be represented;

VB offers the byte as an unsigned value between 0 and 255:

dim i as byte ' i can hold values between 0 and 255 inclusive

Note: If an operation on bytes has a result outside this range, VB will give an ‘overflow’ error. This error is sometimes called a range exception. In some languages, no error is generated, but the result is wrong.

Dim i As Byte, j As Byte, k As Byte

i = 200j = 200k = i + j ' gives a run-time “overflow” error

We can immediately see why this overflow happens if we look at the binary, and remember we only have 8 bits to store the result in:

+ 110010002 200+ 110010002 +200

= 1100100002 =400 Too large; we have ‘overflowed’ into a ninth bit

We can enter binary and hexadecimal values directly in VB as follows:

k = &HA2 ' set k=A2 base 16k = &O75 ' set k=75 base 8

A few other useful VB functions exist for converting decimal to hexadecimal or octal strings.

Dim s as strings = Hex(5) ' Returns 5.s = Hex(10) ' Returns A.s = Hex(459) ' Returns 1CB.s = Oct(4) ' Returns 4.s = Oct(8) ' Returns 10.s = Oct(459) ' Returns 713.

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Signed Integer RepresentationsIn many cases, we wish to represent both negative and positive integer values. In order to represent (signed) integer values we divide the range of available binary patterns into two approximately equal sections, one section to represent positive and the other to represent negative values. There are four standard methods for representing integers: Sign-and-magnitude; One’s complement and Two’s complement. Excess-valueIn most of these, positive integers are represented as if unsigned. These representations mainly differ in how they represent negative numbers.

Sign-and-magnitude for Signed Integers:By convention the most significant bit is set to 0 and 1 for positive and negative numbers, respectively.

Example+4210 = 00101010 2

and so-4210 = 2(Sign&Magnitude)

Under this representation, there are two representations of ....……., namely…In 8 bit binary sign and magnitude these two values are represented by

+0=00000000 and

-0=…

If we have n bits in our representation then, The most positive representable number using

sign-and-magnitude is … The most negative representable number using

sign-and-magnitude is …

Ones complement for Signed Integers:A one’s complement representation of a negative number is obtained from the n-bit binary representation of the positive number by complementing the bits, or replacing all of the 0s to 1s and all of the 1s to 0s.

Example +4210 = 00101010 2

and so-4210 = 2(OnesComplement)

Again, there are two representations of zero. For example, using 8 bit one’s complement;zero is represented

00000000 and

…

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If we have n bits in our representation then, The most positive representable one’s complement number is … The most negative representable one’s complement number is …

Twos Complement for Signed IntegersA two’s complement representation of a negative number is obtained from the n-bit binary representation of the positive number by complementing the bits, and then adding 1.

Example+4210 = 00101010 2

and so-4210 =

+ = 2(TwosComplement)

It is exactly the same to just subtract 1, and then reverse the bits. For example, 42-1= 41,

+4110 = 00101001 2

and so-4210 = 2(TwosComplement)

In twos complement, there is only one representation of 0, being (for 8 bits) 00000000.

If we have n bits in our representation then, The most positive representable two’s complement number is … The most negative representable two’s complement number is …

Two’s complement is the most widely used integer representation on modern computers.

Notes: In two’s complement, we call the most significant bit the ………… bit. If it is set, we

have a negative number. Regardless of the number of bits being used, -1 is given by …

Two’s complement addition and subtractionTwos complement is popular because addition of two’s complement numbers is simple; we can (just about) treat the numbers as they were unsigned as long as we discard any overflow.

Example: Use 8 bit two’s complement to add 4210 and 5610

001010102sComplement = 4210 + 00111000 2sComplement = 5610 = 2sComplement = 9810

Example: Use 8 bit two’s complement to add -4210 and 5610

110101102sComplement =-4210 + 00111000 2sComplement = 5610 = 2sComplement = 1210

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Subtraction involves negating then adding:

Example: Use 8 bit 2’s complement to subtract 5610 from 4210. Now,

56-1=55 =001101112, so

-56 =. So

001010102’s complement =   4210 + 2’s complement = –5610 = 2’s complement = –1410

We said earlier that we could add two’s complement numbers as if they were unsigned. The only exception to this is the correct detection of ….

Example:Consider the following 4 bit examples, and assume we ignore any 5th bit when interpreting the answer.

00012sComplement = +110 + 0100 2sComplement = +410 = 2sComplement = +510 4-bit value=

01112sComplement = +710 + 0111 2sComplement = +710 = 2sComplement = 1410 4-bit value=

11102sComplement = 210 + 1111 2sComplement = 1 10 = 2sComplement = 310 4-bit value=

10012sComplement = 710 + 1001 2sComplement = 7 10 = 2sComplement = -1410 4-bit value=

01112sComplement = +710 + 1111 2sComplement = 1 10 = 2sComplement = +610 4-bit value=

We test for overflow as follows. Let x3x2x1x0 + y3y2y1y0= z4z3z2z1z0 (all base 2) where z4 is a temporary 5th bit in the answer that will be discarded before the result is stored.

