1-2The Intel Microprocessors

The Intel Microprocessors

--from 8086 to Pentium Architecture, Programming and Interfacing

Teach by Zuohang ( 左航 )College of Software

2003-8

• E-mail :

•[email protected]

• Phone:

•66860994

•Content•Arrangement •Reference book•Final score

1. Architecture

2. Programming

3. Interfacing

CPU

1. Architecture

2. Programming

Memory

Printer

I/O

3. Interfacing

1. Architecture

•1.1 The architecture of CPU

•1.2 Addressing modes

Internal Microprocessor Architecture AH (AX) AL

BH (BX) BL

CH (CX) CL

DH (DX) DL

SP

BP

DI

SI

EAX

EBX

ECX

EDX

ESP

EBP

EDI

ESI

Accumulator

Base index

Count

Data

Stack pointer

Base pointer

Destination index

Source index

DR

PR

IR

zuohang

zuohang2003-9-1If the name contains two capitals that means the register is 2 bytes.three capitals means 4 bytes. E** is used in 80386 and aboveHere we always discuss the 8086 and 8088.

Internal Microprocessor Architecture

IP

FLAGS

EIP

EFLAGS

Instruction pointer

Flags

CS

DS

ES

SS

FS

GS

Code

Data

Extra

Stack

Special purpose registers

Segment registers

1. Architecture

2. Programming

3. Interfacing

2. Programming

• 2.1 Data movement instructions

• 2.2 Arithmetic and logic instructions

• 2.3 Program control instructions

program• .DATA • NUM DB 34H• TABLE DW 0012H,0033H,5687H• .CODE• .STARTUP• MOV BX,OFFSET TABLE• MOV AX,[BX+4]• MOV CX,88H• MUL CX• CMP AX,2000H• JAE NEXT• OUT AX,P8• .EXIT• END

Application languages/application program

High-level languages/compiler &interpretative program

Assembly language/ assembly program

Keyboard command and system primitive/ operating system

Machine instruction system/ CPU

1. Architecture

2. Programming

3. Interfacing

3. Interfacing

• 1 8088/8086 hardware specifications

• 2 Memory interface• 3 Basic I/O interface• 4 Interrupts• 5 Direct memory access and

DMA-controlled I/O

8088 hardware specifications

Memory interface

request

•Familiar with•Addressing mode• programming with assembly language

• interfacing of microprocessor

What we can do after learning this

• Programming in assembly language in certain real-time system ,memory limited system or embedded system

• Design interfacing and writing drivers

Reference Books

• 汇编语言»王爽主编

• 微型计算机原理及应用»周明德编著清华大学出版社

• IBM-PC 汇编语言程序设计»沈美明主编，清华大学出版社

• The 80x86 IBM PC and Compatible Computers (Volumes I & II): Assembly Language, Design, and Interfacing (4th Edition)

» 清华大学出版社

Reference Lessons

• Operating system 操作系统• Computer architecture 计算机体系结

构

•Final Exam: 70%•Homework & Attendance:30%

• This may be revised according to the needs.

Now let’s begin our exploration

in microprocessor.

Chapter 1 Introduction to The

Microprocessor and PC

Chapter 1:

1. What mankind has done before the microprocessor finally came out?

2. How many parts are there in the microprocessor ?

Chapter 1:Introduction to The Microprocessor and PC

• 1.1 A Historical Background• 1.2 PC Based on Microprocessor

1.1 A Historical Background


• A. The Mechanical Age– Abacus (Babylonians)– Analytical engine (Babbage, punched

cards, 1823,failure)


• B. The Electrical Age– Motor-driven adding machines, based on mech

anical calculator (Hollerith, set up IBM-International Business Machines Corporation)

– First electronic calculating machine Z3 (German, Konrad Zuse,1942)

– The first general-purpose, programmable electronic computer ENIAC (University of Pennsylvania, 1946)

• ENIAC• Electronic Numerical Integrator And C

alculator [Computer] 电子数字积分计算机

Intel 4004

Intel 8088

Intel Petium

Intel Petium II

The Moore’s Law: the number of transistors integrated in a chip will double very 18 or 24 mouths


• D. The Future of Microprocessors

– The process speed will get more faster

– The memory will get more large

– The bulk will get more smaller

– The width of data bus will increase

– Architecture will get more efficient

1.2 PC Based on Microprocessor


• Question: If we use a computer to figure out an arithmetic expression, how can it finish this work?

