Download Course Outline Preliminaries – 1 Preliminaries – 1b

Course Outline
CMPE328 Microprocessors
(Spring 2008-09)
Preliminaries and
Introduction to Computing
Credits: ( 4 / 1 / 0 ) 4 Prerequisites: CMPE224
Co-requisites: None
0- Introduction to computing: Inside the computer, CPU-RAM-ROM.
1- 80x86 microprocessor: short history, registers,
8086 assembly: mov and add instructions, program segments, logical and physical addresses,
stack, push, pop, flag register, addressing modes.
2- Assembly Language Programming: Directives, .asm, .lst, .obj, .map, linking, and .exe files,
control transfer instructions, data types and data definition.
3- Arithmetic Logic Instructions: unsigned multiplication and division, unsigned, signed, bcd,
packed-bcd and ascii number conversion, rotate and shift instructions.
4- Bios and DOS programming: bios display and keyboard interrupts, int 21h dos function calls.
5- Macro definitions: mouse button and cursor position.
6- Signed Arithmetics: Shift, Rotate, and String Manipulations.
9- 8088 PC/XT expansion slot, 80286 and the ISA bus, Memory and memory interfacing:
EPROM, SRAM and DRAM devices, address decoding circuits,
10- ISA bus memory interfacing. Memory mapped and Isolated I/O methods and device
interfacing: ISA bus I/O address decoding and simple I/O ports,
11-12 Programmable Peripheral Interface 8255 and LED, 7-segment-display, switch, button,
keypad, stepper motor interfacing. D/A converters, A/D converters.
17- Serial Data Communication and 8251 USART.
14- Hardware Interrupts: NMI and INTR pins, interrupt servicing and TSR programs.
Lab 15%, Quiz 25%, MT 25%, Attendance 5%, Final 30%, Total 100%
CMPE328 Spring 2007-08
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Section 0.1
Preliminaries – 1
Preliminaries – 1b
Introduction to computing (reading assignment)
Introduction to computing (reading assignment)
„
Numbering Systems
… Decimal
„
„
and Binary number systems
… Example:
Decimal-Binary and Binary-Decimal Conversion
… Hexadecimal
System
… Unsigned numbers, Signed numbers, Sign-bit,
and two’s complement operation.
Addition and Subtraction with binary and
hexadecimal numbers
„ Overflow condition in signed arithmetics.
„
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
Decimal to Binary Conversion
3
Convert 2510 to binary.
LSB
… Solution:
25 =12×2 + 1
12 = 6×2 + 0
6 = 3×2 + 0
3 = 1×2 + 1 MSB
1 = 0×2 + 1
Result is 2510= 110012.
CMPE328 Spring 2007-08
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25
12
6
3
1
LSB
1
0
0
1
1
MSB
4
Preliminaries – 1c
Preliminaries – 1d
Introduction to computing (reading assignment)
Introduction to computing (reading assignment)
„
Decimal to Binary Conversion of Fractions
… Example:
„
Convert 0.62510 to binary.
… Example:
… Solution:
Convert 110010.1110102 to Hexadecimal.
… Solution:
Split the binary number to 4-bit groups,
starting from the binary-point.
Complete the missing bits with zeros.
0 0 11 0010.1110 10 0 0 Æ 32.E816 .
0.625
1.250 1 Å MSB
.500 0
1.0
1
.0
0 Å LSB
Result is 0.62510= 0.10102
… If a number has both integer and fraction parts,
convert them separately and combine the results.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
Binary to Hexadecimal Conversion.
5
CMPE328 Spring 2007-08
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Section 0.1
Preliminaries – 1e
Arithmetic on Binary Numbers
Introduction to computing (reading assignment)
„
Carry to the Next Digit
„
Hexadecimal to Decimal Conversion.
Addition: 0+0=0, 0+1=1, 1+0=1, 1+1=1 1.
…
… Example:
carry: 1 1 0 0 0 0 1 1 0
End Carry
Convert 32.E816 to Decimal.
… Solution:
32.E816 = 3×16+2 + 14/16+8/256.
= 50.90625
„
…
…
„
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01001011
+ 11000001
= 1 00001100
Negation: For signed binary numbers
…
CMPE328 Spring 2007-08
01001011 + 11000001
Complement
– N = 2-complement of N.
