chapter 2staff.cs.psu.ac.th/sathit/comor/ch02-s.pdf · 10 = (10001.11001) 2 23 2.3 decimal to...

Data Representation in

Computer Systems

Chapter 2

2

Chapter 2 Objectives

• เพอใหเขาใจการแทนตวเลข และการจดการตวเลขใน digital

Computer

• มทกษะในการเปลยนระหวางระบบเลขฐาน (radix system)

• เขาใจ ขอผดพลาดทเกดขน

– overflow

– Truncation / underflow

3

Chapter 2 Objectives

• เขาใจความคดพนฐานการแทนเลขทศนยม (floating point)

• การใชรหสตวอกขระทนยม

• เขาใจความคดเกยวกบการตรวจจบขอผดพลาด และการแกไขรหส

• A bit is the most basic unit of information in a

computer.

• A byte is a group of eight bits.

• A word is a contiguous group of bytes.

• A group of four bits is called a nibble (or nybble). : 4 บตสง 4 บตต า

4

2.1 Introduction

5

2.2 Positional Numbering Systems

• The decimal number 947 in powers of 10 is:

• The decimal number 5836.47 in powers of 10 is:

5 10 3 + 8 10 2 + 3 10 1 + 6 10 0

+ 4 10 -1 + 7 10 -2

9 10 2 + 4 10 1 + 7 10 0

6

2.2 Positional Numbering Systems

• The binary number 11001 in powers of 2 is:

• When the radix of a number is something other

than 10, the base is denoted by a subscript.

– Sometimes, the subscript 10 is added for emphasis:

110012 = 2510

1 2 4 + 1 2 3 + 0 2 2 + 0 2 1 + 1 2 0

= 16 + 8 + 0 + 0 + 1 = 25

7

2.3 Decimal to Binary Conversions

• ในการเปลยนฐานเลข สามารถท าได 2 แบบ

–การลบ (subtraction)

–การหา เอาเศษ

(ไมนยมใช การลบ)

8

• Suppose we want to convert

the decimal number 190 to

base 3.

– We know that 3 5 = 243 so

our result will be less than six

digits wide. The largest

power of 3 that we need is

therefore 3 4 = 81, and

81 2 = 162.

– Write down the 2 and subtract

162 from 190, giving 28.


9

• Converting 190 to base 3...

– The next power of 3 is

3 3 = 27. We’ll need one of

these, so we subtract 27 and

write down the numeral 1 in

our result.

– The next power of 3, 3 2 = 9,

is too large, but we have to

assign a placeholder of zero

and carry down the 1.


10



– 3 1 = 3 is again too large, so

we assign a zero placeholder.

– The last power of 3, 3 0 = 1,

is our last choice, and it gives

us a difference of zero.

– Our result, reading from top

to bottom is:

19010 = 210013

สรปใชวธการลบ

11

เปลยน190 จากฐาน 10 เปน ฐาน 3 : (190)10 = (???)3 5 4 3 2 1 0

2 1 0 0 1

243 81 27 9 3 1

162 = 81 x 2

28

27 = 27 x 1

1 0 = 9 x 0

0

0 = 3 x 0

0 1 = 1 x 1

1

190

12


• Another method of converting integers from

decimal to some other radix uses division.

• This method is mechanical and easy.

• It employs the idea that successive division by a

base is equivalent to successive subtraction by

powers of the base.

• Let’s use the division remainder method to again

convert 190 in decimal to base 3.

13


– First we take the number

that we wish to convert and

divide it by the radix in

which we want to express

our result.

– In this case, 3 divides 190

63 times, with a remainder

of 1.

– Record the quotient and the

remainder.


14


– 63 is evenly divisible by 3.

– Our remainder is zero, and

the quotient is 21.


15


– Continue in this way until

the quotient is zero.

– In the final calculation, we

note that 3 divides 2 zero

times with a remainder of 2.

