TU-Delft TI1400/11-PDS 1 Arithmetic and Data Representation iosup/Courses/2011_ti1400_3.ppt.

TU-DelftTI1400/11-PDS

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Arithmetic and Data Representation

http://www.pds.ewi.tudelft.nl/~iosup/Courses/2011_ti1400_3.ppt


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Lecture 1Making functions

time

A,B Y

A

BYADD

nand gates


Lecture 1Making functions

Circuit Diagramhttp://xkcd.com/730/


Lecture 2Programmable device

2,1 3ProgrammableDevice

input stream output stream

programREAD(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

• READ(X) means read next input value from input stream and store it internally as variable X

• WRITE(X) means put value in variable X on output stream• ADD(X,Y,Z) means assign value of X+Y to Z


Lecture 2Von Neumann Architecture

READ(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

X: 1Y: 2Z: 3 • •

TEMP_A: TEMP_B: RESULT:

IR:

PC:

arithmeticunit

Central Processing Unit

CONTROL

Memory

Input

Output


Problem: How to Represent and Use Data?

1. Representation2. Arithmetic3. Conversion


The Y2K Problem/The Millenium Bug (2000)Representation (Gone Wrong)• Abbreviating the year as two-digits instead of

four- 99 instead of 1999- 00 instead of 2000- Cost saving (space is money)- How to represent 2096?

98 < 99, but 99 < 00?!

• Problem identified by Spencer Bolles in 1985- Y2K acronym used by David Eddy in 1995

• $150B in US (IDC estimate)


The FDIV Problem (1993) Arithmetic (Gone Wrong)

• 4195835.0/3145727.0 = 1.333 820 449 136 241 002 5 4195835.0/3145727.0 = 1.333 739 068 902 037 589 4

• Pentium used the radix-4 SRT algorithm (which is fast)• “The [problem] can be traced to five missing entries in a

lookup table”• Solution: recall defective processors ($350M) or

scale operands and use 80-bit extended-precision internal Pentium format (formats are discussed in this lecture)

• … oh, and hardware errors still happen (latest, Intel’s Cougar/ Sandy Chipsets in February 1, 2011—est. cost $700M)


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Ariane 5 (1996)Conversion (Gone Wrong)

June 4, 1996: Ariane 501 Satellite Launch explodes 40s into the flight sequence

Cost: $370M“The internal SRI* software exception was

caused during execution of a data conversion from 64-bit floating point to 16-bit signed integer value. The floating point number which was converted had a value greater than what could be represented by a 16-bit signed integer.” Inquiry Board Report


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Data Type Representation

1. Integer numbers2. Real numbers3. Booleans4. Characters5. Composite types (e.g. arrays)6. Objects


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Number systems

0-9 8 0 11

277

-15

66 10110

01001110


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Question

• Which decimal number is associated with the bitstring <110> ?


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Definition and Representation

Integers (Z) = Set of natural numbers including 0 and their non-zero negatives

Integer numbers are generally represented by:• An alfabet • A string X of n elements from : Xn n

• Applying Xn to a valuation function F

- signature: F : n Z


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Examples

• R = {I, V, M, ....}- VI = 6- IV = 4

• 10 = {0, 1, 2, 3, .....,9}- 27710 = 2*102 + 7*101 + 7*100

• Many sets and valuation functions F possible

radix


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Question

What value has 267 ?