– 01100101 = 10011010 + 1 = 10011011
– 01101100 = 10010011 + 1 = 10011100
Subtraction: N – M = N + (– M)
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
Copy
Complement
Copy
Sign Bit
8
Section 0.1
Preliminaries – 2
Sign Bit and Overflow
MSB of signed binary numbers is the sign
bit. Negative numbers have Sign-bit=1.
„ When adding two positive numbers, or two
negative numbers, resulting sign may
come out inconsistent to the operands.
„
for 8-bit registers (-96) + (-106) Æ
10100000 + 10010110 = 1 00110110
Sign bit of result is positive Æ overflow.
… Example:
Introduction to Computing (reading assignment)
Coding of Numbers and Text
„ Coding of Decimal Numbers
…
„
Coding of Binary Numbers
…
„
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ASCII coding: binary/decimal/hex ASCII
Other Coding methods
…
CMPE328 Spring 2007-08
8-bit, 16-bit, 32-bit numbers and sign bit
Coding the text strings.
…
„
Packed BCD and Unpacked BCD representation
Coding of positions (Gray coding)
CMPE328 Spring 2007-08
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Section 0.1
Preliminaries – 2b
Introduction to Computing (reading assignment)
Introduction to Computing (reading assignment)
Packed BCD
…
„
120810 Æ 1208pBCD = 120816 = 0001 0010 0000 10002
each decimal digit occupies one nibble.
Unpacked BCD
…
Section 0.1
Preliminaries – 2a
Coding of Decimal Numbers
„
10
120810 Æ 1208uBCD = 0102000816
= 00000001 00000010 00000000 000010002
each decimal digit occupies one byte.
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Coding of Numbers and Text
„
ASCII coding of characters. Each character is
coded by a single byte.
Examples
-0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -A -B -C -D -E -F
“4” Æ 3416 .
2! " # $ % & ' ( ) * + , - . /
3- 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
“a” Æ 6116 .
4- @ A B C D E F G H I J K L M N O
“A” Æ 4116 .
5- P Q R S T U V W X Y Z [ \ ] ^ 6- ` a b c d e f g h i j k l m n o
“[“ Æ 5B16 .
7- p q r s t u v w x y z { | } → ←
backspc Æ7F16
CMPE328 Spring 2007-08
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Section 0.1
Preliminaries – 2c
Preliminaries – 2d
Introduction to Computing (reading assignment)
Introduction to computing (reading assignment)
ASCII Coding of Numbers (for display and print)
„
Binary/decimal/hex ASCII
Example: Convert 50 to decimal and binary
ASCII string.
-0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -A -B -C -D -E -F
„
Decimal Solution
2! " # $ % & ' ( ) * + , - . /
3- 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
“50”=353016
4- @ A B C D E F G H I J K L M N O
„
Binary Solution
5- P Q R S T U V W X Y Z [ \ ] ^ 6- ` a b c d e f g h i j k l m n o
“110010” =
7- p q r s t u v w x y z { | } → ←
31313030313016
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CMPE328 Spring 2007-08
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„
Gray coding.
„
0 Æ 0000, 1 Æ 0001, 2 Æ 0011, 3 Æ 0010,
4 Æ 0110, 5 Æ 0100, 6 Æ 0101, 7 Æ 0111,
8 Æ 1111, 9 Æ 1110, 10 Æ 1100, 11 Æ 1101,
12 Æ 1001, 13 Æ 1011, 14 Æ 1010, 15 Æ 1000,
One bit changing at a time when counting up or down.
…
…
It eliminates the transient glitches in counting.
Mainly used in positional sensors.
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Section 0.2
Preliminaries – 3
Preliminaries – 4
Introduction to computing (reading assignment)
Introduction to computing (reading assignment)
Inside the Computer
„ Data types and registers
„
… Bit,
„
… Instruction
Nibble, Byte, Word
„
Internal Organization of Computers
… Inside
„
„
„
„
CPU:
„
„
Execution of Instructions
Next Instruction is pointed by Program Counter.
… Program
„
Address-bus, Control-bus
Set
Operations defined on Registers and Memory
Locations. Mostly they are carried in ALU.
… Serial
Instruction Decoder,
Registers
Arithmetic Logic Unit
Internal Data/Address buses.