• 133*33+44*14


• 133*33+44*14– First input these numbers.– Do the calculating work.– Store The result – Output the result.

• Control this processing.


• 133*33+44*14Input device

memory

calculator

Outputdevice

controller

Control bus

data bus


Micro-processor

Memory

Interface

Externaldevices

Data bus

Address bus

Control bus

BUS definition p25


• Bus: P25– Address bus requests a memory

location from the memory or an I/O location from the I/O devices.

– Data bus transfers information between the microprocessor and its memory and I/O address space.

– Control bus contains lines that select the memory or I/O and cause them to perform a read or write operation.

1.2 PC Based on MicroprocessorA. Relationship Figure

microprocessor ALU controller

PC register internal memory I/O interface

PC system I/O Devices & external memorysystem softwareapplication softwarepower 、 panel 、 pc frame, etc

Chapter 2

Introduction to Number System & Data Formats

Chapter 2:

• (-133)*33+44*14.5 • How to represent these data?• How to calculate them?• How about other characters?

Chapter 2: Introduction to Number System & Data Formats

• 2.1 Number System ( 数制 )

• 2.2 Computer Data Formats

2.1 Number System ( 数制 )

2.1 Number System( 数制 )

• 2.1.1 Digits (数)• 2.1.2 Positional Notation ( 位计数法）• 2.1.3 Conversion to Decimal• 2.1.4 Converting From Decimal• 2.1.5 Binary-Coded Hexadecimal• 2.1.6 Operation of Binary• 2.1.7 Three Forms of Binary Data

2.1.1 Digits ( 数 )

2.1.1 Digits ( 数 )

• A. Decimal( 十进制 ) (0—9)

• B. Binary (0----1)

• C. Octal( 八进制 ) (0—7)

• D. Hexadecimal( 十六进制 ) • (0—9,A,B,C,D,E,F)

2.1.1 Digits ( 数 )

• Example: A base 13( 以 13 为基， 13 进制 )numb

er contains 13 digits:

0—9,A(10),B(11),C(12)

2.1.2 Positional Notation ( 位计数法）


• Example 1:Decimal 132 1( 百位 ) 3( 十位 ) 2( 个位 )

hundreds tens units position position position

power( 幂 ) 102 101 100

weight( 权 )100 10 1In a base N number system, the radix is N( 基

数 ) and the exponent (i) ( 指数 ) means the position.


• Example 2:Binary 101 1 0 1

power( 幂 ) 22 21 20

weight( 权 )4 2 1Numeric value 1* 22 + 0* 21 + 1* 20 =5 In a base 2 number system, the radix is

2( 基数 ) and the exponent (i) ( 指数 ) means the position.


• Special exampleBinary 1111 1 1 1 1

power( 幂 ) 23 22 21 20

weight( 权 ) 8 4 2 1Numeric value 8+4+2+1 =15

0-15 It can represent 16 different numbers.


• Example 3:Decimal 1.025 1 . 0 2 5

power( 幂 ) 100 10-1 10-2 10-3(Ni)weight( 权 )1 0.1 0.01 0.001In a base N number system, the radix is N and

the exponent (i) ( 指数 ) means the position. If the position is right to the point( 小数点 ),the power is negative (N-i).