– Our result, reading from

bottom to top is:

19010 = 210013


16


การเปลยนเลขฐาน 10 เปนฐาน 2 โดยการหารเอาเศษ

• (37)10 = (100101)2

• (103)10 = (1100111)2

• (317)10 = (100111101)2

17


การเปลยนเลขฐาน 10 เปนฐาน 2 โดยการหารเอาเศษ

• (120)10 = (xxx)2

• (517)10 = (xxx)2

• (1023)10 = (xxx)2

18


การเปลยนเลขฐาน 2 เปนฐาน 10 โดยน า คาต าแหนงนน คณดวยฐานยกก าลงในต าแหนงทอย

• (1001)2 = (9)10

• (110100)2 = (52)10

• (10010010)2 = (146)10

19


การเปลยนเลขฐาน 2 เปนฐาน 10 โดยน า คาต าแหนงนน คณดวยฐานยกก าลงในต าแหนงทอย

• (100101)2 = (x)10

• (110111)2 = (x)10

• (10110110)2 = (x)10

20


• สวนทเปน ทศนยม หรอ เศษสวน ฐาน 10

709.52 = 7x102 + 0x101 + 9x 100

+ 5x10-1 + 2x10-2

• สวนทเปนเศษสวน ฐาน 2 ของ 101.012

101.01 = 1x22 + 0x21 + 1x 20 +

0x2-1 + 1x2-2

• เปลยนเศษสวน 5.25 ฐาน 10 เปน ฐาน 2

(5.25)10 = (101.01)2

101.01

21


.25 x 2 = 0.5

.5 x 2 = 1.0

หยด

• เปลยน 100.62510 เปน ฐาน 2

(100.625)10 = (1100100.101)2

1100100.101

22


.625 x 2 = 1.250

.25 x 2 = 0.5

.5 x 2 = 1.0

หยด

• เปลยน 17.810 เปน ฐาน 2

(17.8)10 = (10001.11001)2

23


.8 x 2 = 1.6 .6 x 2 = 1.2 .2 x 2 = 0.4 .4 x 2 = 0.8 .8 x 2 = 1.6 .6 x 2 = 1.2

หยด ถาตองการ 5ต าแหนง

24


• เปลยนเลขทศนยมฐาน 10 ตอไปนเปน ฐาน 2

(โดยใหมทศนยม 5 ต าแหนง)

51.12510 = (xxx)2

101.910 = (xxx)2

131.562510 = (xxx)2

199.4210 = (xxx)2

25


• แปลง 1011.0112 เปน ฐาน 10

1011.0112 = 11.37510

0 + 1 + 1

2 4 8

0 + 0.25 + 0.125

26


• เปลยนเศษสวน ฐาน 2 เปน ฐาน 10

10.11012 = xxx10

1011.1012 = xxx10

11.11002 = xxx10

27


• เปลยนเลข ฐาน 2 เปน ฐาน 8 (นบ 3 จาก LSB : บตทมนยส าคญนอยสด ทเหลอไมครบ 3 เตม 0 ดานหนา)

(10 100 101 001)2 = (2451)8

010

28


• เปลยนเลข ฐาน 2 เปน ฐาน 16 (นบ 4 จาก LSB : บตทมนยส าคญนอยสด ทเหลอไมครบ 4 เตม 0 ดานหนา)

(101 0010 1001)2 = (529)16

0101

29


• เปลยนจ านวนจรง ฐาน 2 เปน ฐาน 8 (นบ 3)

– หนาจด นบจาก LSB ขาดเตม 0 ดานหนา

– หลงจด นบจาก MSB ขาดเตม 0 ดานหลง

(10011.011)2 = ?8

(010 011.011)2 = (23.3)8

(1111.11)2 = ?8

(001 111.110)2 = (17.6)8

30


• เปลยนจ านวนจรง ฐาน 2 เปน ฐาน 8

(11000110.1)2 = (xxxxx)8

(1.11010)2 = (xxxxx)8

31


• การแปลงเลขจ านวนจรงฐาน 2 เปน ฐาน 16

– หนาจด นบจาก LSB ขาดเตม 0 ดานหนา

– หลงจด นบจาก MSB ขาดเตม 0 ดานหลง

(10011.011)2 = ?16

(0001 0011.0110)2 = (13.6)16

(1111.11)2 = ?16

(1111.1100)2 = (F.C)16

32


• การแปลงเลขจ านวนจรงฐาน 2 เปน ฐาน 16

(11000110.1)2 = (xxxxx)16

(1.11010)2 = (xxxxx)16

33



• ท างานเหมอนกบเปนเปลยนเศษสวน ฐาน 10 เปน ฐาน 2 คอหารดวยฐานไปเรอยๆ

(12.5)10 = ?8

[(12)/8.(58)]10 = (14.4)8

4.0 181+480

34



• ท างานเหมอนกบเปนเปลยนเศษสวน ฐาน 10 เปน ฐาน 2 คอหารดวยฐานไปเรอยๆ

(12.5)10 = ?16

[(12/16).(516)]10 = (C.8)16

8.0 C160

35

2.4 Signed Integer Representation

• The conversions we have so far presented

have involved only positive numbers.