… Data-bus,
Stored Program Concept
Flow control: Jumps and Branches
ALU condition is stored in Flags, CF, ZF, PF, OF
Connects CPU, Memory and I/O devices
CMPE328 Spring 2007-08
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CMPE328 Spring 2007-08
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Section 0.2
Section 1.1
Preliminaries – 5
8086 Architecture – 1
Introduction to computing (reading assignment)
How a CPU works
„ Instruction is fetched from Memory
„ Decoded by Instruction Decoder into control
signals to determine the paths and operations
for several clock cycles.
„
… Their
instruction sets and register architecture
are upward compatible
(4004 programs run in 8008, etc.)
… They are typical Complex Instruction Set
Computers
… Register
to register move,
… Arithmetic Logic operations on Registers
… Move between Registers and Memory.
… Flow Control instructions: Jump and Branches
… Bit and Status manipulation instructions
CMPE328 Spring 2007-08
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Evolution of Intel Processors
4004 Æ 8008 Æ8080 Æ8085 Æ 8086
Æ 8088 Æ 80186 Æ 80286Æ 80386 ...
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CMPE328 Spring 2007-08
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Section 1.2
Section 1.2
8086 Architecture – 2
16-bit
8-bit
8-bit
Registers of 8086
„ General: AX, BX, CX, DX (AXÅAH:AL)
„ Pointer: SP (stack) and BP (base) 16-bit,
„ Index: SI (source) and DI (destination) 16-bit,
„ Segment: CS, DS, SS and ES 16-bit,
„ Program Counter: IP (instruction-ptr) 16-bit,
„ Flags: FR register
(CF, PF, AF, ZF, SF,TF, IF, DF, OF)
Carry
Parity
CMPE328 Spring 2007-08
Zero
Sign
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Overflow
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18
8086 Architecture – 3
Internal Block Diagram of 8086 CPU registers
CMPE328 Spring 2007-08
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Section 1.2
Section 1.2
8086 Data Address Segmentation
„
„
16-bit Registers Æ 64k = 65536 locations
Segmented Addressing Æ 20-bit = 1M locations.
… DS
„
segment: 16-bit,
it points the start of 64k data block
… Physical
Address of a data at Logical Address
DS:DATA1 is obtained by
PAddr = 16×DS+DATA1
Example: DS contains 0100H, and DATA1 = 0123H,
then Physical Address of DS:DATA1 is
01000 H + 0123 H = 01123 H in the Memory.
CMPE328 Spring 2007-08
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8086 Code Address Segmentation
„
CS segment: 16-bit,
… It
points the start of 64k instruction code block
… Physical Address of the code to execute next
is at
CS:IP Æ PA = CS×10h + IP
Example: CS and IP contain 0100H, and 4321H.
Then, Physical Address of next instruction to
be executed is
01000H + 4321H = 05321H in the Memory.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
Section 1.3 and 2.1
Assembly Programming Format
Assembly program source is a series of lines
[label:] [mnemonic [operands] ] [;comments]
„ Each line carries a label, a comment, a directive,
or a program instruction.
„ Directives defines symbols, macros, code
placement and provides conditional assembly .
„ Program instructions and directives are
written in mnemonics and operands,
„ Comments start after a semicolon “;”, and
provide information for humans.
Using symbolic names for registers and constants
makes the program readable.
CMPE328 Spring 2007-08
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22
Section 2.1
Assembly Programming Example
Example Code
.MODEL SMALL
.STACK 64
.DATA
DATA1 DB 52H
DATA2 DB 29H
SUM
DB ?
.CODE
MAIN PROC FAR
MOV AX,@DATA
MOV DS,AX
MOV AL,DATA1
MOV BL,DATA2
ADD AL,BL
MOV SUM,AL
MOV AH,4CH
INT 21H
MAIN ENDP
END MAIN
CMPE328 Spring 2007-08
Directives to assembler
setting the type of the executable,
its stack and data segments
Data placed into Memory, in the Data
Segment.
Directive for the Start of Code Segment
;Program Entry
;Data Seg. Addr
;First Number
;2ND Number
; AL <- AL+BL
; SUM <- AL+BL
; return to DOS
Program starts at this point.
Comments start after a semicolon
End of Program
End of Assembly Source
Dr.Mehmet Bodur, EMU-CMPE
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Section 1.2
Section 1.2
MOV Instruction – 1
„
MOV Instruction – 2
… Examples
MOV destination, source
copies the source into the destination.
