1.
INTERMEDIATE 8086 ASSEMBLY
LANGUAGE PROGRAMMING
Cutajar & Cutajar
© 2012

2.
Machine Code
2
 There are occasions when the programmer must program at the machine’s own
level.
 Machine Code programs are tedious to write and highly error prone.
0000111100001111
In situations where a high-level
0010010101010100
language is inappropriate we
1010101010100101
avoid working in machine code
most of the time by making the computer do more of the work. Thus we write
in assembly language and then the computer converts this assembly language
program into machine code.

3.
Assembly Language
3
 In assembly language, a mneumonic (i.e. memory aid) is used as a short notation
for the instruction to be used.
Assembly Machine Code
Language
SUB AX,BX 001010111000011
MOV CX,AX 100010111001000
MOV DX,0 10111010000000000000000
Assembly language is an intermediate step between high level languages and machine
code. Most features present in HLL are not present in Assembly Language as type
checking etc.

4.
Compilers / Assemblers
4
 High-level Languages such as Pascal Program
Pascal programs are sometimes
Compiler
converted firstly to assembly
language by a computer program Assembler language Program
called compiler and then into
machine code by another Assembler
program called assembler Machine Code Program
This version is
actually loaded and
executed

5.
General Purpose Registers
5
AH AL
 There are 4 general
purpose registers in AX
the 8086.
BH BL
 They are all 16-bit
registers
 Each byte can be BX
addressed individually CH CL
by specifying the High
order or the Low order
byte of the register. CX
DH DL
DX

6.
Some Simple Commands
6
 MOV AX,3 ; Put 3 into register AX
 ADD AX,2 ; Add 2 to the contents of AX
 MOV BX,AX ; Copy the contents of AX in BX
 INC CX ; Add 1 to the contents of CX
 DEC DX ; Subtract 1 from the contents of DX
 SUB AX,4 ; Subtract 4 from the contents of AX
 MUL BX ; Multiply the contents of AX with BX leaving
; the answer in DX-AX
 DIV BX ; Divide the contents of DX-AX by BX leaving
; the quotient in AX and remainder in DX.

7.
Number Formats
7
AX
AH AL
MOV AH,01010101B 0 1 0 1 0 1 0 1
MOV AL,00100111B 0 1 0 1 0 1 0 1 0 0 1 0 0 1 1 1
MOV AX,3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
MOV AH,AL 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1
MOV AL,10D 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0
MOV AL,10H 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0
In case a number is moved (copied) into the register the base of a is specified by a
letter B for Binary, D for Decimal and H for Hex.

8.
AMBIGUITY
8
 Consider the instruction MOV DL, AH
 Does it mean ‘copy the contents of register AH to DL or
 Does it mean ‘copy A in hexadecimal into register DL
 To avoid this ambiguity all hexadecimal numbers must start with a number. This can
always be done by preceding a number starting with A,B,C,D,E and F with a
preceding zero to remove ambiguity.
 Thus MOV DL, AH means copy AH to DL whilst
 MOV DL, 0AH means sore hexadecimal A to DL

9.
The Flags Register
9
 Some of the instructions (but not all) affect the flag register.
 The flag register signals the status of the CPU after the last operation performed.
 For example if SUB AX,2 results in zero the ZF get 1 (lights on) indicating that
the result of the last operation was zero.

10.
JUMPS
10
 Jump instructions allow the 8086 to take decisions according to information provided by
the flag register.
 For example, if AX and BX contain the ASCII code for the same letter then do one
thing, if not then do another.
 `
…
CMP AX,BX ; Compares the contents of BX with that of AX
JE SAME ; Jump if they are equal to the point
; in the code labeled SAME
… ; Obey these instructions if the contents of AX
… ; is not equal to that of BX
SAME: MOV CX,AX ; Program continues from here if AX = BX.
…

11.
Labels
11
 We saw that the jump instruction has a general format JE <label> where <label> is a
facility offered by the assembler.
 These labels are converted by the assembler to exact address where the program is to
continue.
 Labels must start with a letter and can contain thereafter letters, numbers and
underscores (_).
 Spaces and punctuation marks are not permitted
 Avoid using keywords in labels
 Once_again, Next, Name34, this_37 are permitted as labels
 3rdday, tues+wed and semi;colons are not permitted as labels.