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Binary representation

• In computer, we have 2 ={0, 1}

• There are, however, many valuation functions F

277

-15

66

1110

0011

1010

F1

F2

F3


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Criteria for representations

• Let Xn = < xn-1, xn-2 , ....., x0 >• Range of natural numbers of X• Representation of 010

• Efficiency of implementation- Sign inversion : A -A- Addition: A+B- Extension: n bit m bit representation, m>n

X


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1.1. Sign-Magnitude1.2. Two’s Complement (2C)1.3. One’s Complement (1C)1.4. Excess-X


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1. Integer Representation

1.1. Sign Magnitude

• F: sign(xn-1){xn-22n-2 + ....+ x020}

sign(1) = - and sign(0)= +

• Range: [-(2n-1 -1), 2n-1 -1]• Zero: 100... and 000...• Inversion: change only sign bit

1 1 0 V = - 21 = -2

0 1 0 V = +21 = 2sign bit

X


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1.2. Two’s complement (2C)

• F: -xn-12n-1 + xn-22n-2 + ....+ x020

• Range: [-2n-1, 2n-1-1]• Zero: 000... (only a single representation)

1 1 0 V = - 22 +21 = -2

0 1 0 V = +21 = 2sign

X


010

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1. Integer Representation 1.2. 2C 2C negative

100 101 110 111 000 001 010 011-4 -3 -2 -1 0 1 2 3

Inversion rule:- invert all bits- add 1- General: A + Å = 2n

110 0011

+


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1.3. One’s Complement (1C)

• F: -xn-1(2n-1-1) + xn-22n-2 + ....+ x020

• Range: [-(2n-1-1), 2n-1-1]• Zero: 111... and 000...

1 1 0

0 1 0sign

X V = - 22 +1+21 = -1

V = +21 = 2


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1. Integer Representation 1.3. 1C 1C negative

100 101 110 111 000 001 010 011-3 -2 -1 0 0 1 2 3

Inversion rule:- invert all bits- General: A + Å = 2n-1

110 001


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1.4. Excess-x

• F: xn-12n-1 + xn-22n-2 + ....+ x020 - 2n-1

• Range: [-2n-1, 2n-1-1]• Zero: 10000.....

1 1 0

0 1 0

X V = 22 + 21 - 22 = 2

V = +21 - 22= -2


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1.5. Addition1.6. Sign Extension for 2C1.7. Other Representations1.8. Conversion


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1.5. Addition

• SM- Sign and Magnitude are processed separately- Opposite sign operands: subtraction

• 2C- All bits can be added, regardless of position

• 1C- All bits can be added, regardless of position- When carry at position 2n , add 1 to result


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1. Integer Representation 1.5. AdditionExample

X000001010011100101110111

2C0123-4-3-2-1

1C0123-3-2-10

SM01230-1-2-3

Ex-4-4-3-2-10123


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1.6. Sign Extension (for 2C)

• Represent n-bit in k-bit number• If k>n, extend sign bit to the left• If k<n,

- n-k+1 bits equal, remove lefmost n-k bits- if not, overflow

• Example: n=6, k=44 = 000100 -> 01009 = 001001 -> overflow-6 = 111010 -> 1010


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1. Integer Representation 1.7. Other RepresentationsBinary Coded Decimaldecimal

0123456789

X0000000100100011010001010110011110001001


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Radix 2, 8, 10, 16binair

01

1011

100101110111

10001001101010111100110111101111

octal01234567

1011121314151617

dec.0123456789

101112131415

hex0123456789ABCDEF

L33T-Speak

http://www.jayssite.com/stuff/l33t/l33t_translator.html

h4x0r

1007

1337

hacker

money (loot)

elite (leet)

1. Integer Representation 1.7. Other Representations


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1.8. Conversion [1/2]

• Simple, if one radix is power of other radix- binary to octal10100101001 = 010 100 101 001 = 2451oct

- binary to hexadecimal10100101001 = 0101 1010 1001 = 5A9hex =

0x05A9


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1.8. Conversion [2/2]• From binary to decimal

10110101001 = 210 + 28 + 27 + 25+ 23+ 20 = 1449

• From decimal to binary (first method)1449 = 210 +425 = 210 +28 +169 =..=

10110101001

• From decimal to binary (second method)1449 = 2*724 +1 724 = 2*362 +0 ...... =...... 1 = 0*1 +1

LSB


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1.9. Binary Addition1.10. Operation Overflow


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1. Integer Representation 1.9. Binary AdditionAddition of binary digits

• Addition of two bits

• Sum s and Carry c

000

+

101

+

011

+

1 1 10

+

x yc s

+


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1. Integer Representation 1.9. Binary AdditionMore positions

xi

yi

ci

si

ci+1

.