Overflow has occurred if

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Excess-value for Signed Integers (“Move the Zero”)For an n-bit binary number a fixed value of 2n–1 is added to the value to obtain its excess-value (or “excess-2n–1”) representation. Sometimes we use a more general “excess-y” representation, where y is some other value chosen to suit the problem

Example 1 (4 bits):If n=4 bits, 2n–1=…, so we use an excess-… representation. Under this scheme, the representation for x is simply the unsigned representation for x+8.

Example 2 (8 bits):If n=8 bits, 2n–1=……, so we use an excess-… representation. The representations are

+4210 +12810 = 17010

= 101010102(Excess128)

4210 +12810 = 86= 010101102(Excess128)

There is only one representation for zero, …

If we have n bits in our representation then, The most positive excess-2n–1 representable number is … The most negative excess-2n–1 representable number is …

Signed Integers in VBVB uses 2’s complement integers. The following types are available:

dim i as integer ' 2 byte integer, -32,768 to 32,767dim j as long ' 4 byte integer, -2,147,483,648 to 2,147,483,647

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Floating point RepresentationIn the decimal system, the decimal point indicates the start of negative powers of 10. 12.34 = 1*101+2*100+3*10-1+4*10-2

If we are using a system in base (ie the radix is ) the ‘radix point’ serves the same function: 101.1012 = 1*22 +0*21 +1*20 +1*2-1 +0*2-2 +1*2-3

= 410 + 110 + 0.510 + 0.12510

= 5.62510

A floating point representation allows a large range of numbers to be represented in a relatively small number of digits by separating the digits used for precision from the digits used for range.

Example: The following have 4 decimal digits of precision, but together cover a large range of values 1234 = 1.234 x 103

0.000000001234 = 1.234 x 10-9

12,340,000,000 = 1.234 x 109

Note that the same number can be represented in different ways. For example the following are all the same value 3.14159100

0.314159101, 31.415910-1 3141.5910-3

To avoid multiple representations of the same number floating point numbers are usually normalised so that there is only one nonzero digit to the left of the ‘radix’ point, called the leading digit.

In general, a normalised (non-zero) floating-point number will be represented using (–1)sd0·d1d2…dp–1e,

where s is the sign, d0·d1d2…dp–1 - termed the significand - has p significant digits each digit satisfies 0di< e, emineemax , is the exponent is the base (or radix)

Example: If = 10 (base 10) and p = 3, the number 0·1 is represented as … If = 2 (base 2) and p = 24, the decimal number 0·1 cannot be represented exactly but is

approximately 1·100110011001100110011012-4.

Formally, (–1)s d0·d1d2…dp-1e represents the value (–1)s (d0 + d1-1+d2-2 …+dp-1-(p-1)) e

Precision & Range allowed by a RepresentationThe range of numbers that can be represented is determined by the number of digits in the exponent (i.e. by emax ) and the base to which it is raised, while the precision is represented by the number of digits p in the significand and its base .

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Example: If =2 (base 2), p=3 digits in the significand, and the exponent ranges from emin=–1 to emax=2, and the numbers are normalised, then the range of values (most negative to most positive) we can store

is ... the smallest (closed to zero) positive normalised value we can store

is … The full list of normalised positive values that can be represented is

shown in the table and on the number line below.

0 1 2 3 4 5 6 7

Hidden Bits in Binary Representations for Floating Point Numbers For a significand represented as a binary number (=2) the

normalisation condition requires that the leading digit is always ..... There is thus no need to store the leading digit in a binary

significand. This bit is referred to as the .................. eg if we have 1.101101 x 22, for the significand we only need to store ...

IEEE 754 floating point standardThere are many ways to represent floating point numbers. In order to improve portability most computers use the IEEE 754 floating point standard. There are two primary formats: 32 bit single precision; and 64 bit double precision.