12.
JUMP Conditions
12
JA/JNBE (CF and ZF) = 0 Above / Not Below or Equal
JAE/JNB CF = 0 Above or Equal / Not Below
JB/JNAE/JC CF = 1 Below / Not Above or Equal / Carry
JBE/JNA (CF or ZF) = 1 Below or Equal / Not Above
JE/JZ ZF = 1 Equal / Zero
JMP none Unconditionally
JNC CF = 0 No Carry
JNE/JNZ ZF = 0 Not Equal / Not Zero
JNO OF = 0 No Overflow
JNP/JPO PF = 0 No Parity / Parity Odd
JNS SF = 0 No Sign / Positive
JO OF = 1 Overflow
JP/JPE PF = 1 Parity / Parity Even
JS SF = 1 Sign
JG/JNLE ZF = 0 and SF = OF Greater / Not Less nor Equal
JGE/JNL SF = OF Grater or Equal / Not Less
JL / JNGE SF <> OF Less / Not Greater nor Equal
JLE/JNG (ZF = 1) or (SF <> OF) Less or equal / not greater
JCXZ Register CX = 0 CX is equal to zero

13.
Example using Jumps
13
MOV CX, AX ; Keep a copy of AX before modification
SUB AX,BX ; AX := AX – BX
JZ MAKE1 ; This is instruction will cause execution
; to continue from MAKE1 if AX was
; equal to BX (subtraction resulted in Zero)
MOV DX, 0 ; Otherwise store 0 in DX
JMP RESET ; Jump to RESTORE where AX is restored
; thus avoiding the next instruction
MAKE1: MOV DX, 1 ; If AX = BX then we set DX to 1
RESET: MOV AX, CX ; Restore the old value of AX
Note that in the Code a colon ends a
label position

14.
The Logical Family
14
AND NOT (Invert: One’s Complement)
Contents of AX = 0000101011100011 Contents of AX = 0000101011100011
Contents of BX = 1001100000100001 Contents of AX = 1111010100011100
Contents of AX = 0000100000100001 after NOT AX is executed
after AND AX,BX is executed
OR TEST
Contents of AX = 0000101011100011 Contents of AX = 0000101011100011
Contents of BX = 1001100000100001 Contents of BX = 1001100000100001
Contents of AX = 1001101011100011 Contents of AX = 0000101011100011
after OR AX,BX is executed after TEST AX,BX is executed
XOR Similar to AND but the result is not stored in AX
Contents of AX = 0000101011100011 but only the Z-flag is changed
Contents of BX = 1001100000100001 NEG (Two’s Complement)
Contents of AX = 1001001011000010 Contents of AX = 0000101011100011
after XOR AX,BX is executed Contents of AX = 1111010100011101
after NEG AX is executed

15.
Use of Logical Family
15
Symbol ASCII (Dec) ASCII (Hex) By Making an AND between an ASCII value and 0FH
0 48 30 we can obtain the required number.
1 49 31 Say we AND 33H = 00110011B
2 50 32 with 0FH = 00001111B
3 51 33
We obtain = 00000011B (3)
4 52 34
5 53 35 By Making an OR between a number value and 30H we
6 54 36 can obtain its ASCII code.
7 55 37 Say we OR 05H = 00000101B
8 56 38 with 30H = 00110101B
We obtain = 00110101B
9 57 39
(ASCII value for ‘5’)

16.
Masking
16
By the use of masking we can set or test individual bits of a register
Suppose we want to set the 3rd.
bit of AX to 1 leaving the
others unchanged.
Suppose we want to test the if the
AX = 0101010100011001
6th. bit of AX is 1 or 0:
04H = 0000000000000100
AX = 0101010100011001
OR AX,04H = 0101010100011101 20H = 0000000000100000
AND AX,20H = 0000000000000000
Suppose we want to set the 5th. So if the result is 0 then that
bit of AX to 0 leaving the particular bit was 0, 1 otherwise
others unchanged.
AX = 0101010100011001
0FFEFH = 1111111111101111
AND AX,0FFEFH = 0101010100001101

17.
Instructions which affect Memory
17
 Computer memory is best thought of numbered pigeon holes (called locations),
each capable of storing 8 binary digits (a byte)
Data can be retrieved from memory, one or
[0000]
two bytes at a time: [0001]
[0002]
MOV AL, [20H] will transfer the [0003]
Contents of location 20H to AL. [0004]
[0005]
MOV BX, [20H] will transfer the contents of [0006]
[0007]
locations 20H and 21H to BX. [0008]
MOV [20H], AL will transfer the contents of [0009]
[000A]
AL to memory location 20H [000B]
Location ADDRESS [000C]
Location CONTENTS

18.
Changing addresses
18
 Varying an address whilst a program is running involves specifying the locations
concerned in a register.
 From all the general purpose registers BX is the only capable of storing such
addresses.
 Thus MOV AX, [CX] is illegal
 Whilst MOV CL, [BX] copies the contents of memory location whose address is
specified by BX into the register CL.
 And MOV [BX], AL copies the contents of AL in the memory location whose
address is specified in BX