• To represent negative values, computer

systems allocate the high-order bit to indicate

the sign of a value.

– The high-order bit is the leftmost bit in a byte. It

is also called the most significant bit.

• The remaining bits contain the value of the

number.

36


Signed Integer Representation

การแสดงเลขจ านวนเตม แบบมเครองหมาย

37


• There are three ways in which signed binary

numbers may be expressed:

– Signed magnitude,

– One’s complement and

– Two’s complement.

• In an 8-bit word, signed magnitude

representation places the absolute value of

the number in the 7 bits to the right of the

sign bit.

38


การแทนขอมลขนาด 8 บต

LSB MSB

7 6 0

Sign Bit = 1 มคาเปนลบ

= 0 มคาเปนบวก

Sign bit : เครองหมาย Magnitude : ขนาด

39


1 111 1111

:

1 000 0001

1 000 0000

0 000 0000

0 000 0001

:

0 111 1111 -127 … -1 -0 +0 +1 … +127

• Data range

Binary Decimal

0 0001001

1 0110101

1 0100100

0 1111111

+9

-53

-40

+127

• For 8-bit in memory

40


1 111 1111

:

1 000 0000

0 000 0000

0 000 0001

:

0 111 1111

-127 … -1 -0 +0 +1 … +127

Binary Decimal

0 0001001

1 1110110

1 1011000

0 1111111

0 1111111

+9

-9

-40

-127

+127

• For 8-bit in memory for 1’s Complement

• Data range

41


• 2’s Complement

–Invert and +1 1 000 0000

:

1 111 1111

0 000 0000

0 000 0001

:

0 111 1111

-128 … -1 +0 +1 … +127

Binary Decimal

0 0001001

1 1110111

1 1011100

1 0000000

0 1111111

+9

-9

-40

-128

+127

• Data range

42


43


• For example, in 8-bit signed magnitude, positive 3 is: 00000011

• Negative 3 is: 10000011

• Computers perform arithmetic operations on

signed magnitude numbers in much the same

way as humans carry out pencil and paper

arithmetic.

– Humans often ignore the signs of the operands

while performing a calculation, applying the

appropriate sign after the calculation is complete.

44


• Binary addition is as easy as it gets. You need

to know only four rules: 0 + 0 = 0 0 + 1 = 1

1 + 0 = 1 1 + 1 = 10

• The simplicity of this system makes it possible

for digital circuits to carry out arithmetic

operations.

0 + 0 = 00 0 + 1 = 01

1 + 0 = 01 1 + 1 = 10

SUM Carry


• Positive numbers is the same in one’s

complement and two’s complement (as

well as signed-magnitude), the process of

adding two positive binary numbers is the

same.

45


• How to perform addition using two’s

complement notation (and hence

subtraction, because we subtract a number

by adding its opposite).

• Ex: Add 910 to –2310 using two’s

complement arithmetic.

46

Two’s complement arithmetic

operation

• Ex: Find the sum of 2310 and –910 in binary

using two’s complement arithmetic.

47

• Ex: Find the sum of 111010012 (–23) and

111101112 (–9) using two’s complement addition.

Overflow A Simple Rule for Detecting an Overflow

Condition in Signed Numbers:

• If the carry into the sign bit equals the carry out of

the bit, no overflow has occurred.

• If the carry into the sign bit is different from the

carry out of the sign bit, overflow (and thus an

error) has occurred. See Ex.

48

Overflow

Consider 4-bit unsigned and signed numbers.

• If we add the two unsigned values 0111 (7)

and 0001 (1), we get 1000 (8).

– There is no carry (out), and thus no error.

• If we add the two unsigned values 0111 (7)

and 1011 (11), we get 0010 with a carry,

indicating that there is an error (indeed, 7 + 11

is not 2).

49

Overflow in signed numbers

• Adding the two’s complement integers 0101

(+5) and 0011 (+3) equal to 1000 (–8),

which is clearly incorrect.

– The problem is that we have a carry in to the

sign bit, but no carry out, which indicates that

we have an overflow.

• However, add 0111 (+7) and 1011 (–5), we

get the correct result: 0010 (+2).