• Summarize– A base N number system Xm-1Xm-2Xm-3……X0

. Y1Y2Y3……Yn

– Numeric value( 数值 ): Xm-1* Nm-1+Xm-2 * Nm-2+ ……X0

* N0

+Y1 *N-1 +Y2 *N-2 + …… Yn * N-n

2.1.3 Conversion to Decimal

2.1.3 Conversion to Decimal

• Example:• Hexadecimal: E6A.C2• Decimal: E*162+6*161+6*160+C*16-1+2*16-2

=14*162+6*161+6*160+12*16-

1+2*16-2

=3690+0.75+0.0078125 =3690.7578125

2.1.4 Converting From Decimal


• Summarize– A base N number system Xm-1Xm-2Xm-3……X0

. Y1Y2Y3……Yn

– Decimal Xm-1* Nm-1+Xm-2 * Nm-2+ ……X0

* N0

+Y1 *N-1 +Y2 *N-2 + …… Yn * N-n


• From conversion to decimal we can assume that if we can figure out how many Ni are there in a decimal number, we may convert it back to a base N number system.


• Principle A:– To convert a decimal whole numb

er portion( 整数部分 ) to another number system, divide by the radix N ( 基数 ) and save the remainders( 余数 ).


• Principle A:• Step 1: Divide the decimal number by

the radix (number base).

• Step 2: Save the remainder (first remainder is the least significant digit －－最先得到的余数是最低有效位 )

• Step 3: Repeat steps 1 and 2 until the quotient( 商 ) is zero.

2.1.4 Converting From Decimal• Principle A:

•Example convert decimal number 14 to a base 4 number system

4 14 remainder = 2 (least significant)

4 3 remainder = 3 0 result = 32

2.1.4 Converting From Decimal• Principle A:

•Example convert decimal number 14 to a base 2 number system

2 14 remainder = 0 (least significant)

2 7 remainder = 1 2 3 remainder = 1 2 1 remainder = 1 0 result = 1110


• Principle B:– To convert a decimal fractional po

rtion( 小数部分 ) to another number system, multiply it by the radix ( 基数 ) and the whole number portion( 整数部分 ) of the result is saved as a significant digit of the result.


• Principle B:• Step 1: Multiply the decimal fraction by the

radix (number base)

• Step 2: Save the whole number portion of the result (even if zero) as a digit. The first result is written immediately to the right of the radix point ( 得到的第一个整数写在最靠近小数点左边的位置 ).

• Step 3: Repeat 1 and 2 until the fraction part of step 2 is zero

2.1.4 Converting From Decimal• Principle B:

•Example convert decimal number 0.125 to a base 4 number system

.125* 4 0.5 digit is 0

0.5* 4 2.0 digit is 2 result is 0.02


•Principle B:–Note: Some numbers are never-ending . That is, a zero is never a remainder.

– For example converting 0.15 to a base 4 number system.

–Why ?


• Example: convert 0.15 to a base 4 number system.

.15 0.4 * 4 * 4

0.6 digit is 0 1.6 digit is 1

0.6 * 4

2.4 digit is 2 result is 0.0212……


• 0.0212……– Numeric value:– 0*4-1 + 2*4-2 + 1*4-3 + 2*4-4

– =0 + 0.125 + 0.015625 + 0.0078125

– = 0.1484375– ≈ 0.15

2.1.5 Binary-Coded Hexadecimal


• How many different numbers can 4 binary numbers form ?


• BCH codehexadecimal digit BCH code 0 1 0000 0001 2 3 0010 0011 4 5 0100 0101 6 7 0110 0111 8 9 1000 1001 A B 1010 1011 C D 1100 1101 E F 1110 11119E = 1001 1110

BCD


• Principle:•We just use 4-bit binary to replace each hexadecimal digit and write down them in the same sequence. Note that there must be a space between each 4-bit binary.


• Example:Hexadecimal :2 A C . 5BCH :0010 1010 1100 . 0101

BCH :0010 1110 . 0101 1011Hexadecimal : 2 E . 5 B

2.1.6 Operation of Binary

All the numbers are unsigned.


