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Help for Emu8086

Help I ndex | Overv iew | Tu to r ia l s | Em u 80 86 re fe rence |

Docu m e n t a t i o n f o r Em u 8 0 8 6

q W h e r e t o s t a r t ?

q Tuto r ia l s

q Em u8 08 6 re fe rence

q Com ple te 8086 in s t ru c t ion set

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Overview of Emu8086

Emu8086 Overview

verything for learning assembly language in one pack! Emu8086 combines andvanced source editor, assembler, disassembler, software emulator (Virtual PC)ith debugger, and step by step tutorials.

his program is extremely helpful for those who just begin to study assemblynguage. It compiles the source code and executes it on emulator step by step.

isual interface is very easy to work with. You can watch registers, flags andmemory while your program executes.

rithmetic & Logical Unit (ALU) shows the internal work of the central processornit (CPU).

mulator runs programs on a Virtual PC, this completely blocks your programom accessing real hardware, such as hard-drives and memory, since yourssembly code runs on a virtual machine, this makes debugging much easier.

086 machine code is fully compatible with all next generations of Intel's micro-rocessors, including Pentium II and Pentium 4, I'm sure Pentium 5 will support086 as well. This makes 8086 code very portable, since it runs both on ancientnd on the modern computer systems. Another advantage of 8086 instruction set ishat it is much smaller, and thus easier to learn.

mu8086has a much easier syntax than any of the major assemblers, but will stillenerate a program that can be executed on any computer that runs 8086 machineode; a great combination for beginners!

ote: If you don't useEmu8086to compile the code, you won't be able to stephrough your actual source code while running it.

Where to start?

1. StartEmu8086by selecting its icon from the start menu, or by runningEmu8086.exe.

2. Select "Samples" from "File" menu.

3. Click [Compile and Emulate] button (or press F5 hot key).

4. Click [Single Step] button (or press F8 hot key), and watch how the code

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Overview of Emu8086

is being executed.

5. Try opening other samples, all samples are heavily commented, so it's agreat learning tool.

6. This is the right time to see the tutorials.

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utorials

Tuto r ia l s

08 6 Assem b le r Tu t o r ia l s

q Num ber i ng Sys tem s

q Par t 1 : Wh at i s an assem b ly language?

q Par t 2 : Mem ory Access

q Par t 3 : Var iab les

q Pa r t 4 : I n t e r r u p t s

q Par t 5 : L i b ra r y o f com m on func t i ons - em u808 6 . inc

q Par t 6 : A r i t h m e t i c and Log i c I ns t r uc t i ons

q Par t 7 : P rog ram Flow Con t r o l

q Par t 8 : Procedur es

q Par t 9 : The St ack

q

Par t 10 : Macros

q Par t 1 1 : M ak ing you r ow n Opera t i ng System

q Par t 12 : Con t ro l l i ng Ex t e rna l Dev ices ( Robo t , Stepper -

Motor . . . )

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Reference for Emu8086

Emu8086reference

q Source Code Editor

q Compiling Assembly Code

q Using the Emulator

q Complete 8086 instruction set

q List of supported interrupts

q Global Memory Table

q Custom Memory Map

q MASM / TASM compatibility

q I/O ports

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086 instructions

om p let e 8086 i ns t ru ct i on set

uick reference:

AA

ADAM

ASDC

DDND

ALL

BW

LCLDLI

MCMP

CMPSBCMPSW

CWDDAA

DASDEC

DIVHLT

IDIV

IMUL

ININCINT

INTOIRET

JA

JAE

JBJBE

JCJCXZ

JEJG

JGE

JL

JLEJMPJNA

JNAEJNB

JNBE

JNCJNE

JNGJNGE

JNLJNLE

JNO

JNP

JNSJNZJO

JPJPE

JPO

JSJZ

LAHFLDS

LEALES

LODSB

LODSW

LOOPLOOPELOOPNE

LOOPNZLOOPZ

MOVMOVSB

MOVSWMUL

NEGNOP

NOTOR

OUT

POP

POPAPOPFPUSH

PUSHAPUSHF

RCL

RCR

REPREPE

REPNEREPNZ

REPZRET

RETF

ROL

RORSAHFSAL

SARSBB

SCASB

SCASWSHL

SHRSTC

STDSTI

STOSB

STOSW

SUBTESTXCHG

XLATBXOR

perand types:

EG: AX, BX, CX, DX, AH, AL, BL, BH, CH, CL, DH, DL, DI, SI, BP, SP.

REG: DS, ES, SS, and only as second operand: CS.

e m o r y: [BX], [BX+SI+7], variable, etc...(see Mem ory Access).

m m e d i a t e: 5, -24, 3Fh, 10001101b, etc...

otes:

q When two operands are required for an instruction they are separated bycomma. For example:

REG, memory

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086 instructions

q When there are two operands, both operands must have the same size(except shift and rotate instructions). For example:

AL, DLDX, AXm1 DB ?

AL, m1m2 DW ?AX, m2

q Some instructions allow several operand combinations. For example:

memory, immediateREG, immediate

memory, REG

REG, SREG

q Some examples contain macros, so it is advisable to use Sh i f t + F8 hot kto Step Over(to make macro code execute at maximum speed set stepde lay to zero), otherwise emulator will step through each instruction of amacro. Here is an example that uses PRINTN macro:

#make_COM#

include 'emu8086.inc'ORG 100hMOV AL, 1MOV BL, 2PRINTN 'Hello World!' ; macro.MOV CL, 3PRINTN 'Welcome!' ; macro.RET

hese marks are used to show the state of the flags:

- instruction sets this flag to 1.- instruction sets this flag to 0.- flag value depends on result of the instruction.- flag value is undefined (maybe 1 or 0).


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086 instructions

ome instructions generate exactly the same machine code, so disassembler may have a problem

ecoding to your original code. This is especially important for Conditional Jump instructions

Program Flow Control" in Tutorials for more information).

structions in alphabetical order:

Instruction Operands Description

AAA No operands

ASCII Adjust after Addition.Corrects result in AH and AL after addition when workingwith BCD values.

It works according to the following Algorithm:

if low nibble of AL > 9 or AF = 1 then:

q AL = AL + 6q AH = AH + 1q AF = 1q CF = 1

else

q AF = 0q CF = 0

in both cases:clear the high nibble of AL.

Example:

MOV AX, 15 ; AH = 00, AL = 0FhAAA ; AH = 01, AL = 05RET


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086 instructions

C Z S O P A

r ? ? ? ? r

AAD No operands

ASCII Adjust before Division.

