Transcript بسم الله الرحمن الرحيم

‫بسم هللا الرحمن الرحيم‬
INTEL 8086
Microprocessor
1
8086 Memory Segmentation
Instruction Set
Assembly Language
8086 Addressing Modes
First Assembly Program
Assembly: Step by Step
2

Introduction

Memory Interface

8086 Microprocessor

8086 Registers

8086 Addressing Modes

8086 Instruction Set

Assembly Program Examples
3

The 8086 was Intel’s first 16-bit
microprocessor

This means that the 8086 has a 16-bit ALU.
The 8086 contains 20 address pins

Therefore, it has a main (directly
addressable) memory of 1 megabyte (220
bytes)
4

The 8086 was Intel’s first 16-bit
microprocessor. This means that the 8086
has a 16-bit ALU

The 8086 contains 20 address pins

It has a main (directly addressable) memory
of 1 megabyte (220 bytes)
5

The 16-bit word at the even address 0200016
is A10216

The 16-bit word stored at the odd address
3015116 is 462E316
6

The 8086 always reads a 16-bit word from
memory

This means that a word instruction
accessing a word starting at an even
address can perform its function with one
memory read
7

A word instruction starting at an odd
address, however, must perform two
memory accesses to two consecutive
memory even addresses, discarding the
unwanted bytes of each
8

For byte read starting at odd address N, the
byte at the previous even address N – 1 is
also accessed but discarded

Similarly, for byte read starting at even
address N, the byte with odd address N + 1
is also accessed but discarded
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
The 8086 systems have two “0.5 Mega x 8
bit” memory banks

Data-bytes associated with even-addresses
reside in low bank and odd-address reside
in high bank

Address pins A1 to A19 selects the storage
locations, whereas A0 and BHE’ pins are
used to enable high or low memory banks
10
.
11
.
12
.
13
.
14
.
15
.
16

A word instruction accessing a word
starting at an even address can perform its
function with one memory read

A word instruction starting at an odd
address, however, must perform two
memory accesses to two consecutive
memory even addresses
17

In a 8086-based microcomputer, Identify
each of the following instructions as
aligned or misaligned transfer for 16-bit
memory

Briefly explain for each case how data will
be transferred via which data pins (D15 D0) using A0 and BHE’ control pin
18

MOV
AH, [1015H]

MOV
BL, [1020H]

MOV
BX, [0077H]

MOV
CX, [0156H]
19

For the 8086, register names followed by
the letters X, H, or L in an instruction for
data transfer between register and memory
specify whether the transfer is 16-bit or 8bit
20
.
21
.
22

The 8086 is packaged in a 40-pin chip. A
single +5 V power supply is required

The clock input signal is generated by the
8284 clock generator driver chip

Instruction execution times vary between 2
and 30 clock cycles
23

The 8086 family consists of two types of
16-bit microprocessors, the 8086 and 8088

The main difference is how the processors
communicate with the outside world

The 8088 will have to do two READ
operations to read a 16-bit word from
memory
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
The 8086 uses a segmented memory
There are some advantages to working
with the segmented memory
 First, the 8086 has to deal with only 16-bit
effective addresses
 Second, because of memory segmentation,
the 8086 can be effectively used in timeshared systems

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
For example, in a time-shared system,
several users may share one 8086. Suppose
that the 8086 works with one user’s
program for, say, 5 milliseconds

Segmentation makes it easy to switch from
one user program to another
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
The 8086’s main memory can be divided
into 16 segments of 64K bytes each (16 x
64 KB = 1 MB)

A segment may contain codes or data

The 8086 uses 16-bit registers to address
segments
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
The 8086 includes on-chip hardware to map
or translate the two 16-bit components of a
memory address into a 20-bit address
called a “physical address”

Note that the 8086 contains 20 address
pins, so the physical address size is 20 bits
wide
28
.
29
.
30

The 8086 is divided internally into two
independent units: the bus interface unit
(BIU) and the execution unit (EU). The BIU
reads (fetches) instructions, reads
operands, and writes results

The EU executes instructions already
fetched by the BIU
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
The 8086 prefetches up to 6 instruction
bytes from external memory into a FIFO
(first-in-first-out) memory in the BIU and
queues them in order to speed up
instruction execution

The BIU contains a dedicated adder to
produce the 20-bit address
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
The bus control logic of the BIU generates
all the bus control signals, such as the
READ and WRITE signals

The BIU also has four 16-bit segment
registers: the code segment (CS), data
segment (DS), stack segment (SS), and
extra segment (ES) registers
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
All program instructions must be located in
main memory, pointed to by the 16-bit CS
register with a 16-bit offset contained in
the 16-bit instruction pointer (IP)

Note that immediate data are considered
as part of the code segment
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
The 20-bit physical stack address is
calculated from the SS and SP for stack
instructions such as PUSH and POP

The programmer can create a
programmer’s stack with the BP instead of
the SP for accessing the stack using the
based addressing mode
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
The DS register points to the current data
segment; operands for most instructions
are fetched from this segment

The 16-bit contents of a register such as the
SI or DI or a 16-bit displacement are used as
offsets for computing the 20-bit physical
address
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
The ES register points to the extra segment
in which data is stored

String instructions always use the ES and
DI to determine the 20-bit physical address
for the destination

The segments can be contiguous, partially
overlapped, fully overlapped, or disjointed
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
Every segment must start on 16-byte
memory boundaries

Typical examples physical addresses
starting at 0000016, 0001016, 0002016,
0003016, . . .