• Add Operation Rules:• 0+0 = 0• 0+1 = 1• 1+0 = 1• 1+1 = 0 submit( 进位 ) 1• 1+1+1 = 1 submit( 进位 ) 1


• Add Operation• 1101 + 1011

1 1 1 1 submit bit 1 1 0 1 add bit

+ 1 0 1 1 added bit 1 1 0 0 0 result


• Subtract Operation Rules• 0-0 = 0• 1-1 = 0• 1-0 = 1• 0-1 = 1 borrow( 借位 ) 1

• Binary addition and subtraction get the same results on the unit position


• Subtract Operation• 11000100 – 00100101• 1 1 1 1 1 1 ( 借位 )

1 0 1 1 1 0 1 ( 借位后的被减数 ) 1 1 0 0 0 1 0 0 ( 被减数 )

- 0 0 1 0 0 1 0 1 ( 减数 ) 1 0 0 1 1 1 1 1


• Multiply Operation Rules• 0*0 = 0• 0*1 = 0• 1*0 = 0• 1*1 = 1


• Multiply Operation 1111 * 1101 1 1 1 1 * 1 1 0 1

1 1 1 1

0 0 0 0 1 1 1 1 1 1 1 1

1 1 0 0 0 0 1 1


• Divide Operation 0 0 0 1 1 1 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 1 0 0 1 0

1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 0 1 1 0 1 1 1 1 1 0 1 1 0 1 0 1 0


• All the numbers are unsigned , so they are positive.

• How can we represent negative number in binary in a computer?

2.1.7 Three Forms of Binary Data


• Signed data can be represented in three forms.

•Sign magnitude( 原码 )•Radix-1 complements ( 反码 )•Radix complements ( 补码 )

• We use byte-sized data in our examples.

2.1.7 Three Forms of Binary Data• Sign magnitude ( 原码 )

• 0 0 0 0 0 0 0 0

105 ---- 0 1101001-105 ---- 1 1101001

0 positive1 negative Numerical value (-127~+127)


• sign magnitude ( 原码 ) 105 ---- 0 1101001-105 ---- 1 1101001

• Radix-1 complements ( 反码 )• 105 ---- 0 1101001• -105 ---- 1 0010110• 0 ---- 0 0000000• -0 ---- 1 1111111

2.1.7 Three Forms of Binary Data• Radix complements ( 补码 )

反码 radix-1 complement 补码 radix complement 105 0 1101001 0 1101001-105 1 0010110 1 00101110 0 0000000 0 0000000-0 1 1111111 0 0000000


• Example: -8• +8=00001000• 1000 (write number to

first 1)• 1111 (invert the remaining

bits)• -8= 1111 1000


• － 25

• ＋ 25 ＝ 0001 1001• － 25 （补）＝ 1110 0111

• － 39

• ＋ 39 ＝ 0010 0111• － 39 （补）＝ 1101 1001


• Negative signed numbers are stored in the two’s complement form. ( 负数以 2 进制的补码形式存储 )

• A simpler technique for two’s implement:– Write the number exactly as it appears from r

ight to left until the first one. Write down the first one, and then invert all the remaining bits.


• If we use radix complement to represent a negative data, we can change subtract operation to add operation.

Z = X – Y equal to Z 补 =X 补 + (-Y) 补


• Z = X – Y equal to Z 补 =X 补 + (-Y) 补

• Example 1: 64 -10 64-10 = 64 补 + (-10) 补

64 补 = 01000000 (-10) 补 =11110110 01000000+ 11110110 (right) 100110110 result = 00110110 = 54

Nature lost

2.1.7 Three Forms of Binary Data• Example 2: 34-68

• 34-68 = 34 补 +( -68) 补• 34 补 =00100010 ( -68) 补 =10111100 00100010 + 10111100 11011110 result = 11011110 = -34(right)

11011110( 补 )= 11011101( 反 ) = - 00100010( 原 ) =-34-1

• 10001101( 补 ) • = 10001100( 反 ) = - 01110011( 原 )• =-115


• You can turn to the reference book “Microcomputer Principle and Application” 《微机原理和应用》 page 9 for details.