IEEE 754 Single precision floating point standardSingle precision consists of: a single sign bit, 0 for positive and 1 for negative; an 8 bit base-2 (=2) excess-127 exponent, with emin=–126 (stored as 12710-

12610=110=000000012) and emax=127 (stored as 12710+12710=25410=111111102). Note that stored values 000000002=010 and 111111112=25510 are reserved for special numbers

a 23 bit base-2 (=2) significand, with a hidden bit giving a precision of 24 bits (i.e. 1.d1d2…d23);

31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

significandexponent±

1 bit 8 bits 23 bitsNotes Single precision has 24 bits precision, equivalent to about 7.2 decimal digits. The largest representable non-infinite number is almost 221273.4028231038

The smallest representable non-zero normalised number is 12–1261.1754910–38

Denormalised numbers (eg 0.01x2-126) can be represented. There are two zeros, ±0. There are two infinities, ±. A NaN (not a number) is used for results from undefined operations such as …

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Representation Base 101.002 x2-1 0.51.012 x2-1 0.6251.102 x2-1 0.751.112 x2-1 0.8751.002 x20 11.012 x20 1.251.102 x20 1.51.112 x20 1.751.002 x21 21.012 x21 2.51.102 x21 31.112 x21 3.51.002 x22 41.012 x22 51.102 x22 61.112 x22 7

All positive values we can represent when =2 with a p=3 digit significand &

e=–1,0,1,2

Text RepresentationThe ASCII standard was developed in 1963. ASCII -- American Standard Code for Information Interchange -- permitted machines from different manufacturers to exchange data. ASCII consists of 128 binary values (0 to 127), each associated with a character or command. ASCII was developed a long time ago and now the non-printing characters are rarely used for their original purpose. ASCII was actually designed for use with teletypes and so the descriptions are somewhat obscure. If someone says they want your file in ASCII format, all this means is they want 'plain' text with no formatting such as different fonts, different font sizes, bold or underline. They want the raw ASCII format that any computer can understand. This is usually so they can easily import the file into their own applications without issues. Notepad.exe creates ASCII text, or in MS Word you can save a file as “text only”.

Non-Printing Characters   Printing Characters

Name Ctrlchar Dec Hex Char Dec Hex Char   Dec Hex Char   Dec Hex Char

null ctrl-@ 0 00 NUL 32 20 Space   64 40 @   96 60 `start of heading ctrl-A 1 01 SOH 33 21 !   65 41 A   97 61 astart of text ctrl-B 2 02 STX 34 22 "   66 42 B   98 62 bend of text ctrl-C 3 03 ETX 35 23 #   67 43 C   99 63 cend of transmit ctrl-D 4 04 EOT 36 24 $   68 44 D   100 64 denquiry ctrl-E 5 05 ENQ 37 25 %   69 45 E   101 65 eacknowledge ctrl-F 6 06 ACK 38 26 &   70 46 F   102 66 fbell ctrl-G 7 07 BEL   39 27 '   71 47 G   103 67 gbackspace ctrl-H 8 08 BS 40 28 (   72 48 H   104 68 hhorizontal tab ctrl-I 9 09 HT 41 29 )   73 49 I   105 69 iline feed ctrl-J 10 0A LF 42 2A *   74 4A J   106 6A jvertical tab ctrl-K 11 0B VT 43 2B +   75 4B K   107 6B kform feed ctrl-L 12 0C FF 44 2C ,   76 4C L   108 6C lcarriage feed ctrl-M 13 0D CR 45 2D -   77 4D M   109 6D mshift out ctrl-N 14 0E SO 46 2E .   78 4E N   110 6E nshift in ctrl-O 15 0F SI 47 2F /   79 4F O   111 6F odata line escape ctrl-P 16 10 DLE 48 30 0   80 50 P   112 70 pdevice control 1 ctrl-Q 17 11 DC1 49 31 1   81 51 Q   113 71 qdevice control 2 ctrl-R 18 12 DC2 50 32 2   82 52 R   114 72 rdevice control 3 ctrl-S 19 13 DC3 51 33 3   83 53 S   115 73 sdevice control 4 ctrl-T 20 14 DC4 52 34 4   84 54 T   116 74 tnegative acknowledge ctrl-U 21 15 NAK 53 35 5   85 55 U   117 75 usynchronous idel ctrl-V 22 16 SYN 54 36 6   86 56 V   118 76 vend of transmit block ctrl-W 23 17 ETB 55 37 7   87 57 W   119 77 wcancel ctrl-X 24 18 CAN 56 38 8   88 58 X   120 78 xend of medium ctrl-Y 25 19 EM 57 39 9   89 59 Y   121 79 ysubstitute ctrl-Z 26 1A SUB 58 3A :   90 5A Z   122 7A zescape ctrl-[ 27 1B ESC 59 3B ;   91 5B [   123 7B {file separator ctrl-\ 28 1C FS 60 3C <   92 5C \   124 7C |group separator ctrl-] 29 1D GS 61 3D =   93 5D ]   125 7D }record separator ctrl-^ 30 1E RS 62 3E >   94 5E ^   126 7E ~unit separator ctrl-_ 31 1F US 63 3F ?   95 5F _   127 7F DEL

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