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Lecture 5: Computer Architecture

• Computer Architecture– Registers– Flags– Address Calculation

Computer Organization

Control Unit

ALU

Registers

Memory

CPU

Input Output

Components

• Control Unit –

• Arithmetic logical unit (ALU) –

• Registers –

• Buses –

8086 Block Diagram

CSDSSSES

SIDIBPSP

AXBXCXDX

AHBHCHDH

ALBLCLDL

Data Registers

Index Registers

ALU

Bus Interface

Unit

IP

Flags R

Control Unit

Instruction

Data

DataBus

AddressBus

Control BusClock

Interrupt

+5V

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RegistersGeneral Purpose

Registers

• Data registers, also known as general purpose registers: AX, BX, CX, DX

• Used for arithmetic operations and data movement

• Can be addressed as 16 bit or 8 bit values. For AX, upper 8 bits are AH, lower 8 bits are AL.

• Remember: when when a 16 bit register is modified, so are the corresponding 8 bit registers!

Example

• move 0001 1001 1110 0100 to AX

AX AH AL

15 0

7 0 7 0

AX

AH AL

0000 0000 0000 0000

AX

AH AL

0001 1001 1110 0100

• move 0011 1101 to AH

AX

AH AL

Special Attributes of GP Registers

• AX – accumulator

• BX – base

• CX – counter

• DX – data

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Segment Registers

• Segment registers are used as base locations for program instructions, data, and the stack.

• All references to memory involve a segment register as the base location.

Segment Registers, cont.

• CS – code segment

• DS – data segment

• SS – stack segment

• ES – extra segment

Index Registers

• Index registers contain the offsets of data and instructions.

• Offset:

• Index registers are used when processing strings, arrays, and other data structures.

Index Registers, cont.• BP – base pointer

• SP – stack pointer

• SI – source index

• DI - destination index

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Status and Control Registers

• IP – instruction pointer

• Flags –

Status Flags

• Carry flag (CF)

• Overflow flag (OF)

• Sign flag (SF)

Status Flags, cont.

• Zero flag (ZF)

• Parity

Addressing

• Address: a number referring to an 8-bit memory location

• Logical addresses go from 0 to the highest location

• Logical addresses require translation into physical addresses

• For Intel (8086):

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Pins on the 8086

1

2

3

4

5

6

7

8

9

10

11

12

14

13

15

16

17

18

19

20

40

39

38

37

36

35

34

33

32

31

30

29

27

28

26

25

24

23

22

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8086CPU

GND

D14/A14

D13/A13

D12/A12

D11/A11

D10/A10

D9/A9

D8/A8

D7/A7

D6/A6

D5/A5

D4/A4

D3/A3

D2/A2

D1/A1

D0/A0

NMI

INTR

CLK

GND

VCC

A15/D15

A16/S3

A17/S4

A18/S5

A19/S6

MN/MX

RD

RQ/GT0

R/GT1

LOCK

S2

S1

S0

QS0

INTA

TEST

READY

RESET

Addressing, cont.

• How do you have a 20-bit address with 16 bit registers?

Why Segment-Offset? Data Segmentaddressable memory on 8086

FFFFF

00000Example:DS = 0100h

data seg. start?data seg. end?

If BX contains theoffset:

BX = 005AhEA = ?

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Segment Register Combinations

• Code Segment – the CS register and IP (instruction pointer) are used to point to the next instruction.

• Stack – the SS register is used with the SP (stack pointer) or BP (base pointer)

• Data Segment – DS with BX, SI, or DI

• Extended Segment – BX, SI, or DI

More on Effective Addresses

• There’s more than one way to get the same effective address!

• Example:– CS = 147Bh– IP = 131Ah– EA = 147B0 + 131A = 15ACAh

or– CS = 15ACh– IP = 000Ah– EA = 15AC0 + 000A = 15ACAh

• If CS = 147B, what range of effective addresses can be referenced without changing the value in CS?

Homework 2• Two parts:

– Part 1: Use Debug to enter and run a simple machine code program

• convert input data into 2’s complement hex

• enter data at the correct address• enter program at the correct address• run the program

– Part 2: Write a simple machine code program, given pseudo-code

• these instructions should be similar to those in the Part 1 problem.

• enter and run the resulting program.

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Part I - Example Program Given below is a machine code program that calculates the sum of all the words in a given range of addresses in memory. The code expects that the lower bound of this range is specified in the BX register and the upper bound in the DX register. (BX holds the offset of the beginning of the data to be summed from the beginning of the data segment (DS). DX holds the offset of the last data element from the beginning of the data segment.) The sum gets stored in AX. The first 4 hex digits given on each linebelow represent the offset of the instruction from the beginningof the code segment. The digits after the dash are the machine code instructions. To the right are English explanations of the instructions.

0000 - 2BC0 subtract AX from itself (to make it 0) 0002 - 0307 add the word pointed to by BX to AX 0004 - 83C302 add 2 to BX (to point to the next word) 0007 - 3BD3 compare BX to DX

(compare sets internal flags that are used by subsequent jump instructions)

0009 - 7DF7 if DX >= BX, then jump back to the instruction at 0002

000B - B8004C this instruction and the next one return control to DOS

000E - CD21

Part 1

• Convert input data into 2’s complement hex– use the techniques you used on HW2

• Enter data at the correct address– DS holds the segment– Look at how you did this in lab!

• Enter program at the correct address– CS holds the segment– IP holds the offset– Again – look at the lab!

• Run your program!