• Z = X – Y equal to Z 补 =X 补 + (-Y) 补

• Example 1: 64 -10 64-10 = 64 补 + (-10) 补

64 补 = 01000000 (-10) 补 =11110110 01000000+ 11110110 (right) 100110110 result = 00110110 = 54

Nature lost


• Example 3: 127 +8127= 01111111 8=00001000

01111111+ 00001000

10000111 result = 135

OverflowError!

• 加法运算溢出– 当次高位向最高位产生进位，而最高位不向

前产生进位时， O = 1( 溢出）– 当次高位向最高位无进位，而最高位向前有

进位时， O ＝ 1 （溢出）– 当最高位向前有进位时， C=1, 否则 C ＝ 0

– 0 1 1 0 0 1 0 0 – ＋ 0 1 1 0 0 1 0 0– 1 1 0 0 1 0 0 0 O=1 C=0




– 0 1 1 0 0 1 0 0 – ＋ 0 1 1 0 0 1 0 0– 1 1 0 0 1 0 0 0 O=1 C=0

次高位最高位




– 1 0 1 0 1 0 1 1 – ＋ 1 1 1 1 1 1 1 1– 1 1 0 1 0 1 0 1 0 O=0 C=1

2.2 Computer Data Formats

2.2 Computer Data Formats

• 2.2.1 ASCII Data• 2.2.2 BCD Data• 2.2.3 Byte-Sized Data• 2.2.4 Word-Sized Data• 2.2.5 Doubleword-Sized Data• 2.2.6 Real Numbers

2.2.1 ASCII Data

2.2.1 ASCII Data

• ASCII (American Standard Code for Information Interchange) represent alphanumeric characters in the memory of a computer system.

• ASCII use 8-bit(00H—FFH) represent alphanumeric characters in computer.

2.2.1 ASCII Data

• Table 1-7 on page 35 shows what is 00H—7FH representing.

• Table 1-8 on the same page shows extended ASCII code. Including what is 8FH—FFH representing.

• How to represent characters in other countries?

2.2.1 ASCII Data

• Many Windows-based applications use the Unicode system to store alphanumeric data and thus resolve the problem.

• Unicode system use 16-bit data to represent character.

• 0000H—00FFH is the same as ASCII code

• 0100H—FFFFH is used to store other special characters from other countries.

2.2.2 BCD Data

2.2.2 BCD Data

• BCD codedecimal digit BCD code 0 1 0000 0001 2 3 0010 0011 4 5 0100 0101 6 7 0110 0111 8 9 1000

1001

• 1MB 1M BYTE

8 Bit

8 Bit

2.2.2 BCD Data

• BCD data has two forms. One is packed and the other is unpacked.

• In all case, store the least-significant data first.( 规则 : 先存储最低有效位 )

• Packed BCD data are stored as two digits per byte 37-----0011 0111-----37H

• Unpacked BCD data are stored as one digit per byte

37-----0000 0011 0000 0111 -----0307H

2.2.3 Byte-Sized Data


• The weights of each binary-bit position

• Unsigned byte

• Signed byte

128

64 32 16 8 4 2 1

-128 64 32 16 8 4 2 1

Radix complements ( 补码 )

• -1 1111 1111• -2 1111 1110• -3 1111 1101• -4 1111 1100• ……• -127 1000 0001• -128 1000 0000


• One byte contains 8 bits.• Byte-sized data are stored as

unsigned and signed integers.• Unsigned integers range in value

from 00H to FFH (0-255)• Signed integers range in value

from -128 to +127.


• Example of defining a byte-sized data:– unsigned– 0000 FE DATA1 DB 254– 0001 87 DATA2 DB 87H– 0002 47 DATA3 DB 71– Signed– 0003 9C DATA4 DB -100– 0004 64 DATA5 DB +100

Store address

Store value

All the data formats are based on a

8088 microprocessor, whose data width

is 8 bits.