19.
Examples Affecting Memory
19
 Consider the checkerboard memory test where a section of memory is filled with
alternate 01010101 and 10101010.
 The following program does the checkerboard test on locations 200H-300H
inclusive.
MOV BX,200H
MOV AX,1010101001010101B
NEXT: MOV [BX],AX
INC BX
CMP BX,300H
JLE NEXT

20.
The Instruction Pointer (IP)
20
 The computer keeps track of the next line to
be executed by keeping its address in a special START
. This is the
register called the Instruction Pointer (IP) or line which is
Program Counter. .
. executing
 This register is relative to CS as segment
register and points to the next instruction to MOV AX,BX
be executed. MOV CX,05H
 The contents of this register is updated with MOV DX,AX IP
every instruction executed.
.
 Thus a program is executed sequentially line
.
by line
.

21.
The Stack
21
 The Stack is a portion of memory
which, like a stack of plates in a
canteen, is organized on a Last-
In-First-Out basis.
 Thus the item which was put last
on the stack is the first to be
withdrawn

22.
The Stack Pointer
22
[0000]
[0002]
 The Stack pointer keeps track of the [0004]
position of the last item placed on the [0006]
[0008]
stack (i.e. the Top Of Stack) [000A]
[000C]
SP [000E]
[0010]
[0012]
[0014]
 The Stack is organized in words, (i.e. two [0016]
bytes at a time). Thus the stack pointer is [0018]
incremented or decremented by 2. Note that on placing items on the
 The Stack Pointer points to the last stack the address decreases
occupied locations on the stack

23.
PUSH & POP
23
PUSH AX AX
 The two set of instructions which [0000]
[0002]
[0004]
explicitly modify the stack are the [0006]
[0008] NEW SP
PUSH (which places items on the OLD SP [000A]
[000C]
stack) and the POP (which [000E]
[0010]
retrieves items from the stack). In [0012]
[0014]
both cases, the stack pointer is [0016]
[0018]
adjusted accordingly to point
POP AX
always to the top of stack. [0000]
AX
[0002]
 Thus PUSH AX means SP=SP-2 [0004]
[0006]
and AX -> [SP] [0008]
[000A]
OLD SP NEW SP
[000C]
 POP AX means [SP] -> AX and [000E]
[0010]
SP=SP+2. [0012]
[0014]
[0016]
[0018]

24.
Subroutines
24
 In high-level languages, procedures START SUB1 PROC
.
make it possible to break a large . .
program down into smaller pieces so . .
. RET
that each piece can be shown to work
independently. In this way the final CALL SUB1
program is built up of a number of
.
trusty bricks and is easier to debug .
because the error is either localized to .
one subprogram or its interlinking.
This has also the advantage of re-
usability of bricks.

25.
The CALL Mechanism
25
 Although at first sight the CALL START SUB1 PROC
and RET mechanism can be .
. .
implemented by using two JMP’s. . .
In fact this cannot be done since . RET
the CALL mechanism remembers CALL SUB1
the place where it was called from 1
.
and returns to the line following it. .
Thus this is not a fixed address. .
CALL SUB1
.
2
.
.

26.
The Return Mechanism
26
 When a CALL is encountered the current value of the instruction pointer is pushed
on the stack and the it is filled with the address stated by the call.
 Since the fetch cycle goes to search for the instruction pointed at by the instruction
pointer, the program continues it’s execution from the first statement in the
subroutine.
 On encountering the RET instruction the contents of the IP is popped from the stack
thus continuing the execution where it was suspended.
 Thus care must be taken to leave the return address intact before leaving a
subroutine. (i.e. a symmetrical number of pushes and pops within the subroutine)

27.
Software Interrupts
27
 Software interrupts are like hardware interrupts which are generated by the program
itself. From the interrupt number, the CPU derives the address of the Interrupt service
routine which must be executed.
 Software interrupts in assembly language can be treated as calls to subroutines of
other programs which are currently running on the computer.
 One of the most famous software interrupt is Interrupt No. 21H, which branches in
the operating system, and permits the use of PC-DOS functions defined there.
 The function required to be performed by DOS is specified in AH prior to the the
interrupt.
 The functions return and accept values in various registers.
 AN interrupt is called using the instruction INT followed by the interrupt number
. For example: INT 21H