… Examples
of MOV instruction with 8-bit data
Immediate values cannot be moved into segment registers
(CS, DS, ES or SS). Only nonsegment registers can
move into segment registers.
8-bit data can be moved between all registers,
except the flag register FR.
CMPE328 Spring 2007-08
of MOV instruction with 16-bit data
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CMPE328 Spring 2007-08
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Section 1.2
MOV Instruction – 3
… ILLEGAL
„
MOV instructions
Moving too large values into a register
MOV CL, 3B7H ; immediate value larger than 8-bit
MOV AX, 54321H ; immediate value over 16-bit
„ Moving immediate value into a segment register
MOV DS, 1020H ; not allowed
MOV CS, 4321H ; not allowed
MOV AX, 1020H ; immediate value goes into AX
MOV DS, AX ; then it is moved into DS
„ Register cannot moved into a value
MOV 1234H, AX ; destination cannot be a value.
„
CMPE328 Spring 2007-08
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ADD Instruction – 1
27
ADD destination, source
Adds source onto destination.
… Examples
Executing the code results
AL = 25H+34H=59H.
BL = 34H
Register AX is called Accumulator, but ADD
accepts all general purpose registers for
source and destination.
CMPE328 Spring 2007-08
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Flags after 8-bit ADD
ADD effects CF, PF, AF, ZF, SF and OF.
Flags after 16-bit ADD
ADD instruction changes CF, PF, AF, ZF, SF and OF.
Show how the flag register is affected by
Solution:
1
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CMPE328 Spring 2007-08
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CMPE328 Spring 2007-08
Looping on Zero Flag ZF
„
„
„
„
„
INC reg ; increments contents of reg.
DEC reg ; decrements contents of reg.
INC or DEC instructions effect zero flag ZF.
JZ Label ; gives jump to Label on ZF=1.
JNZ Label ; makes jump to Label on ZF=0.
MOV CX, 1000
LOOP1000: DEC CX
JNZ LOOP1000
loops for DEC CX and JNZ for 1000 times, making a
small delay in execution.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
„
„
„
31
30
Looping on Zero Flag ZF
„
… Example:
Dr.Mehmet Bodur, EMU-CMPE
ZF is changed by DEC CX before the JNZ instruction tests it.
CX is initially 5.
After the fifth looping DEC CX makes CX = 0, and ZF=1.
Consequently, JNZ ADD_LP does not result in a jump, and
looping comes to end.
CMPE328 Spring 2007-08
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Memory Organization
„
A microprocessor system needs several kinds of
memory devices.
Only Memory (ROM) keeps data even if power turns
off. BIOS and Boot up program shall stay in ROM.
… Read-Write Memory (RAM) is a large block of memory
to load and execute the operating system and user
programs, and to keep data before saving to the mass
storage devices. Data in a RAM device is lost whenever
power is turned off.
… Mass Storage Memory is peripheral memory system,
usually interfaced through I/O ports, or external buses.
… Read
CMPE328 Spring 2007-08
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Memory System Addressing
„
„
„
„
8086 has 20-bit address space (=1024 kBytes).
A 16-bit Instruction Register (IR) allows
addressing 64k byte effective address range.
A 16-bit Code Segment registers (CS)
determine the start address of this 64k block.
The same segmentation mechanism works also
for data addressing, and for the stack
CMPE328 Spring 2007-08
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Section 1.5
8086 Stack
Memory Organization in DOS – 2
„
„
„
CPU keeps the return addresses and temporary data in a
section of Read-Write Memory.
A stack pointer SP, and the stack-segment SS together
form the physical address of the top-of-stack.
Examples:
PUSH AX stores AX into the stack.
Let’s denote the initial contents of SP by SP0.
After the execution of PUSH AX,
Mem-Address(SS:[SP0] – 1) Å AH ;
Mem-Address(SS:[SP0] – 2) Å AL ;
SP Å SP – 2
(i.e., AX is saved to stack in little-endian convention,
and stack pointer is decremented by 2. Stack has
grown from higher addresses to lower addresses.)
PC-DOS Operating
system uses 640k
Bytes main memory.
A0000H to BFFFFH is
reserved for video
display.
C0000H to FFFFFH is
occupied by BIOS
ROM.