2.2.4 Word-Sized Data


• One word contains 16 bits. That is it is formed with two bytes of data.

• Word-sized data are stored as unsigned and signed integers. The difference between them is the weight of the highest position. For unsigned integers it is 32768 while signed is -32768.

• In Intel microprocessor, the least significant byte is stored in the lowest-numbered memory location, and the most significant byte is stored in the highest-numbered memory location.

• DATA DW 1234H 3002H

3001H3000H2FFFH


12H34H

Memory width :

8 bit


• Example:• Unsigned• 0000 09F0 DATA1 DW 2544• 0002 87AC DATA2 DW 87ACH• 0004 02C6 DATA3 DW 710• Signed• 0006 CBA8 DATA4 DW -13400• 0008 00C6 DATA5 DW +198

2.2.5 Doubleword-Sized Data


• One doubleword contains 32 bits. That is it is formed with two words or four bytes of data.

• Doubleword-sized data are stored as unsigned and signed integers. The difference between them is the weight of the highest position.

• In Intel microprocessor, the least significant byte is stored in the lowest-numbered memory location, and the most significant byte is stored in the highest-numbered memory location.

• DATA DD 12345678H

3003H

3002H3001H3000H

DATA DD 5678H ???


12H

34H

56H

78H


• Example :

• Unsigned• 0000 0003E1C0 DATA1 DD 254400• 0004 87AC1234 DATA2 DD

87AC1234H• 0008 00000046 DATA3 DD 70• Signed• 000C FFEB8058 DATA4 DD -1343400• 0010 000000C6 DATA5 DD +198


• Note ： Byte-sized, word-sized, doubleword-sized data are standard form. We can use DB to define those nonstandard data.

• Example : Define 123456H• DB 56H,34H,12H• DD 123456H

What is the difference?

2.2.6 Real Numbers

2.2.6 Real Numbers

• In high-level language we often encounter number like 24.5. We call it real number or floating-point number( 浮点数 ).

• As we have noticed, 24.5 (2.45*101) contained two parts: significand ( 尾数 ) and exponent ( 指数 ).

• 4-byte real number is called single-precision and 8-byte double-precision.

2.2.6 Real Numbers

• Single-precision using a bias of 7FH

• Double-precision using a bias of 3FFH

S31

Exponent30-23

Significant22-0

S63

Exponent62-52

Significant51-0

2.2.6 Real Numbers

• The exponent is stored as bias exponent.

• For single-precision: bias exponent = exponent + 7FH

• For double-precision:bias exponent = exponent + 3FFH

2.2.6 Real Numbers

• Table 1.10 single-precision number(P41)

decimal binary normalized sign Biased exponent

significant

+12 1100 1.1*23 0 0111,1111+11

100 0000 0000 0000 0000 0000H

-12 1100 -1.1*23 1 10000010

+100 1100100

1.1001*26

0 01111111+110

100 1000 0000 0000 0000 0000H

2.2.6 Real Numbers

• Define single-precision data(32 bit)•DD / REAL4

• Define double-precision data(64 bit)

•DQ / REAL8

• Define extended precision data•REAL10

2.2.6 Real Numbers

• Examples:• Single-precision 0000 3F9DF3B6 NUMB1 DD 1.2340004 C1BB3333 NUMB2 REAL4 -23.4

• Double-precision0008 405ED9999999999A NUMB3 DQ(REAL8)

123.4

• Extended-precision0010 4005F6CCCCCCCCCCCCCD NUMB4

REAL10 123.4

2.2.6 Real Numbers

• Two exceptions to the rules for floating-point numbers.

• 0.0 is stored as all zeros.

• Positive infinity.

• Negative infinity.

S0

Exponent30-23 (ALL ONES)

Significant 22-0 (ALL ZEROS)

S1

Exponent30-23 (ALL ONES)

Significant 22-0 (ALL ZEROS)

S0

Significant 22-0 (ALL 0)

Exponent30-23 (ALL 0)

1-2The Intel Microprocessors

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