Prepares two BCD values for division.

Algorithm:

q AL = (AH * 10) + ALq AH = 0

Example:

MOV AX, 0105h ; AH = 01, AL = 05AAD ; AH = 00, AL = 0Fh (15)RET

C Z S O P A

? r r ? r ?

AAM No operands

ASCII Adjust after Multiplication.Corrects the result of multiplication of two BCD values.

Algorithm:

q AH = AL / 10q AL = remainder

Example:

MOV AL, 15 ; AL = 0FhAAM ; AH = 01, AL = 05RET

C Z S O P A


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086 instructions

? r r ? r ?

AAS No operands

ASCII Adjust after Subtraction.Corrects result in AH and AL after subtraction whenworking with BCD values.

Algorithm:


q AL = AL - 6q AH = AH - 1q AF = 1q CF = 1

else

q AF = 0q CF = 0

in both cases:clear the high nibble of AL.

Example:

MOV AX, 02FFh ; AH = 02, AL = 0FFhAAS ; AH = 01, AL = 09RET

C Z S O P A

r ? ? ? ? r


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086 instructions

ADC

REG, memorymemory, REGREG, REGmemory, immediateREG, immediate

Add with Carry.

Algorithm:

operand1 = operand1 + operand2 + CF

Example:

STC ; set CF = 1MOV AL, 5 ; AL = 5ADC AL, 1 ; AL = 7RET

C Z S O P A

r r r r r r

ADD


Add.

Algorithm:

operand1 = operand1 + operand2

Example:

MOV AL, 5 ; AL = 5ADD AL, -3 ; AL = 2RET

C Z S O P A

r r r r r r


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086 instructions

AND


Logical AND between all bits of two operands. Result isstored in operand1.

These rules apply:

1 AND 1 = 11 AND 0 = 00 AND 1 = 00 AND 0 = 0

Example:

MOV AL, 'a' ; AL = 01100001bAND AL, 11011111b ; AL = 01000001b ('A')RET

C Z S O P

0 r r 0 r

CALL procedure namelabel4-byte address

Transfers control to procedure, return address is (IP) ispushed to stack. 4-byte address may be entered in this form1234h:5678h, first value is a segment second value is anoffset (this is a far call, so CS is also pushed to stack).

Example:

#make_COM#ORG 100h ; for COM file.

CALL p1

ADD AX, 1

RET ; return to OS.

p1 PROC ; procedure declaration.MOV AX, 1234hRET ; return to caller.

p1 ENDP


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086 instructions

C Z S O P A

unchanged

CBW No operands

Convert byte into word.

Algorithm:

if high bit of AL = 1 then:

q AH = 255 (0FFh)

else

q AH = 0

Example:

MOV AX, 0 ; AH = 0, AL = 0MOV AL, -5 ; AX = 000FBh (251)CBW ; AX = 0FFFBh (-5)

RET

C Z S O P A

unchanged

CLC No operands

Clear Carry flag.

Algorithm:

CF = 0

C

0


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086 instructions

CLD No operands

Clear Direction flag. SI and DI will be incremented bychain instructions: CMPSB, CMPSW, LODSB, LODSW,MOVSB, MOVSW, STOSB, STOSW.

Algorithm:

DF = 0

D

0

CLI No operands

Clear Interrupt enable flag. This disables hardwareinterrupts.

Algorithm:

IF = 0

I

0

CMC No operands

Complement Carry flag. Inverts value of CF.

Algorithm:

if CF = 1 then CF = 0if CF = 0 then CF = 1

C

r


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086 instructions

CMP


Compare.

Algorithm:

operand1 - operand2

result is not stored anywhere, flags are set (OF, SF, ZF, AF

PF, CF) according to result.

Example:

MOV AL, 5MOV BL, 5CMP AL, BL ; AL = 5, ZF = 1 (so equal!)RET

C Z S O P A

r r r r r r

CMPSB No operands

Compare bytes: ES:[DI] from DS:[SI].

Algorithm:

q DS:[SI] - ES:[DI]q set flags according to result:

OF, SF, ZF, AF, PF, CFq if DF = 0 then

r SI = SI + 1r DI = DI + 1

elser SI = SI - 1r DI = DI - 1

Example:see cmpsb.asm in Samples.

C Z S O P A

r r r r r r


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086 instructions

CMPSW No operands

Compare words: ES:[DI] from DS:[SI].

Algorithm:

q DS:[SI] - ES:[DI]q set flags according to result:


r SI = SI + 2r DI = DI + 2


Example:

see cmpsw.asm in Samples.

C Z S O P A

r r r r r r

CWD No operands

Convert Word to Double word.

Algorithm:

if high bit of AX = 1 then:

q DX = 65535 (0FFFFh)

else

q

DX = 0

Example:

MOV DX, 0 ; DX = 0MOV AX, 0 ; AX = 0MOV AX, -5 ; DX AX = 00000h:0FFFBhCWD ; DX AX = 0FFFFh:0FFFBh


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086 instructions

RET

C Z S O P A

unchanged

DAA No operands

Decimal adjust After Addition.Corrects the result of addition of two packed BCD values.

Algorithm:


q AL = AL + 6q AF = 1

if AL > 9Fh or CF = 1 then:

q AL = AL + 60hq CF = 1

Example:

MOV AL, 0Fh ; AL = 0Fh (15)DAA ; AL = 15hRET

C Z S O P A

r r r r r r


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086 instructions

DAS No operands

Decimal adjust After Subtraction.Corrects the result of subtraction of two packed BCDvalues.

Algorithm:


q AL = AL - 6q AF = 1

if AL > 9Fh or CF = 1 then:

q AL = AL - 60hq CF = 1

Example:

MOV AL, 0FFh ; AL = 0FFh (-1)DAS ; AL = 99h, CF = 1RET

C Z S O P A

r r r r r r

DECREGmemory

Decrement.

Algorithm:

operand = operand - 1

Example:

MOV AL, 255 ; AL = 0FFh (255 or -1)DEC AL ; AL = 0FEh (254 or -2)RET

Z S O P A


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086 instructions

r r r r r

CF - unchanged!

DIVREGmemory

Unsigned divide.

Algorithm:

when operand is a byte:AL = AX / operandAH = remainder (modulus)

when operand is a word:AX = (DX AX) / operandDX = remainder (modulus)

Example:

MOV AX, 203 ; AX = 00CBhMOV BL, 4DIV BL ; AL = 50 (32h), AH = 3RET

C Z S O P A

? ? ? ? ? ?