– We have both a carry in to and a carry out of

the leftmost bit, so there is no error. 50

Overflow summation

• The rule of thumb used to determine when

carry indicates an error depends on

whether we are using signed or unsigned

numbers.

– Unsigned numbers: a carry (out of the leftmost

bit) indicates the total number of bits was not

large enough to hold the resulting value, and

overflow has occurred.

– Signed numbers: if the carry in to the sign bit

and the carry (out of the sign bit) differ, then

overflow has occurred. 51

Examples of Carry and Overflow

in Signed Numbers

52

2.4.4 Unsigned Versus Signed Numbers

• Unsigned numbers represent values that are guaranteed

not to be negative.

• A computer programmer must be able to manage both

signed and unsigned numbers.

– For instance, the C language has int and unsigned int for

integer variables, defining signed and unsigned integers,

respectively.

– Different type declarations, many languages have different

arithmetic operations with signed and unsigned numbers.

• Good programmers understand this condition and

make appropriate plans to deal with the situation

before it occurs. 53

• วงจรเกทสรางการบวก บทแรก (LSB)

54


A B Sum Carry

0 0 0 0

0 1 1 0

1 0 1 0

1 1 0 1

XOR AND

2.4.7 Binary Multiplication and

Division Using Shifting

55

56


signed magnitude

• มบตบอกเครองหมาย

0 : บวก 1 : ลบ

• บตบอกขนาด 57


58


• Example:

– Using signed magnitude

binary arithmetic, find the

sum of 75 and 46.

• First, convert 75 and 46 to

binary, and arrange as a sum,

but separate the (positive)

sign bits from the magnitude

bits.

59


• Example:



sum of 75 and 46.

• Just as in decimal arithmetic,

we find the sum starting with

the rightmost bit and work left.

60


• Example:



sum of 75 and 46.

• In the second bit, we have a

carry, so we note it above the

third bit.

61


• Example:



sum of 75 and 46.

• The third and fourth bits also

give us carries.

62


• Example:

– Using signed magnitude binary

arithmetic, find the sum of 75

and 46.

• Once we have worked our way

through all eight bits, we are

done.

In this example, we were careful careful to pick two

values whose sum would fit into seven bits. If that is not

the case, we have a problem.

63


• Example:

– Using signed magnitude binary


and 46.

• We see that the carry from the

seventh bit overflows and is

discarded, giving us the

erroneous result: 107 + 46 = 25.

64


• The signs in signed

magnitude representation

work just like the signs in

pencil and paper arithmetic.

– Example: Using signed

magnitude binary arithmetic,

find the sum of - 46 and - 25.

• Because the signs are the same, all we do is

add the numbers and supply the negative sign

when we are done.

65


• Mixed sign addition (or

subtraction) is done the

same way.

– Example: Using signed

magnitude binary arithmetic,

find the sum of 46 and - 25.

• The sign of the result gets the sign of the number

that is larger.

– Note the “borrows” from the second and sixth bits.

66


• Signed magnitude representation is easy for

people to understand, but it requires complicated computer hardware.

• Another disadvantage of signed magnitude is

that it allows two different representations for

zero: positive zero and negative zero.

67


• In complement systems, negative values are

represented by some difference between a number and its base.

• In diminished radix complement systems, a

negative value is given by the difference between

the absolute value of a number and one less than

its base.

• In the binary system, this gives us one’s

complement. It amounts to little more than flipping

the bits of a binary number.

one’s complement.

•คาลบเกบแบบสลบบต

ตองการเกบในหนวยความจ าขนาด 8 บต (5)10 = (0000 0101)2

(-5)10 = (1111 1010)2

68


69


• one’s complement.

• For example, in 8-bit one’s complement, positive 3 is: 00000011

• Negative 3 is: 11111100

• In one’s complement, as with signed

magnitude, negative values are indicated by a

1 in the high order bit.

70


• With one’s complement

addition, the carry bit is

“carried around” and added

to the sum.

– Example: Using one’s

complement binary arithmetic,

find the sum of 48 and - 19

We note that 19 in one’s complement is 00010011,

so -19 in one’s complement is: 11101100.

71


• Although the “end carry around” adds some

complexity, one’s complement is simpler to implement than signed magnitude.

• But it still has the disadvantage of having two

different representations for zero: positive

zero and negative zero.

• Two’s complement solves this problem.

• Two’s complement is the radix complement of

the binary numbering system.

Two’s complement.