...

. .

.

xi

001

yi

011

ci

011

si

001

ci+1

011

truth table (partial)


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1. Integer Representation 1.9. Binary AdditionBasic Adder

Full Adder

xi yi

ci

si

ci+1


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1. Integer Representation 1.9. Binary AdditionRipple Carry Adder

FA FA FA


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1. Integer Representation 1.10. Overflow Addition/subtraction• 2C

- Simple, all bits can be treated the same- Carry last position can be ignored- Subtraction by first complementing the

subtrahend(minuend – subtrahend = difference)

• Example: 0111 (7) 1101 (-3)------- +10100 (4)


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1. Integer Representation 1.10. Overflow Addition/subtraction

• 1C- all bits can be treated the same- when carry in last position add +1 to

Least Significant Bit (LSB)

0111 (7) 1100 (-3)------- +10011 1-------- + 0100 (4)


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1. Integer Representation 1.10. Overflow Overflow detection

• Suppose we have 2C notation- Carry-out MSB is not sufficient to indicate

overflow

0111 (7) 0100 (4)-------- +01011 (-5?)


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1. Integer Representation 1.10. Overflow Overflow

• Overflow occurs in 2C iff- Both operand have the same sign- The sum bit in the MSB position has the opposite

sign

OVFL xn 1yn 1s n 1 x n 1y n 1sn 1


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2.1. Fixed Point2.2. Floating Point2.3. IEEE 754


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2.1. Fixed Point (Fractions)

• Real numbers (R) = value in a continuous space (vs. Integers—discrete space)

• F: -xn-120 + xn-22-1 + ....+ x 02-n+1

• Range: [-1, 1-2-n+1]• But: very small numbers cannot be represented,

…


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Question

Can conversion between decimal and binary numbers always be done exacly in

a. integers ?b. fractions ?


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Answer

a) Can conversion between decimal and binary numbers always be done exacly in integers?-> Yes

b) Can conversion between decimal and binary numbers always be done exacly in fractions?-> No- Counter-example: 0.210 =0.00110011001100.....2


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2.2. Floating Point (1)

• Need a represention for:- very large numbers- real numbers

• With a limited number of bits

• Floating point notation: G = m.re

r = radixm = mantissa, significant, or fractione = exponent


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2.2. Floating Point (2)

• r is normally 10 or 2• mantissa m is normalized: (1/r) ≤ m < 1

- e.g., if r = 2 then 0.5 ≤ m < 1- e.g., if r = 10 then

6.543 x 1010 is normalized, 654,300 x 105 is not

• Normalization is used to save as much as possible significant digits


2.2. Floating PointFixed or Floating Point?

• Metrics- Range = ratio largest:smallest number- Precision = how many significant bits for representation

• Floating point idea: separate the exponent from the mantissa. Vs Fixed point:- Wider range (through exponent)- Lower precision (mantissa has fewer bits)

• There’s no free lunch in computer science!


Birth of a Standard (IEEE 754)

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2.3. FP Representation/IEEE 754

• Need- Many competing hardware makers

IBM, Cray, CDC, Intel, …- Various range and accuracy

• Work Group initiated by Intel 1976, led by W. Kahan- Correct FP- Developers can write portable code AND prove correctness

• Result: IEEE 754

Read: IEEE 754: An Interview with William Kahan, IEEE Computer 1998


IEEE Floating point standard (IEEE 754)

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e’

ms

32 bit23 bit8 bit

value = ±1s x 2e’-127 x m



Mantissa

• Implicit 1 bit just left of the point, so 24-bit total

• Mantissa m is normalized: (1/r) ≤ m < 1- So, including implicit 1., 1 ≤ m’ < 2

value = ±1s x 2e’-127 x m’

m’=1.m, m=[01][01]…



Exponent

• e: [1,254] in excess-127 representation->[-126,127]• e=0 and e=255 are reserved values• Floating point (radix point is “floating”)