HLT No operands

Halt the System.

Example:

MOV AX, 5HLT

C Z S O P A

unchanged


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086 instructions

IDIVREGmemory

Signed divide.

Algorithm:

when operand is a byte:AL = AX / operand

AH = remainder (modulus)

when operand is a word:AX = (DX AX) / operandDX = remainder (modulus)

Example:

MOV AX, -203 ; AX = 0FF35h

MOV BL, 4IDIV BL ; AL = -50 (0CEh), AH = -3 (0FDh)RET

C Z S O P A

? ? ? ? ? ?

IMULREGmemory

Signed multiply.

Algorithm:

when operand is a byte:AX = AL * operand.

when operand is a word:(DX AX) = AX * operand.

Example:

MOV AL, -2MOV BL, -4IMUL BL ; AX = 8RET


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086 instructions

C Z S O P A

r ? ? r ? ?

CF=OF=0 when result fits into operand of IMUL.

IN

AL, im.byteAL, DXAX, im.byteAX, DX

Input from port into AL or AX.Second operand is a port number. If required to access port

number over 255 - DX register should be used.Example:

IN AX, 4 ; get status of traffic lights.IN AL, 7 ; get status of stepper-motor.

C Z S O P A

unchanged

INCREGmemory

Increment.

Algorithm:

operand = operand + 1

Example:

MOV AL, 4INC AL ; AL = 5RET

Z S O P A

r r r r r

CF - unchanged!


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086 instructions

INT immediate byte

Interrupt numbered by immediate byte (0..255).

Algorithm:

Push to stack:r flags registerr

CSr IP

q IF = 0q Transfer control to interrupt procedure

Example:

MOV AH, 0Eh ; teletype.

MOV AL, 'A'INT 10h ; BIOS interrupt.RET

C Z S O P A I

unchanged 0

INTO No operands

Interrupt 4 if Overflow flag is 1.

Algorithm:

if OF = 1 then INT 4

Example:

; -5 - 127 = -132 (not in -128..127)

; the result of SUB is wrong (124),; so OF = 1 is set:MOV AL, -5SUB AL, 127 ; AL = 7Ch (124)INTO ; process error.RET


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086 instructions

IRET No operands

Interrupt Return.

Algorithm:

Pop from stack:r IPr

CSr flags register

C Z S O P A

popped

JA label

Short Jump if first operand is Above second operand (as se

by CMP instruction). Unsigned.

Algorithm:

if (CF = 0) and (ZF = 0) then jump

Example:

include 'emu8086.inc'

#make_COM#ORG 100hMOV AL, 250CMP AL, 5JA label1PRINT 'AL is not above 5'JMP exit

label1:PRINT 'AL is above 5'

exit:RET

C Z S O P A

unchanged


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086 instructions

JAE label

Short Jump if first operand is Above or Equal to secondoperand (as set by CMP instruction). Unsigned.

Algorithm:

if CF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 5CMP AL, 5JAE label1

PRINT 'AL is not above or equal to 5'JMP exitlabel1:

PRINT 'AL is above or equal to 5'exit:

RET

C Z S O P A

unchanged

JB label

Short Jump if first operand is Below second operand (as seby CMP instruction). Unsigned.

Algorithm:

if CF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 1CMP AL, 5


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086 instructions

JB label1PRINT 'AL is not below 5'JMP exit

label1:PRINT 'AL is below 5'

exit:RET

C Z S O P A

unchanged

JBE label

Short Jump if first operand is Below or Equal to secondoperand (as set by CMP instruction). Unsigned.

Algorithm:

if CF = 1 or ZF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 5CMP AL, 5JBE label1PRINT 'AL is not below or equal to 5'JMP exit

label1:PRINT 'AL is below or equal to 5'

exit:RET

C Z S O P A

unchanged


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086 instructions

JC label

Short Jump if Carry flag is set to 1.

Algorithm:

if CF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 255ADD AL, 1JC label1PRINT 'no carry.'

JMP exitlabel1:PRINT 'has carry.'

exit:RET

C Z S O P A

unchanged

JCXZ label

Short Jump if CX register is 0.

Algorithm:

if CX = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV CX, 0JCXZ label1PRINT 'CX is not zero.'JMP exit


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086 instructions

label1:PRINT 'CX is zero.'

exit:RET

C Z S O P A

unchanged

JE label

Short Jump if first operand is Equal to second operand (asset by CMP instruction). Signed/Unsigned.

Algorithm:

if ZF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 5CMP AL, 5JE label1PRINT 'AL is not equal to 5.'JMP exit

label1:PRINT 'AL is equal to 5.'

exit:RET

C Z S O P A

unchanged


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086 instructions

JG label

Short Jump if first operand is Greater then second operand(as set by CMP instruction). Signed.

Algorithm:

if (ZF = 0) and (SF = OF) then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 5CMP AL, -5JG label1

PRINT 'AL is not greater -5.'JMP exitlabel1:

PRINT 'AL is greater -5.'exit:

RET

C Z S O P A

unchanged

JGE label

Short Jump if first operand is Greater or Equal to secondoperand (as set by CMP instruction). Signed.

Algorithm:

if SF = OF then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, -5


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086 instructions

JGE label1PRINT 'AL < -5'JMP exit

label1:PRINT 'AL >= -5'

exit:RET

C Z S O P A

unchanged

JL label

Short Jump if first operand is Less then second operand (asset by CMP instruction). Signed.

Algorithm:

if SF OF then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, -2CMP AL, 5JL label1PRINT 'AL >= 5.'JMP exit

label1:PRINT 'AL < 5.'

exit:RET

C Z S O P A

unchanged


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086 instructions

JLE label

Short Jump if first operand is Less or Equal to secondoperand (as set by CMP instruction). Signed.

Algorithm:

if SF OF or ZF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, -2CMP AL, 5JLE label1

PRINT 'AL > 5.'JMP exitlabel1:

PRINT 'AL

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086 instructions

MOV AL, 5JMP label1 ; jump over 2 lines!PRINT 'Not Jumped!'MOV AL, 0

label1:PRINT 'Got Here!'RET

C Z S O P A

unchanged

JNA label

Short Jump if first operand is Not Above second operand(as set by CMP instruction). Unsigned.