•คาลบเกบแบบสลบบต +1

ตองการเกบในหนวยความจ าขนาด 8 บต (5)10 = (0000 0101)2

(-5)10 = (1111 1011)2

72


73


• To express a value in two’s complement:

– If the number is positive, just convert it to binary and

you’re done.

– If the number is negative, find the one’s complement of

the number and then add 1.

• Example:

– In 8-bit one’s complement, positive 3 is: 00000011

– Negative 3 in one’s complement is: 11111100

– Adding 1 gives us -3 in two’s complement form: 11111101.

74


• With two’s complement arithmetic, all we do is add

our two binary numbers. Just discard any carries

emitting from the high order bit.

We note that 19 in one’s complement is: 00010011,

so -19 in one’s complement is: 11101100,

and -19 in two’s complement is: 11101101.

– Example: Using one’s

complement binary

arithmetic, find the sum of

48 and - 19.

75


• When we use any finite number of bits to

represent a number, we always run the risk of

the result of our calculations becoming too large

to be stored in the computer.

• While we can’t always prevent overflow, we can

always detect overflow.

• In complement arithmetic, an overflow condition

is easy to detect.

76


• Example:

– Using two’s complement binary


and 46.

• We see that the nonzero carry

from the seventh bit overflows into

the sign bit, giving us the

erroneous result: 107 + 46 = -103.

Rule for detecting signed two’s complement overflow: When

the “carry in” and the “carry out” of the sign bit differ,

overflow has occurred.

77


0110

78


• Signed and unsigned numbers are both useful.

– For example, memory addresses are always unsigned.

• Using the same number of bits, unsigned integers

can express twice as many values as signed

numbers.

• Trouble arises if an unsigned value “wraps around.”

– In four bits: 1111 + 1 = 0000.

• Good programmers stay alert for this kind of

problem.

79


80


81


82


83


84


85


86


87


88


89

2.5 Floating-Point Representation

Floating-Point Representation

การแสดง Floating-Point

90


• Floating-point numbers allow an arbitrary

number of decimal places to the right of the

decimal point.

– For example: 0.5 0.25 = 0.125

• They are often expressed in scientific notation.

– For example:

0.125 = 1.25 10-1

5,000,000 = 5.0 106

91


• Computers use a form of scientific notation for

floating-point representation

• Numbers written in scientific notation have three

components:

92

• Computer representation of a floating-point

number consists of three fixed-size fields:

• This is the standard arrangement of these fields.


93

• The one-bit sign field is the sign of the stored value.

• The size of the exponent field, determines the

range of values that can be represented.

• The size of the significand determines the precision

of the representation.


94

• The IEEE-754 single precision floating point

standard uses an 8-bit exponent and a 23-bit

significand.

• The IEEE-754 double precision standard uses an

11-bit exponent and a 52-bit significand.


For illustrative purposes, we will use a 14-bit model

with a 5-bit exponent and an 8-bit significand.

Example IEEE-754 Single-Precision

Floating-Point Numbers

95

Range of IEEE-754

Double-Precision Numbers

96

2127 × 1.1111111111 11111111111112 (let’s call this value MAX)

2–127 × .000000000000000000000012 (let’s call this value MIN).

Single precision floating-point numbers cannot represent:

- negative numbers less than –MAX (negative overflow);

- negative numbers greater than –MIN (negative underflow);

- positive numbers less than +MIN (positive underflow); and

- positive numbers greater than +MAX (positive overflow).

98

• Example:

– Express 3210 in the simplified 14-bit floating-point

model.

• We know that 32 is 25. So in (binary) scientific

notation 32 = 1.0 x 25 = 0.1 x 26.

• Using this information, we put 110 (= 610) in the

exponent field and 1 in the significand as shown.


99

• Another problem with our system is that we have

made no allowances for negative exponents. We

have no way to express 0.5 (=2 -1)! (Notice that

there is no sign in the exponent field!)


All of these problems can be fixed with no

changes to our basic model.

100

• To provide for negative exponents, we will use a

biased exponent.

• 5 bits แทนไดถง 32 เพอเกบคาลบดวย ถงแบงครง 16 เรยกวา excess-

16

– In our case, we have a 5-bit exponent. We will use 16 for

our bias. This is called excess-16 representation.


16 31 0 - +

101

• Example:

– Express 3210 in the revised 14-bit floating-point model.

• We know that 32 = 1.0 x 25 = 0.1 x 26.

• To use our excess 16 biased exponent, we add 16 to

6, giving 2210 (=101102).