- e=35 => m+1 means +235 (+34 359 738 368)- e=0 => m+1 means +20 (+1)




Sign

• Sign is 0 for positive, 1 for negative• Floating-point numbers that in value differ only in

sign differ in IEEE 754 representation in the s bit



Recap


Lecture 2Von Neumann Architecture

READ(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

X: 1Y: 2Z: 3 • •

TEMP_A: TEMP_B: RESULT:

IR:

PC:

arithmeticunit

Central Processing Unit

CONTROL

Memory

Input

Output


60

Definition and Representation

Integers (Z) = Set of natural numbers including 0 and their non-zero negatives

Integer numbers are generally represented by:• An alfabet • A string X of n elements from : Xn n

• Applying Xn to a valuation function F

- signature: F : n Z


61

Criteria for representations

• Let Xn = < xn-1, xn-2 , ....., x0 >• Range of natural numbers of X• Representation of 010

• Efficiency of implementation- Sign inversion : A -A- Addition: A+B- Extension: n bit m bit representation, m>n

X


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1.1. Sign-Magnitude1.2. Two’s Complement (2C)1.3. One’s Complement (1C)1.4. Excess-X


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1. Integer Representation 1.9. Binary AdditionBasic Adder

Full Adder

xi yi

ci

si

ci+1


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2.1. Fixed Point (Fractions)

• Real numbers (R) = value in a continuous space (vs. Integers—discrete space)

• F: -xn-120 + xn-22-1 + ....+ x 02-n+1

• Range: [-1, 1-2-n+1]• But: very small numbers cannot be represented,

…


IEEE Floating point standard (IEEE 754)

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e’

ms

32 bit23 bit8 bit



Recap over


Memory Jog 1: Select the correct answer

1 0011 0010 1101 1011 1101 11012 is …1.321BE0010

2.123EBDD16

3.231DB9916

4.132DBDD16


Memory Jog 2: Select the correct answer

10510 is to 101012 as 07FD16 is to …1.40910

2.204510

3.90410

4.204511



An Example: Number to IEEE 754• Input: 8.75• Step 1 (binary fixed-point): +1000.11 (.75=1x2-1+1x2-2)• Step 2 (normalize): 1.0001100… x 23

• Step 3 (bias the exponent): 3 + 127 = 130 (10000010)

Source: C. Hecker, Let’s Go to the (Floating) Point, Game Developer, 1996



An Example: IEEE 754 to Number

• Step 1: Split into IEEE 754 components

• Step 2 (Sign): 0 is positive, 1 is negative. Sign is …?• Step 3 (exponent): 128 (10000000)-127 (bias)=1• Step 4 (mantissa): 1.01010000… = 1.3125

(1x2-1=.5 1x2-2 =.25 1x2-3 =.125 1x2-4 =.0625)• Step 5 (result): (-1)1 x 1.3125 x 21 = -2.625

1 10000000 01010000 00000000 0000000

110000000 01010000 00000000 0000000In:


Excercise 1: Select the correct answer6.510 in IEEE 754 Single Precision is …

1.0 10000100 10101000000000000000000

2.0 10000001 10100000000000000000000

3.1 01000010 00101010100000000000000

4.0 01000010 00101010100000000000002


Excercise 2: Select the correct answer

0 1000 0100 0101 0000 0000 0000 0000 000 is the IEEE 754 Single Precision for …1.3210

2.4216

3.4210

4.3216Hint: http://www.h-schmidt.net/FloatApplet/IEEE754.html


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FP exceptions [see wiki ieee 754]

0 00000000 00000000000000000000000 = +01 00000000 00000000000000000000000 = - 00 11111111 00000000000000000000000 = +Infinity1 11111111 00000000000000000000000 = - Infinity0 11111111 00000000000001000001000 = NaN