Algorithm:

if CF = 1 or ZF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#

ORG 100hMOV AL, 2CMP AL, 5JNA label1PRINT 'AL is above 5.'JMP exit

label1:PRINT 'AL is not above 5.'

exit:RET

C Z S O P A

unchanged


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086 instructions

JNAE label

Short Jump if first operand is Not Above and Not Equal tosecond operand (as set by CMP instruction). Unsigned.

Algorithm:

if CF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, 5JNAE label1

PRINT 'AL >= 5.'JMP exitlabel1:

PRINT 'AL < 5.'exit:

RET

C Z S O P A

unchanged

JNB label

Short Jump if first operand is Not Below second operand (set by CMP instruction). Unsigned.

Algorithm:

if CF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 7CMP AL, 5


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086 instructions

JNB label1PRINT 'AL < 5.'JMP exit

label1:PRINT 'AL >= 5.'

exit:RET

C Z S O P A

unchanged

JNBE label

Short Jump if first operand is Not Below and Not Equal tosecond operand (as set by CMP instruction). Unsigned.

Algorithm:

if (CF = 0) and (ZF = 0) then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 7CMP AL, 5JNBE label1PRINT 'AL 5.'

exit:RET

C Z S O P A

unchanged


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086 instructions

JNC label

Short Jump if Carry flag is set to 0.

Algorithm:

if CF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2ADD AL, 3JNC label1PRINT 'has carry.'

JMP exitlabel1:PRINT 'no carry.'

exit:RET

C Z S O P A

unchanged

JNE label

Short Jump if first operand is Not Equal to second operand(as set by CMP instruction). Signed/Unsigned.

Algorithm:

if ZF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, 3JNE label1


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086 instructions

PRINT 'AL = 3.'JMP exit

label1:PRINT 'Al 3.'

exit:RET

C Z S O P A

unchanged

JNG label

Short Jump if first operand is Not Greater then secondoperand (as set by CMP instruction). Signed.

Algorithm:

if (ZF = 1) and (SF OF) then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, 3JNG label1PRINT 'AL > 3.'JMP exit

label1:PRINT 'Al

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086 instructions

JNGE label

Short Jump if first operand is Not Greater and Not Equal tosecond operand (as set by CMP instruction). Signed.

Algorithm:

if SF OF then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, 3JNGE label1

PRINT 'AL >= 3.'JMP exitlabel1:

PRINT 'Al < 3.'exit:

RET

C Z S O P A

unchanged

JNL label

Short Jump if first operand is Not Less then second operan(as set by CMP instruction). Signed.

Algorithm:

if SF = OF then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, -3


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JNL label1PRINT 'AL < -3.'JMP exit

label1:PRINT 'Al >= -3.'

exit:RET

C Z S O P A

unchanged

JNLE label

Short Jump if first operand is Not Less and Not Equal tosecond operand (as set by CMP instruction). Signed.

Algorithm:

if (SF = OF) and (ZF = 0) then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 2CMP AL, -3JNLE label1PRINT 'AL -3.'

exit:RET

C Z S O P A

unchanged


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086 instructions

JNO label

Short Jump if Not Overflow.

Algorithm:

if OF = 0 then jump

Example:

; -5 - 2 = -7 (inside -128..127); the result of SUB is correct,; so OF = 0:

include 'emu8086.inc'#make_COM#ORG 100h

MOV AL, -5SUB AL, 2 ; AL = 0F9h (-7)JNO label1

PRINT 'overflow!'JMP exitlabel1:

PRINT 'no overflow.'exit:

RET

C Z S O P A

unchanged

Short Jump if No Parity (odd). Only 8 low bits of result arechecked. Set by CMP, SUB, ADD, TEST, AND, OR, XORinstructions.

Algorithm:

if PF = 0 then jump

Example:



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086 instructions

JNP label

#make_COM#ORG 100hMOV AL, 00000111b ; AL = 7OR AL, 0 ; just set flags.JNP label1PRINT 'parity even.'JMP exit

label1:PRINT 'parity odd.'

exit:RET

C Z S O P A

unchanged

JNS label

Short Jump if Not Signed (if positive). Set by CMP, SUB,ADD, TEST, AND, OR, XOR instructions.

Algorithm:

if SF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 00000111b ; AL = 7OR AL, 0 ; just set flags.JNS label1PRINT 'signed.'JMP exit

label1:PRINT 'not signed.'

exit:RET

C Z S O P A

unchanged


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086 instructions

JNZ label

Short Jump if Not Zero (not equal). Set by CMP, SUB,ADD, TEST, AND, OR, XOR instructions.

Algorithm:

if ZF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 00000111b ; AL = 7OR AL, 0 ; just set flags.JNZ label1PRINT 'zero.'JMP exit

label1:PRINT 'not zero.'

exit:RET

C Z S O P Aunchanged

Short Jump if Overflow.

Algorithm:

if OF = 1 then jump

Example:

; -5 - 127 = -132 (not in -128..127); the result of SUB is wrong (124),; so OF = 1 is set:



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086 instructions

JO label#make_COM#org 100h

MOV AL, -5SUB AL, 127 ; AL = 7Ch (124)

JO label1PRINT 'no overflow.'

JMP exit

label1:PRINT 'overflow!'

exit:RET

C Z S O P A

unchanged

JP label

Short Jump if Parity (even). Only 8 low bits of result arechecked. Set by CMP, SUB, ADD, TEST, AND, OR, XORinstructions.

Algorithm:

if PF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 00000101b ; AL = 5OR AL, 0 ; just set flags.JP label1PRINT 'parity odd.'JMP exit

label1:PRINT 'parity even.'

exit:RET

C Z S O P A


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086 instructions

unchanged

JPE label

Short Jump if Parity Even. Only 8 low bits of result arechecked. Set by CMP, SUB, ADD, TEST, AND, OR, XORinstructions.

Algorithm:

if PF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#

ORG 100hMOV AL, 00000101b ; AL = 5OR AL, 0 ; just set flags.JPE label1PRINT 'parity odd.'JMP exit

label1:PRINT 'parity even.'

exit:

RET

C Z S O P A

unchanged


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086 instructions

JPO label

Short Jump if Parity Odd. Only 8 low bits of result arechecked. Set by CMP, SUB, ADD, TEST, AND, OR, XORinstructions.

Algorithm:

if PF = 0 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 00000111b ; AL = 7OR AL, 0 ; just set flags.

