( 25 )育教日八十月二十年午庚豆

1991 中學會考預習專欄

明德出版社

MILL & DALE PRESS.

Computer (18)

the number of digits is eight. (2) The nineth bit is.ignored. because the length of ward is limited to 8-bit.

:

(3) 00001000 2 81

''' 21-13 = 8

The answer is verified.

*(4) Only negative number.

represented by 2's complement!

(b) 13qg = 000011012

(4) If

報日僑

more bits' are used for the mantissa which will then be more accurate. If:more bits are used for the exponent then the range of number will be larger.

Floating Point Representation

Flood Point Representation

Range

small

000101012 2110 -2110 11101011.

-32768 to 32767

(16-bit word)

Jargo -38

38

(Accuracy Exact

110 - 10

(8-bit exponent)]

The number of

(12's complement)

(except overflow significant lar underflow)

1310-2110

digits depends on the length of mantisse.

Choy

1340

+ (-2110 000011012 +11101011, 11.1110002

00001101 11101011 11111000

in

e.g.

Subtraction of numbers in two complement representation

A-, B.

(-8)

As negative. numbers are represented in 2's complement format. Therefore suttraction of numbers can be done by adding negative numbers in complement form..

Example:

two's

Perform the following subractions. complement 8-bit word

in

representation:

-(a) 21-13

(b) 13 21

Solution:

(*) 21,0 = 000101011⁄2

1340

= 00001

201701

131011110

100112

Remarks:

(1) The "1" in the first bit from the left may indicate that it is a negative number. It is the sign bit.

(2).The decimal equivalent of the

answer

-128+64+32+16+8:

Floating Point Representation inside. computer:

(1) Numbers

form.

eg.

are

changed to normalized.

1000:1-0-1000110101 (6inary)

0.100001

101

exponent:

Mantissa

(2). Suppose the word

computer word to point humbers are 12 bits. The format to store the above now may be

Tength of the store floating

(5) Truncation and rounding off

floating point numbers

3.141592654...

If 4 decimal places are required for the number, then

1.= 3.1415 (under truncatlon} ii. w=.3.14.16. (after rounding). (a) Both numbers . contain cut off- .errors. But it is worthwhile

in

to notice. the fact that Error in rounding off & Error truncation (b) Overflow error:

e.g. Use 8-bit word tocompute

20810

+ 103 10

11010000

01100111. 100110111*

Carry. bit ignored.

55:0

=

001101112130

The answer is obviously wrong. Similiarly, underflow may also

cause error.

2110

(-1310)

00010101 +11110011.

000101012 + 1111001100001000 000010002

ignored

Remarks:

(1) Preceding zeros are added until

CHOIRE S

I(a) The

̇六期蠱:日二月二 ( 一九九一年十八國民中

excess-16 is used for the exponent. to represent negative exponents.

(b) The actual value of the exponent.

should be compute. by less 16.

e.g. The zerd exponent: P100002

the actual exponent = 15 - 16

e.g. The exponent 10011

the actual exponent = 000112

| (c) If 6 bits are -used for the representation, then excess-32 are

used.

Representation information

Character

ASCII

.48 49

67

Alphanumeric

store, and process both numbers and characters.

Alphanumeric character set (a):10. decimal digits. (b) 26 upper-case.

alphabets.

(c) 26 Tower-case

alphabets..

(d) special characters.

($, +, -, *, ?}

Jetters

letters

- distinctive pattern of bits

(a) EBCDIC (Extended Binary coded.

'B'bits

Decimal Interchange code) BM

(b) ASCII (American Standard Code for

Information Interchange)

widely used in Communication and microcomputer.

7 bits (126 characters)

MSB parity: bit for checking

purpose.

A partial Character List for ASCII

Character

ASCII

73

74

ARARNE25.......ORENAJAR

This table will be given in both papers in the HKCES Characters. aneshown.

You just have to know how to use it.

As are represented by seven bits.:

165 bit pattern > {11101010003

parity ASCII 66 bit pattern is 1000010 B: ASCII = * ajsebattern is 1000000

Excess-16 aresresentation exponent.

the

1.0.0.0.0.1.0

wulf 5. bits nekarteux for then representations are

01.0

oxponent

00000

00001

00010

negative exponent.

The ASOL

:blt Linery

paint

10000

Zero exponent

point representation may

1.1-101

dependent. Different

positive exponent

11110

different formats.

11111

Exercise: Convert

超入門

DOS

LOTUS 123

DBASED O

WORDSTAR

中文(衢/倚天)

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into ASCII code.

Character

ASCIE

98 99

100

1,011

102

103

104

105

106

107

:108

109

110

111

112

113

114

115

116

117

118.

119

120

121

122

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·術長心矣。於遒上更有虧,

Fac

生滔滔而談;這,太過有遼學 用 - 在演講會上對若一班大學 不應該。何況岑作家避據爲己 · 被盜舂來

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手十八國民蕺中

作員

中心誠聘幼 或未註册均

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