• Graphically:


102

• Example:

– Express 0.062510 in the revised 14-bit floating-point

model.

• We know that 0.0625 is 2-4. So in (binary) scientific

notation 0.0625 = 1.0 x 2-4 = 0.1 x 2 -3.


-3, giving 1310 (=011012).


103

• Example:

– Express -26.62510 in the revised 14-bit floating-point

model.

• We find 26.62510 = 11010.1012. Normalizing, we

have: 26.62510 = 0.11010101 x 2 5.


5, giving 2110 (=101012). We also need a 1 in the sign

bit.


104

• Calculations aren’t useful until their results can

be displayed in a manner that is meaningful to

people.

• We also need to store the results of calculations,

and provide a means for data input.

• Thus, human-understandable characters must be

converted to computer-understandable bit

patterns using some sort of character encoding

scheme.

2.6 Character Codes

รหสแทนขอมลภายในเครองคอมพวเตอร

• รหสแทนขอมลทเปนตวเลข – Sign and Magnitude

– 1’s Complement

– 2’s Complement

• รหสแทนขอมลทเปนตวอกษร – BCD : Binary Code Decimal

– EBCDIC : Extended Binary Code Decimal Interchange

Code

– ASCII : American Standard Code Interchange International

จ านวนแบบตวอกขระในระบบคอมพวเตอร นอยสด

• ตวเลข 10 ตว ( 0, 1 ,2 … 9)

• ตวอกษรภาษาองกฤษ 26 ตว ( A,B,…, Z)

• เครองหมายพเศษ 16 ตว ( + - * / > < ; ,!@#$%&() )

รวมทงหมด 52 ตว สามารถใชกลมเลขฐานสองเพยง 6 bits

ในระบบ BCD เพม Parity bit อก 1 bit เพอการตรวจสอบความถกตอง

รหสแทนขอมลทเปนตวอกษร แบบ BCD

• ใน 1 Byte ม 7 Bits

BA Numeric แทน 00 0 - 9 0 - 9 11 1 - 9 A - I 10 1 - 9 J - R 01 2 - 9 S - Z

• อกขระพเศษแตละบรษทก าหนดเอง

C B A 8 4 2 1

Numeric Zone

Parity

Binary-Coded Decimal

108

Digit BCD

0

1

2

3

4

5

6

7

8

9

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

Zones

1111

1100

1101

Unsigned

Positive

Negative

BCD

• For example, to encode 146, the decimal

digits are replaced by 0001, 0100, and 0110,

respectively.

• Computers use bytes as the smallest unit of

access, most values are stored in 8 bits, not

4. Using this approach, padding with zeros,

146 would be stored as 00000001 00000100

00000110.

109

packed BCD

110

รหสแทนขอมลทเปนตวอกษร แบบ EBCDIC

• 1 Byte ใช 8 bits

• แทนอกขระได 256 ตว 8 4 2 1 8 4 2 1

Zone Numeric

EBCDIC Code

112

American Standard Code

for Information Interchange (ASCII)

• ออกโดย สมาคมรกษามาตรฐานของ สหรฐอเมรกา • เพอใหการตดตอสอสารระหวางเครองคอมพวเตอร มความสะดวก

และเปนมาตรฐานเดยวกน

• 1 Byte ใช 8 bit

• มใชในเครองไมโครคอมพวเตอร ทดสอบการแทนรหสไดโดยการกด Alt + Numeric Key Pad แทนรหส ASCII

114

รหสแทนขอมลทเปนตวอกษร แบบ ASCII

Parity bit

ใชในการตรวจสอบความผดพลาด โดยกระท าภายใตขอสมมตฐาน 2 ขอ • คาของความผดพลาดทจะเกดม Percent นอยมาก • ความผดพลาดทเกดขน เกดไดเพยง bit เดยวเทานนใน 1 byte

Check Parity Bit ม 2 แบบ

• Odd Parity จ านวน bit ค • Even Parity จ านวน bit ค

116

Parity bit

รหสแทนขอมลทเปนตวอกษร แบบ ASCII

117

Unicode

• In 1991, to establish a new international

information exchange code called Unicode.

• Unicode is a 16-bit alphabet that is downward

compatible with ASCII and the Latin-1

character set.

• Conformant with the ISO/IEC 10646-1

international alphabet.

• Capacity to encode the majority of characters

used in every language of the world. 118

Unicode Codespace

119

120

End of Chapter 2

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