JPO label1PRINT 'parity even.'JMP exit

label1:PRINT 'parity odd.'

exit:RET

C Z S O P A

unchanged

JS label

Short Jump if Signed (if negative). Set by CMP, SUB,ADD, TEST, AND, OR, XOR instructions.

Algorithm:

if SF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 10000000b ; AL = -128


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086 instructions

OR AL, 0 ; just set flags.JS label1PRINT 'not signed.'JMP exit

label1:PRINT 'signed.'

exit:

RET

C Z S O P A

unchanged

JZ label

Short Jump if Zero (equal). Set by CMP, SUB, ADD,TEST, AND, OR, XOR instructions.

Algorithm:

if ZF = 1 then jump

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV AL, 5CMP AL, 5JZ label1PRINT 'AL is not equal to 5.'JMP exit

label1:PRINT 'AL is equal to 5.'

exit:RET

C Z S O P A

unchanged


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086 instructions

LAHF No operands

Load AH from 8 low bits of Flags register.

Algorithm:

AH = flags register

AH bit: 7 6 5 4 3 2 1 0[SF] [ZF] [0] [AF] [0] [PF] [1] [CF]

bits 1, 3, 5 are reserved.

C Z S O P A

unchanged

LDS REG, memory

Load memory double word into word register and DS.

Algorithm:

q REG = first wordq DS = second word

Example:

#make_COM#ORG 100h

LDS AX, m

RET

m DW 1234hDW 5678h

END

AX is set to 1234h, DS is set to 5678h.


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086 instructions

C Z S O P A

unchanged

LEA REG, memory

Load Effective Address.

Algorithm:

q REG = address of memory (offset)

Generally this instruction is replaced by MOV whenassembling when possible.

Example:

#make_COM#ORG 100h

LEA AX, m

RET

m DW 1234h

END

AX is set to: 0104h.LEA instruction takes 3 bytes, RET takes 1 byte, we start a100h, so the address of 'm' is 104h.



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086 instructions

LES REG, memory

Load memory double word into word register and ES.

Algorithm:

q REG = first wordq ES = second word

Example:

#make_COM#ORG 100h

LES AX, m

RET

m DW 1234hDW 5678h

END

AX is set to 1234h, ES is set to 5678h.

C Z S O P A

unchanged

Load byte at DS:[SI] into AL. Update SI.

Algorithm:

q AL = DS:[SI]q if DF = 0 then

r SI = SI + 1else

r SI = SI - 1

Example:


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086 instructions

LODSB No operands

#make_COM#ORG 100h

LEA SI, a1MOV CX, 5MOV AH, 0Eh

m: LODSBINT 10hLOOP m

RET

a1 DB 'H', 'e', 'l', 'l', 'o'


LODSW No operands

Load word at DS:[SI] into AX. Update SI.

Algorithm:

q AX = DS:[SI]q if DF = 0 then

r SI = SI + 2else

r SI = SI - 2

Example:

#make_COM#

ORG 100h

LEA SI, a1MOV CX, 5

REP LODSW ; finally there will be 555h in AX.

RET


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086 instructions

a1 dw 111h, 222h, 333h, 444h, 555h

C Z S O P A

unchanged

LOOP label

Decrease CX, jump to label if CX not zero.

Algorithm:

q CX = CX - 1q if CX 0 then

r jumpelse

r no jump, continue

Example:

include 'emu8086.inc'#make_COM#ORG 100hMOV CX, 5

label1:PRINTN 'loop!'LOOP label1RET

C Z S O P A

unchanged


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086 instructions

LOOPE label

Decrease CX, jump to label if CX not zero and Equal (ZF =1).

Algorithm:

q CX = CX - 1q

if (CX 0) and (ZF = 1) thenr jump

elser no jump, continue

Example:

; Loop until result fits into AL alone,; or 5 times. The result will be over 255

; on third loop (100+100+100),; so loop will exit.

include 'emu8086.inc'#make_COM#ORG 100hMOV AX, 0MOV CX, 5

label1:

PUTC '*'ADD AX, 100CMP AH, 0LOOPE label1RET

C Z S O P A

unchanged


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086 instructions

LOOPNE label

Decrease CX, jump to label if CX not zero and Not Equal(ZF = 0).

Algorithm:

q CX = CX - 1q

if (CX 0) and (ZF = 0) thenr jump

elser no jump, continue

Example:

; Loop until '7' is found,; or 5 times.

include 'emu8086.inc'#make_COM#ORG 100hMOV SI, 0MOV CX, 5

label1:PUTC '*'MOV AL, v1[SI]

INC SI ; next byte (SI=SI+1).CMP AL, 7LOOPNE label1RETv1 db 9, 8, 7, 6, 5

C Z S O P A

unchanged


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086 instructions

LOOPNZ label

Decrease CX, jump to label if CX not zero and ZF = 0.

Algorithm:

q CX = CX - 1q if (CX 0) and (ZF = 0) then

r

jumpelser no jump, continue

Example:

; Loop until '7' is found,; or 5 times.

include 'emu8086.inc'#make_COM#ORG 100hMOV SI, 0MOV CX, 5

label1:PUTC '*'MOV AL, v1[SI]INC SI ; next byte (SI=SI+1).

CMP AL, 7LOOPNZ label1RETv1 db 9, 8, 7, 6, 5

C Z S O P A

unchanged


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086 instructions

LOOPZ label

Decrease CX, jump to label if CX not zero and ZF = 1.

Algorithm:

q CX = CX - 1q if (CX 0) and (ZF = 1) then

r

jumpelser no jump, continue

Example:

; Loop until result fits into AL alone,; or 5 times. The result will be over 255; on third loop (100+100+100),

; so loop will exit.

include 'emu8086.inc'#make_COM#ORG 100hMOV AX, 0MOV CX, 5

label1:PUTC '*'

ADD AX, 100CMP AH, 0LOOPZ label1RET

C Z S O P A

unchanged


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086 instructions

MOVSB No operands

Copy byte at DS:[SI] to ES:[DI]. Update SI and DI.

Algorithm:

q ES:[DI] = DS:[SI]q if DF = 0 then

r

SI = SI + 1r DI = DI + 1


Example:

#make_COM#

ORG 100h

LEA SI, a1LEA DI, a2MOV CX, 5REP MOVSB

RET

a1 DB 1,2,3,4,5a2 DB 5 DUP(0)

C Z S O P A

unchanged


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086 instructions

MOVSW No operands

Copy word at DS:[SI] to ES:[DI]. Update SI and DI.