28.
Some INT 21H functions
28
Function Description Explanation
Number
1 Keyboard Waits until a character is typed at the keyboard and then puts the ASCII
Input code for that character in register AL and echoed to screen
(echoed)
2 Display Prints the character whose ASCII code is in DL
Output
8 Keyboard Waits until a character is typed at the keyboard and then puts the ASCII
Input code for that character in register AL and NOT echoed to screen
(No echo)
9 Display Prints a series of characters stored in memory starting with the one in the
String address given in DX (relative to DS).Stop when the ASCII code for $ is
encountered

29.
INT 21H Example
29
Prompt DB ‘Please enter 1 or 2: ‘,13D,10D,’$’
Song1 DB ‘So you think you can tell heaven from hell’
Song2 DB ‘Blue Sky is in pain’,13D,10D,’$’
ASK: MOV DX, OFFSET Prompt
MOV AH,09H
This is only a
INT 21H
program fragment to
illustrate the use of
GET: MOV AH,01H
interrupt 21H – For
INT 21H
full details consult the
MASM notes
CMP AL,01H
JE NEXT
MOV DX, OFFSET Song1
MOV AH,09H
INT 21H

30.
Addition and Subtraction with carry or
30
borrow
 In assembly language there are two versions
of addition and two versions of subtraction. CF CF
 ADD - Simple addition of two numbers
0
 ADC - Adds two numbers together with
the carry flag 0
 SUB – Simple subtraction of two
numbers
 SBB – Subtracts the second number and Last 0 1 1
the carry flag (borrow) addition in 00 01 98 41 +
 This provides a means of adding numbers case of an 00 02 71 64
outgoing 00 04 70 05
greater than 32-bits.
carry
 CLC clears the carry for the first digit
addition

31.
The Compare Instruction
31
 The compare instruction does not change the contents of the registers involved but only
sets the flag register accordingly.
 The actual operation performed by the compare is a subtraction, leaving the source and
destination registers intact
 Consider CMP AX,BX : Flags are set according to the result of subtracting BX from AX:
 If AX = BX then the ZF is set to 1
 If AX > BX then the ZF is set to 0 and CF is set to 0 too
 If AX < BX then we need an external borrow, which is reflected in CF = 1
 These flags are tested in the ABOVE or BELOW jumps which test unsigned numbers
 The GREATER and LESS jumps are for signed numbers and work on the SF, OF
and the ZF instead

32.
Addressing Modes
32
 The addressing modes deal with the source and destination of the data required by
the instruction. This can be either a register or a location in memory, or even a port.
 Various addressing modes exist:
 Register Addressing
 Immediate and Direct Addressing
 Indirect Addressing
 Indexed Addressing
 Based Addressing
 Based-Indexed Addressing
Computer Logic II

33.
Register Addressing
33
 This addressing mode General Purpose Segment Registers
involves the contents of AX AH AL CS
the register directly as
BX BH BL DS
for example:
CX CH CL SS
 MOV AX, BX
DX DH DL ES
 MOV CL, DL
 Note that the IP and SI FLAGS
Flags register cannot be DI
IP
accessed directly by the SP
programmer BP
AX BX
Ex. MOV AX,BX

34.
Immediate and Direct Addressing
34
 In Immediate addressing – for example Ex. MOV CL,61H
MOV CL,61H – the immediate operand CL
61H
61H is stored as part of the instruction.
Thus the number 61H is loaded directly in
CL.
 Direct addressing is similar except that in Ex. MOV AL,[210H]
this case the effective address of one of the AL
operands is taken directly from the
instruction. Thus in MOV AL, [210H] the
contents of location 210H relative to DS is
put in AL (DS:210H) 75H

35.
Default Segment Register
35
Note that the default segment register can be changed
using the segment override, i.e. stating the whole
General Purpose
address in the form DS: Offset
AX AH AL
BX BH BL Relative to DS by default
CX CH CL
DX DH DL NORMALLY FOR STRINGS
SI Relative to DS by default DS
DI Relative to DS by default ES
SP Relative to SS by default
BP Relative to SS by default
IP Relative to CS by default

36.
Some other useful Instructions
36
 CLC: Clear Carry Flag (CF = 0)
 STC: Set Carry Flag (CF = 1)
 CMC : Complement Carry Flag (CF = CF)
 CBW: Convert Byte to Word
 CWD: Convert Word to Double-Word
 NEG: Negate (2’s Complement)
 NOT: Compliment (1’s Complement)

37.
Reference Books
37
 Programming the 8086/86 for the IBM PC and Compatibles . Michael Thorne
 Microprocessors and Interfacing – Programming and Hardware – Douglas V.Hall
 Microsoft Macro Assembler – for the MS-DOS Operating Systems – Reference
Manual