CMPE328 Spring 2007-08
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CMPE328 Spring 2007-08
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Section 1.5
Section 1.5
8086 Stack – 2
Example:
Assume that SP=1236, AX=24B6, DI=85C2,
and DX=5F93, show the contents of the
stack after each of the following push
instructions is executed.
8086 Stack – 3
Example
POP AX restores AX from the stack, and shrinks the stack
2-bytes .
PUSH AX
PUSH DI
PUSH DX
If we denote the initial contents of SP by SP0.
After the execution of POP AX,
Even if not written
explicitly an address
AL Å SS:[SP0] ;
denotes the contents
of the addressed
AH Å SS:[SP0] + 1 ;
memory location
SP Å SP + 2
(i.e., AX is restored from the stack in little-endian
convention, and stack pointer is decremented by 2. Stack
has shrunk from lower addresses to higher addresses.)
CMPE328 Spring 2007-08
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CMPE328 Spring 2007-08
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Section 1.5
Section 1.6
8086 Stack – 4
Example:
Assume that SP=18FA, AX=24B6,
DI=85C2, and DX=5F93, show the
contents of the stack after each of the
following pop instructions is executed.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
POP CX
POP DX
POP BX
8086 Addressing Modes – 1
Register
„ Immediate
„ Direct
„ Register Indirect
„ Based Relative
„ Indexed Relative
„ Based Indexed Relative
„
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CMPE328 Spring 2007-08
AL, AH, BX, DI, SP
0123H, 100, 10010100B
[0123H]
[BX]
Memory
[BX] + 12
Locations
[DI] + 12
[BX] [DI ] + 4
Dr.Mehmet Bodur, EMU-CMPE
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Section 1.6
8086 Addressing Modes – 2
Register
„ Immediate
„ Direct
„ Register Indirect
„ Based Relative
„ Indexed Relative
„ Based Indexed Relative
„
CMPE328 Spring 2007-08
MOV AH, BL
MOV AX, 0123H
MOV DL, [0123H]
MOV AL, [BX]
MOV AL, [BX] + 12
MOV AL, [DI] + 12
MOV AL, [BX][DI] + 12
Dr.Mehmet Bodur, EMU-CMPE
Register Addressing
Registers are addressed by their names.
„ Register names are keywords.
„
Source Register
Destination Register
Examples
… MOV BX,DX ; copy the contents of DX into BX
… MOV ES,AX ; copy the contents of AX into ES
Segment registers are loaded only through AX.
… ADD AL,BH
;add the contents of BH to the contents of AL
Source and destination registers must match in size.
Register name addresses the contents of that register.
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CMPE328 Spring 2007-08
Immediate Addressing
A constant is used as the source operand. This
constant number is immediate in the instruction.
Examples
… MOV CX,625
; load the decimal value 625 into CX
… MOV BL,40H
; load 40H into BL.
Use AX to load immediate numbers into a segment register.
… MOV AX,2550H ; load hexadecimal 2550 into AX
MOV DS, AX
; move AX into DS.
CMPE328 Spring 2007-08
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42
Direct Addressing
„
„
Dr.Mehmet Bodur, EMU-CMPE
A memory address is directly specified
immediate in the instruction.
Examples
… MOV DL, [2400] ; moves contents of DS:2400H into DL
… MOV AL,40H
; loads 40H into AL.
MOV [2800], AL ; moves AL to location DS:2800H
… MOV AX,1234H ; loads the value 1234H into AX.
MOV [2800], AX ; moves AX to location DS:2800H
; in the LITTLE ENDIAN convention, i.e.,
; DS:2800H Å 34H , DS:2801H Å12H
CMPE328 Spring 2007-08
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44
Register Indirect Addressing
„
A memory address is directly specified by a
register instruction.
Based Relative Addressing
„
DSx10h+BX
Examples
… MOV AL, [BX+10] ; DS:BX+10 Æ AL
… MOV AL, [BX]+10 ; DS:BX+10 Æ AL (alternate form)
… MOV CX, [BP]+4 ; SS:BP+ 4 Æ CL and
; SS:BP+ 5 Æ CH
Examples
… MOV AL, [BX] ; moves contents of the memory location
; pointed by DS:BX into AL
… MOV [SI], AX ; DS:SI Å AL ; and
; DS:SI+1 ÅAH .