= ±1s x 2e’-126 x 0.fraction = ±1s x 2e’-127 x 1.fraction



Smallest Denormalized Positive Number


• Step 2 (Sign): 0 is positive, 1 is negative. Sign is +

• Step 3 (exponent): 0 (00000000)-126 (bias)=-126• Step 4 (mantissa): 0.00…1 = 2-23(23 bits)• Step 5 (result): (-1)0 x 2-23 x 2-126 = 2-149 ~ 10-45

0 00000000 00000000 00000000 0000001

000000000 00000000 00000000 0000001In:



Smallest Normalized Positive Number


• Step 2 (Sign): 0 is positive, 1 is negative. Sign is +• Step 3 (exponent): 1 (00000001)-127 (bias)=-126• Step 4 (mantissa): 1.00…0 = 1• Step 5 (result): (-1)0 x 1 x 2-126 = 2-126 ~ 10-38

(smallest denormalized was 2-149 ~ 10-45)

0 00000001 00000000 00000000 0000000

000000001 00000000 00000000 0000000In:


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FP Operations

• Multiplication f1 x f2:

• Addition f1 + f2 (e2>e1, shift f1 by e1-e2):

m1 re1 m2 re2 m1 m2 re1 e2

m1 re1 m2 re2 m1 r shift re2 m2 re2

(m1 r shift m2)re2


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FP accuracy

• Rounding: danger of false accuracy• e.g., 1.5*106 + 0.1 *10-3

1.5000000.000000 0001------------------- +1.500000


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FP rounding

• Rounding = convert to lower precision (w/ error)

• Operations use guard bits -> used for rounding

0.1010.110--------- *0.011 110


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FP-Rounding [1/2]

• Chopping: cut at guards bits (Round to zero)- error: 0 to almost -1 in LSB + Bias towards zero

• 0.011110 0.011• Von Neumann Rounding/Round to Nearest

- all guard bits 0 cut; else LSB:=1- error (-1,1) in LSB

• 0.01000 0.010 error: 0• 0.01110 0.011 error: 1/16• 0.0101110 0.011

Truncation error grows with the #ops


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FP-Rounding [2/2]

• Round to nearest EVEN number (default IE3754)- Add 1 to LSB if truncated MSB is 1- error [-1/2,1/2] in LSB

• 0.001000 0.001 • 0.001011 0.001• 0.000111 0.001• 0.001111 0.010• 0.001100 0.010 (to nearest EVEN number)

• Round downward/upward/to nearest ODD number

Decimal:• 10.5 10• 11.5 12• -10.5 -10• -11.5 -12


2.2. Floating PointFixed or Floating Point? (Revisited)

• Problem- What range is needed?- What precision is needed?- Need to replicate results later? (standard format needed)

• Platform AND Algorithm AND Input- How to optimize instructions/memory requirements?- What can the Platform do for this Algorithm AND Input?

• There’s no free lunch in computer science!

staff.ee.sun.ac.za/whsteyn/Papers2/CONPRA2299_final.pdf


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3. Booleans

• Booleans have two values: true/false• Used in programming languages

- if COND then ACTION1 else ACTION2

• Storage takes usually a complete byte- false is often represented as 00000000- true by any other pattern


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4.1. ASCII4.2. Unicode4.3. Processing Characters


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4. Characters

4.1. ASCII

• American Standard Code for Information Interchange

• ASCII set 128 Characters (7 bit)

• Alphabet {a-zA-Z}• Numbers {0,...,9}• Special symbols, eg. <, =, >, $• Punctuation marks, ! ?, .• Data transmission codes (non printable

characters)


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4. Characters

4.1. ASCII• Designed 1960—1963 • Underlies the WWW (actually, as common part of UTF-8)


4. Characters 4.1. ASCII An Example

The Demo, http://sloan.stanford.edu/MouseSite/1968Demo.html


4. Characters 4.1. ASCII Raster Images as ASCII (Example)P3 # RGB.ppm 3 3 255 0 0 255 255 255 0 0 0 255 255 255 0 255 0 255 255 2550 0 0 255 255 255 0 0 255