Algorithm:

q ES:[DI] = DS:[SI]q if DF = 0 then

r

SI = SI + 2r DI = DI + 2


Example:

#make_COM#

ORG 100h

LEA SI, a1LEA DI, a2MOV CX, 5REP MOVSW

RET

a1 DW 1,2,3,4,5a2 DW 5 DUP(0)

C Z S O P A

unchanged


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086 instructions

MULREGmemory

Unsigned multiply.

Algorithm:

when operand is a byte:AX = AL * operand.

when operand is a word:(DX AX) = AX * operand.

Example:

MOV AL, 200 ; AL = 0C8hMOV BL, 4MUL BL ; AX = 0320h (800)

RET

C Z S O P A

r ? ? r ? ?

CF=OF=0 when high section of the result is zero.

NEGREGmemory

Negate. Makes operand negative (two's complement).

Algorithm:

q Invert all bits of the operandq Add 1 to inverted operand

Example:

MOV AL, 5 ; AL = 05hNEG AL ; AL = 0FBh (-5)NEG AL ; AL = 05h (5)RET

C Z S O P A

r r r r r r


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086 instructions

NOP No operands

No Operation.

Algorithm:

q Do nothing

Example:

; do nothing, 3 times:NOPNOPNOPRET

C Z S O P A

unchanged

NOTREGmemory

Invert each bit of the operand.

Algorithm:

q if bit is 1 turn it to 0.q if bit is 0 turn it to 1.

Example:

MOV AL, 00011011bNOT AL ; AL = 11100100bRET



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086 instructions

OR


Logical OR between all bits of two operands. Result isstored in first operand.

These rules apply:

1 OR 1 = 11 OR 0 = 10 OR 1 = 10 OR 0 = 0

Example:

MOV AL, 'A' ; AL = 01000001bOR AL, 00100000b ; AL = 01100001b ('a')RET

C Z S O P A

0 r r 0 r ?

OUT

im.byte, ALim.byte, AXDX, ALDX, AX

Output from AL or AX to port.First operand is a port number. If required to access portnumber over 255 - DX register should be used.

Example:

MOV AX, 0FFFh ; Turn on allOUT 4, AX ; traffic lights.

MOV AL, 100b ; Turn on the thirdOUT 7, AL ; magnet of the stepper-motor.

C Z S O P A

unchanged


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086 instructions

POPREGSREGmemory

Get 16 bit value from the stack.

Algorithm:

q operand = SS:[SP] (top of the stack)q SP = SP + 2

Example:

MOV AX, 1234hPUSH AXPOP DX ; DX = 1234hRET


POPA No operands

Pop all general purpose registers DI, SI, BP, SP, BX, DX,CX, AX from the stack.SP value is ignored, it is Popped but not set to SP register)

Note: this instruction works only on 80186 CPU and later!

Algorithm:

q POP DIq POP SIq POP BPq POP xx (SP value ignored)q POP BX

q POP DXq POP CXq POP AX

C Z S O P A

unchanged


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086 instructions

POPF No operands

Get flags register from the stack.

Algorithm:

q flags = SS:[SP] (top of the stack)q SP = SP + 2

C Z S O P A

popped

PUSH

REGSREGmemory

immediate

Store 16 bit value in the stack.

Note: PUSH immediate works only on 80186 CPU and

later!

Algorithm:

q SP = SP - 2q SS:[SP] (top of the stack) = operand

Example:

MOV AX, 1234hPUSH AXPOP DX ; DX = 1234hRET

C Z S O P A

unchanged


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086 instructions

PUSHA No operands

Push all general purpose registers AX, CX, DX, BX, SP,BP, SI, DI in the stack.Original value of SP register (before PUSHA) is used.

Note: this instruction works only on 80186 CPU and later!

Algorithm:

q PUSH AXq PUSH CXq PUSH DXq PUSH BXq PUSH SPq PUSH BPq PUSH SIq PUSH DI

C Z S O P A

unchanged

PUSHF No operands

Store flags register in the stack.

Algorithm:

q SP = SP - 2q SS:[SP] (top of the stack) = flags

C Z S O P A

unchanged


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086 instructions

RCL


memory, CLREG, CL

Rotate operand1 left through Carry Flag. The number ofrotates is set by operand2.When immediate is greater then 1, assembler generatesseveral RCL xx, 1 instructions because 8086 has machinecode only for this instruction (the same principle works forall other shift/rotate instructions).

Algorithm:

shift all bits left, the bit that goes off is set to CF anprevious value of CF is inserted to the right-mostposition.

Example:

STC ; set carry (CF=1).MOV AL, 1Ch ; AL = 00011100bRCL AL, 1 ; AL = 00111001b, CF=0.RET

C O

r r

OF=0 if first operand keeps original sign.

RCR


memory, CLREG, CL

Rotate operand1 right through Carry Flag. The number ofrotates is set by operand2.

Algorithm:

shift all bits right, the bit that goes off is set to CFand previous value of CF is inserted to the left-most

position.

Example:

STC ; set carry (CF=1).MOV AL, 1Ch ; AL = 00011100bRCR AL, 1 ; AL = 10001110b, CF=0.RET


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086 instructions

r exit from REPE cycle

else

q exit from REPE cycle

Example:see cmpsb.asm in Samples.

Z

r

REPNE chain instruction

Repeat following CMPSB, CMPSW, SCASB, SCASW

instructions while ZF = 0 (result is Not Equal), maximumCX times.

Algorithm:

check_cx:

if CX 0 then

q do following chain instructionq CX = CX - 1q if ZF = 0 then:

r go back to check_cxelse

r exit from REPNE cycle

else

q exit from REPNE cycle

Z

r


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086 instructions

else

q exit from REPZ cycle

Z

r

RETNo operandsor even immediate

Return from near procedure.

Algorithm:

q Pop from stack:r IP

q if immediate operand is present: SP = SP + operand

Example:

#make_COM#ORG 100h ; for COM file.

CALL p1

ADD AX, 1

RET ; return to OS.

p1 PROC ; procedure declaration.MOV AX, 1234hRET ; return to caller.

p1 ENDP

C Z S O P A

unchanged


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086 instructions

RETF

No operands

or even immediate

Return from Far procedure.

Algorithm:

q Pop from stack:r IPr

CSq if immediate operand is present: SP = SP + operand

C Z S O P A

unchanged

ROL


memory, CLREG, CL

Rotate operand1 left. The number of rotates is set by

operand2.