A physical memory location may be
mapped into one or more logical segments
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
A segment can be pointed to by more than
one segment register

For example, the DS and ES may point to
the same segment in memory
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.
40
.
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
The EU decodes and executes instructions

It has a 16-bit ALU for performing
arithmetic and logic operations

The EU has nine 16-bit registers: AX, BX,
CX, DX, SP, BP, SI, and DI, and the flag
register
42

The 8086 has 12 addressing modes, which
can be classified into five groups:
 Register and immediate modes
 Memory addressing modes (six modes)
 Port addressing mode (two modes)
 Relative addressing mode (one mode)
 Implied addressing mode (one mode)
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
Register Mode:
MOV AX, BX
MOV AH, BL

Immediate Mode:
MOV CX, 5062H
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

Memory Direct Addressing Mode:
MOV
BX , [1875H]
MOV
[1257H] , CX
Register Indirect Addressing Mode:
MOV
CX , [BX]
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
Based Addressing Mode:
 MOV
AX , 4 [ BX ]
 MOV
-4 [BP] , BX

The segment register is DS or SS

The content of BX or BP is unchanged
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
Indexed Addressing Mode:
 The effective address is calculated from the
sum of a displacement value and the
contents of register SI or DI
 For example,
MOV AX , VALUE [ SI ]
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
Based Indexed Addressing Mode:

The effective address is computed from the
sum of a base register (BX or BP), an index
register (SI or DI), and a displacement

For example,
MOV AX, 4 [ BX ] [ S I ]
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
String Addressing Mode:

This mode uses index registers

SI is assumed to point to the first byte or
word of the source string, and DI is
assumed to point to the first byte or word
of the destination when a string instruction
is executed
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
The SI or DI is automatically incremented
or decremented to point to the next byte or
word depending on DF

The default segment register for source is
DS, and it may be overridden; the segment
register used for the destination must be
ES, and can not be overridden
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
An example is:
MOVS WORD

If (DF) = 0, (DS) = 3000H, (SI) = 0020H, (ES)
5000H, (DI) = 0040H, (30020) = 30H, (30021)
= 05H, (50040) = 06H, and (50041) = 20H

After this MOVS, (50040) = 30H, (50041) =
05H, (SI) = 0022H, and (DI) = 0042H
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
Two I/O port addressing modes can be
used: direct port and indirect port
addressing

In either case, 8- or 16-bit I/O transfers
must take place via AL or AX respectively
IN AL , 02
-
IN AL , DX
OUT 25 , AL
-
OUT DX , AL
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
ADD a, b

ADC a, b

SUB a, b

SBB a, b

DEC reg/mem

NEG reg/mem

CMP a, b
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
MUL
reg/mem

IMUL
reg/mem

DIV
reg/mem

IDIV
reg/mem

CBW

CWD
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
NOT mem/reg

AND a, b

OR a, b

XOR a, b

TEST a, b
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
SHL mem/reg, CNT

SHR Mem/reg, CNT

ROL mem/reg, CNT

ROR mem/reg, CNT

RCL mem/reg, CNT

RCR mem/reg, CNT
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
What is the difference between MOV AX ,
[1200] , MOV AX , 1200 , and MOV [1200] ,
AX?

Initialize register CL to the value 3EH and
memory location 43H to the value 55D, and
write a program to exchange their contents
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
Write a 8086 assembly program to add AX
and BX and store the result in CX register

Write a 8086 assembly program to add AH
and CL registers, and store the result in
memory location 1040H
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
Write an assembly program to find the
larger of two unsigned numbers stored in
DX and BP registers and store the result in
AX register
63

Write an assembly program to find the
smaller of two signed numbers stored in DL
register and memory location 120H, and
store the result in DH register
64

Write an assembly program to multiply an
unsigned number stored in BX register by 8
without using MUL instruction (use shift
instructions)
65

Write a 8086 assembly program to clear 10
consecutive bytes of memory starting at
offset TABLE

Write a 8086 assembly program to initialize
10 consecutive bytes of memory starting at
offset TABLE with the value 44H
66

Write a 8086 assembly program to find the
maximum value of a string of 10 unsigned
8-bit numbers

Assume that offset TABLE contains the first
number
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
Write a 8086 assembly program to find the
minimum value of a string of 10 signed 8-
bit numbers

Assume that offset TABLE contains the first
number
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