PPM Format Specification, http://netpbm.sourceforge.net/doc/ppm.html


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4. Characters

4.2. Unicode

• Large set of international alphabets represented• Subset ISO 10646• 16 bits representation (see Joseph Becker, Unicode 1988, http://www.unicode.org/history/unicode88.pdf)

“a simple unambiguous fixed-length character encoding”


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4. Characters

4.2. Unicode/UTF-8

• Variable-width encoding for Unicode (1-4 bytes)

• 1 byte: First 128 characters are ASCII (US-ASCII)• 2 bytes: Next 1,920 characters encode Latin+diacritics,

Greek, Cyrillic, Coptic, …• 3 bytes: Most others (Chinese, Hindi, tagalog/PH, …)• 4 bytes: Historical/Exotic variants (Klingon? Actually,

Klingon has an unofficial 2-byte Unicode encryption, see www.evertype.com/standards/csur/klingon.html )


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4. Characters 4.2. Unicode/UTF-8 An Example: The Romanian Alphabet

• HTML:Ă orĂ

http://bucovina.chem.tue.nl/romanian.utf8.htm


92

4. Characters

4.3. Processing of a number [1/3]

• User types a number, e.g. 123 (decimal)• Computer receives ASCII codes

- ASCII-code of 1: 00110001- ASCII-code of 2: 00110010- ASCII-code of 3: 00110011


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4. Characters


• Program converts codes to binary values1*100 011001002*10 000101003*1 00000011123 01111011

• Perform operation (e.g. subtract 50)123 0111101150 0011001073 01001001


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4. Characters


• The number 73 must be fed back to the screen0 (*100) 000000007 (*10) 000001113 (*1) 00000011

• Computer sends following ASCII codes to screen- ASCII code of 0: 00110000- ASCII code of 7: 00110111


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5. Arrays [1/3]

• Most programming languages support arraysA: array[0..N,0..M] of integer;

• Must be mapped onto linear memory• Row major order• Column major order


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5. Arrays [2/3]

row major

column major

array


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5. Arrays [3/3]

• Storage method is stored as part of array descriptor: so called dope vector.

begin addressno of dimensionsdim #1: low bounddim #1: high bounddim #1: multiplierdim #2: low boundetc

address fixed elementno of dimensionsstride (skipped memory

elements)


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6. Objects

Records

Record

field 1: type 1;

field 2: type 2;

end record• Records can be stored with fixed off-set • Records can contain pointers to other data

structures. Results in linked lists


6. Objects

Raster Images as Records (Example)• Simple, fixed representation:

- Height, Width- RGB(Alpha) data

03 03 FF 00 00 FF FF FF 00 00 00 FF FF FF 00 FF 00 FF FF FF00 00 00 FF FF FF 00 00 FF


6. Objects

Raster Images as Records (Example)• Fixed-format representation with look-up

table:- Height, Width- Color map/look-up table (here, RGB)- Image = index in color map per pixel03 03

FF FF FF 00 00 00 FF 00 00 00 FF 00 00 00 FF

02 00 0100 03 0001 00 04

Map[02] = FF 00 00


6. Objects

Raster Images as Records (Example)• Variable-format representation with look-up

table:- Image:

Height, Width, number of bits per color, …

- Color map structure: RGB/ARGB, number of entries, …

- Color map/look-up table- Image = index in color map per pixel


Fun With Hexadecimal NumbersColor Codes [1/2] http://en.wikibooks.org/wiki/LaTeX/Colors


Fun With Hexadecimal NumbersColor Codes [2/2] http://johncfish.com/bggallery/otherchart/index.htm

http://www.visibone.com/

TU-Delft TI1400/11-PDS 1 Arithmetic and Data Representation iosup/Courses/2011_ti1400_3.ppt.

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