Algorithm:

shift all bits left, the bit that goes off is set to CF anthe same bit is inserted to the right-most position.

Example:

MOV AL, 1Ch ; AL = 00011100bROL AL, 1 ; AL = 00111000b, CF=0.RET

C O

r r



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086 instructions

ROR


memory, CLREG, CL

Rotate operand1 right. The number of rotates is set byoperand2.

Algorithm:

shift all bits right, the bit that goes off is set to CF

and the same bit is inserted to the left-most position

Example:

MOV AL, 1Ch ; AL = 00011100bROR AL, 1 ; AL = 00001110b, CF=0.RET

C O

r r


SAHF No operands

Store AH register into low 8 bits of Flags register.

Algorithm:

flags register = AH

AH bit: 7 6 5 4 3 2 1 0[SF] [ZF] [0] [AF] [0] [PF] [1] [CF]

bits 1, 3, 5 are reserved.

C Z S O P A

r r r r r r


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086 instructions

SAL


memory, CLREG, CL

Shift Arithmetic operand1 Left. The number of shifts is setby operand2.

Algorithm:

q Shift all bits left, the bit that goes off is set to CF.q

Zero bit is inserted to the right-most position.

Example:

MOV AL, 0E0h ; AL = 11100000bSAL AL, 1 ; AL = 11000000b, CF=1.RET

C O

r r


SAR


memory, CLREG, CL

Shift Arithmetic operand1 Right. The number of shifts is sby operand2.

Algorithm:

q Shift all bits right, the bit that goes off is set to CF.q The sign bit that is inserted to the left-most position

has the same value as before shift.

Example:

MOV AL, 0E0h ; AL = 11100000bSAR AL, 1 ; AL = 11110000b, CF=0.

MOV BL, 4Ch ; BL = 01001100bSAR BL, 1 ; BL = 00100110b, CF=0.

RET

C O

r r


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086 instructions


SBB


Subtract with Borrow.

Algorithm:

operand1 = operand1 - operand2 - CF

Example:

STCMOV AL, 5SBB AL, 3 ; AL = 5 - 3 - 1 = 1

RET

C Z S O P A

r r r r r r

SCASB No operands

Compare bytes: AL from ES:[DI].

Algorithm:

q ES:[DI] - ALq set flags according to result:


r DI = DI + 1else

r DI = DI - 1

C Z S O P Ar r r r r r


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086 instructions

SCASW No operands

Compare words: AX from ES:[DI].

Algorithm:

q ES:[DI] - AXq set flags according to result:


r DI = DI + 2else

r DI = DI - 2

C Z S O P A

r r r r r r

SHL


memory, CLREG, CL

Shift operand1 Left. The number of shifts is set byoperand2.

Algorithm:

q Shift all bits left, the bit that goes off is set to CF.q Zero bit is inserted to the right-most position.

Example:

MOV AL, 11100000bSHL AL, 1 ; AL = 11000000b, CF=1.

RET

C O

r r



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086 instructions

SHR


memory, CLREG, CL

Shift operand1 Right. The number of shifts is set byoperand2.

Algorithm:

q Shift all bits right, the bit that goes off is set to CF.q

Zero bit is inserted to the left-most position.

Example:

MOV AL, 00000111bSHR AL, 1 ; AL = 00000011b, CF=1.

RET

C O

r r


STC No operands

Set Carry flag.

Algorithm:

CF = 1

C

1


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086 instructions

STD No operands

Set Direction flag. SI and DI will be decremented by chaininstructions: CMPSB, CMPSW, LODSB, LODSW,MOVSB, MOVSW, STOSB, STOSW.

Algorithm:

DF = 1

D

1

STI No operands

Set Interrupt enable flag. This enables hardware interrupts.

Algorithm:

IF = 1

I

1

STOSB No operands

Store byte in AL into ES:[DI]. Update DI.

Algorithm:

q ES:[DI] = ALq if DF = 0 then

r DI = DI + 1else

r DI = DI - 1

Example:

#make_COM#ORG 100h

LEA DI, a1MOV AL, 12hMOV CX, 5


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086 instructions

REP STOSB

RET

a1 DB 5 dup(0)

C Z S O P A

unchanged

STOSW No operands

Store word in AX into ES:[DI]. Update DI.

Algorithm:

q ES:[DI] = AXq if DF = 0 then

r DI = DI + 2else

r DI = DI - 2

Example:

#make_COM#ORG 100h

LEA DI, a1MOV AX, 1234hMOV CX, 5

REP STOSW

RET

a1 DW 5 dup(0)

C Z S O P A

unchanged


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086 instructions

SUB


Subtract.

Algorithm:

operand1 = operand1 - operand2

Example:

MOV AL, 5SUB AL, 1 ; AL = 4

RET

C Z S O P A

r r r r r r

TEST


Logical AND between all bits of two operands for flagsonly. These flags are effected: ZF, SF, PF. Result is notstored anywhere.

These rules apply:

1 AND 1 = 1

1 AND 0 = 00 AND 1 = 00 AND 0 = 0

Example:

MOV AL, 00000101bTEST AL, 1 ; ZF = 0.

TEST AL, 10b ; ZF = 1.RET

C Z S O P

0 r r 0 r


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086 instructions

XCHGREG, memorymemory, REGREG, REG

Exchange values of two operands.

Algorithm:

operand1 < - > operand2

Example:

MOV AL, 5MOV AH, 2XCHG AL, AH ; AL = 2, AH = 5XCHG AL, AH ; AL = 5, AH = 2RET

C Z S O P A

unchanged

XLATB No operands

Translate byte from table.Copy value of memory byte at DS:[BX + unsigned AL] toAL register.

Algorithm:

AL = DS:[BX + unsigned AL]

Example:

#make_COM#ORG 100hLEA BX, datMOV AL, 2XLATB ; AL = 33h

RET

dat DB 11h, 22h, 33h, 44h, 55h

C Z S O P A

unchanged


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086 instructions

XOR


Logical XOR (Exclusive OR) between all bits of twooperands. Result is stored in first operand.

These rules apply:

1 XOR 1 = 01 XOR 0 = 10 XOR 1 = 10 XOR 0 = 0

Example:

MOV AL, 00000111b

XOR AL, 00000010b ; AL = 00000101bRET

C Z S O P A

0 r r 0 r ?