Even if not written
explicitly, an address
denotes the contents of
; Byte ordering is Little Endian.
the addressed memory
Pointers BP and SP use the segment SS, BX uses DS.
Byte ordering for words and double-words is Little Endian.
BP+4 is called the effective address
SS:BP+4 forms the physical address PA=SSx10h+BP+4
location
CMPE328 Spring 2007-08
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CMPE328 Spring 2007-08
A memory address is specified by an index
register and a relative immediate address.
Even if not written
DI and SI are index registers. explicitly
an address
„
denotes the contents
of the addressed
Examples
memory location
… MOV AL, [DI+10] ; DS:DI+10 Æ AL
… MOV AL, [SI]+10 ; DS:SI+10 Æ AL (alternate form)
… MOV CX, [DI]+4 ; DS:DI+4 Æ CL and
; DS:DI+5 Æ CH
; BX, DI and SI holds offset address, and uses DS.
; Byte ordering is Little Endian.
CMPE328 Spring 2007-08
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A memory address is specified by an index
register and a relative immediate address.
DI and SI are index registers.
Examples
… MOV CL, [BX] [DI]+8 ; DS:BX+DI+8 Æ CL
… MOV CL, [BX+DI+8] ; DS:BX+DI+8 Æ CL (alternate)
… MOV CX, [BP] [SI]+4 ; SS:BP+SI+4 Æ CL, SS:BP+SI+5 Æ CH
; Byte ordering is Little Endian.
„
47
Dr.Mehmet Bodur, EMU-CMPE
Based Indexed Relative Addressing
Indexed Relative Addressing
„
A memory address is specified by a base register
and a relative immediate address.
BX and BP are the only base registers.
By default, BP and SP uses SS.
BX, DI and SI holds offset address, and uses DS.
CMPE328 Spring 2007-08
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Section 1.6
Section 1.6
8086 Default Segments
Even if not written explicitly an address denotes
the contents of the addressed memory location
Default Segments for addressing registers.
CS
DS
SS
Segment Registers:
IP
BX, SI, DI BP, SP
Offset Registers
Examples,
MOV AL, [BX] ;
AL <- DS:BX
MOV AL, [BX]+16 ;
AL <- DS:BX +16
MOV AX, [BP][SI]+24 ; AX <- SS:BP +SI +24
i.e., two bytes starting from the address
(SS×10h+BP+SI+24) moves into AX
in little-endian convention
(Lower address carries LSB, higher keeps MSB).
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
8086 Segment Override
Examples,
MOV [0122H], AL
MOV CS:[0122H], AL
MOV AL, [BX]
MOV AL, ES:[BX]
MOV AL, [BX]+16
MOV AL, CS:[BX]+16
MOV AX, ES:[BP][SI]+24
MOV SS:[BX]+16, AH
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CMPE328 Spring 2007-08
; DS:0122H <- AL
; CS:0122H <- AL
; AL <- DS:BX
; AL <- ES:BX
; AL <- DS:BX+16
; AL <- CS:BX+16
; AX <- ES:BP +SI+24
; SS:BX+16 <- AH
Dr.Mehmet Bodur, EMU-CMPE
Section 1.5
Section 1.5
8086 Segment Overlapping
A segment points the starting address of a
64kBytes memory window.
Three segments, CS for code, DS for data,
and SS for stack, are usual components of
physical addressing of programs.
These segments can be fully or partially
overlapping.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
51
50
8086 Segment Wrap-Around
„
What happens to segment windows
if DS=FF20?
… FF200
… FFFFF is the first section of
segment window.
… 00000 … 0FF1F is the second section.
„
This kind of segment splits are a result of
segment wrap-around over 20-bit address
space.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
52
Section 1.5
8086 Flag Register
ALU Condition Flags (CF, PF, AF, ZF, SF, OF),
and CPU Control Flags are collected in the Flag
(or Status) Register. (See slides 29 and 30)
Summary
„
„
„
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
53
We have seen the architecture of Intel processor
family and the preliminary concepts for assembly
programming.
The addressing modes of Intel processors, the
program segments, and operations on the stack
are essential concepts in assembly
programming.
Next, we will see the tools for assembling a
program, and write simple assembly programs.
CMPE328 Spring 2007-08
Dr.Mehmet Bodur, EMU-CMPE
54