Copyright 2003 Emu8086, Inc.All rights reserved.

http://www.emu8086.com

http://www.emu8086.com/http://www.emu8086.com/

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Numbering Systems Tutorial

umbering Systems Tutorial

What is it?

here are many ways to represent the same numeric value. Long ago, humans used sticks to count, and later learned draw pictures of sticks in the ground and eventually on paper. So, the number 5 was first represented as: | | | (for five sticks).

ater on, the Romans began using different symbols for multiple numbers of sticks: | | | still meant three sticks, V now meant five sticks, and an X was used to represent ten of them!

sing sticks to count was a great idea for its time. And using symbols instead of real sticks was much better. One ost ways to represent a number today is by using the modern decimal system. Why? Because it includes the majoreakthrough of using a symbol to represent the idea of counting nothing. About 1500 years ago in India, zero (0) wst used as a number! It was later used in the Middle East as the Arabic, sifr. And was finally introduced to the Wethe Latin,zephiro. Soon you'll see just how valuable an idea this is for all modern number systems.

ecimal System

ost people today use decimal representation to count. In the decimal system there are 10 digits:

1, 2, 3, 4, 5, 6, 7, 8, 9

hese digits can represent any value, for example:4.

he value is formed by the sum of each digit, multiplied by the base (in this case it is 10 because there are 10 digits icimal system) in power of digit position (counting from zero):

osition of each digit is very important! for example if you place "7" to the end:7

will be another value:

mportant note: any number in power of zero is 1, even zero in power of zero is 1:

le:///D|/Heep/Assem/SW/Emu8086v3.07/Help/numbering_systems_tutorial.html (1 of 7)01/05/2006 12:16:43

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inary System

omputers are not as smart as humans are (or not yet), it's easy to make an electronic machine with two states: on anf, or 1 and 0.omputers use binary system, binary system uses 2 digits:

1

nd thus the base is 2.

ach digit in a binary number is called a BIT, 4 bits form a NIBBLE, 8 bits form a BYTE, two bytes form a WORDo words form a DOUBLE WORD (rarely used):

here is a convention to add "b" in the end of a binary number, this way we can determine that 101b is a binary numth decimal value of 5.

he binary number 10100101b equals to decimal value of 165:

Hexadecimal System

exadecimal System uses 16 digits:

1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F


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nd thus the base is 16.

exadecimal numbers are compact and easy to read.is very easy to convert numbers from binary system to hexadecimal system and vice-versa, every nibble (4 bits) cannverted to a hexadecimal digit using this table:

Decimal(base 10)

Binary(base 2)

Hexadecimal(base 16)

0 0000 0

1 0001 1

2 0010 2

3 0011 3

4 0100 4

5 0101 5

6 0110 67 0111 7

8 1000 8

9 1001 9

10 1010 A

11 1011 B

12 1100 C

13 1101 D

14 1110 E

15 1111 F

here is a convention to add "h" in the end of a hexadecimal number, this way we can determine that 5Fh is axadecimal number with decimal value of 95.e also add "0" (zero) in the beginning of hexadecimal numbers that begin with a letter (A..F), for example 0E120h

he hexadecimal number 1234h is equal to decimal value of 4660:


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onverting from Decimal System to Any Other

order to convert from decimal system, to any other system, it is required to divide the decimal value by the base osired system, each time you should remember the result and keep the remainder, the divide process continues unte result is zero.

he remainders are then used to represent a value in that system.

et's convert the value of39 (base 10) toHexadecimal System (base 16):

s you see we got this hexadecimal number: 27h.l remainders were below 10 in the above example, so we do not use any letters.

ere is another more complex example:'s convert decimal number 43868 to hexadecimal form:

he result is 0AB5Ch, we are using the above table to convert remainders over 9 to corresponding letters.

sing the same principle we can convert to binary form (using 2 as the divider), or convert to hexadecimal number, a

http://-/?-http://-/?-

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en convert it to binary number using the above table:

s you see we got this binary number: 1010101101011100b

igned Numbers

here is no way to say for sure whether the hexadecimal byte 0FFh is positive or negative, it can represent both decilue "255" and "- 1".

bits can be used to create 256 combinations (including zero), so we simply presume that first 128 combinations..127) will represent positive numbers and next 128 combinations (128..256) will represent negative numbers.

order to get "- 5", we should subtract 5 from the number of combinations (256), so it we'll get: 256 - 5 = 251.

sing this complex way to represent negative numbers has some meaning, in math when you add "- 5" to "5" you sht zero.

his is what happens when processor adds two bytes 5 and 251, the result gets over 255, because of the overflowocessor gets zero!

hen combinations 128..256 are used the high bit is always 1, so this maybe used to determine the sign of a number

he same principle is used for words (16 bit values), 16 bits create 65536 combinations, first 32768 combinations..32767) are used to represent positive numbers, and next 32768 combinations (32767..65535) represent negativembers.

here are some handy tools inEmu8086to convert numbers, and make calculations of any numerical expressions, alu need is a click on Math menu:

http://-/?-http://-/?-

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umber Convertor allows you to convert numbers from any system and to any system. Just type a value in any textx, and the value will be automatically converted to all other systems. You can work both with 8 bit and 16 bit valu

xpression Evaluator can be used to make calculations between numbers in different systems and convert numbersom one system to another. Type an expression and press enter, result will appear in chosen numbering system. Youork with values up to 32 bits. When Signed is checked evaluator assumes that all values (except decimal and doub

ords) should be treated as signed. Double words are always treated as signed values, so 0FFFFFFFFh is converted

or example you want to calculate: 0FFFFh * 10h + 0FFFFh (maximum memory location that can be accessed by 80PU). If you check Signed and Word you will get -17 (because it is evaluated as (-1) * 16 + (-1) . To make calculatith unsigned values uncheck Signed so that the evaluation will be 65535 * 16 + 65535 and you should get 1114095

ou can also use the Number Convertor to convert non-decimal digits to signed decimal values, and do the calculath decimal values (if it's easier for you).

hese operation are supported:

not (inverts all bits).

multiply./ divide.

modulus.sum.subtract (and unary -).

< shift left.> shift right.

bitwise AND.bitwise XOR.

bitwise OR.


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nary numbers must have "b" suffix, example:011011b

exadecimal numbers must have "h" suffix, and start with a zerohen first digit is a letter (A..F), example:ABCDh

ctal (base 8) numbers must have "o" suffix, example:o

>>> Next Tutorial >>>


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086 Assembler Tutorial for Beginners (Part 1)

0 86 Assem b le r Tu t o r i a l f o r Beg inn e rs ( Pa r t 1 )

his tutorial is intended for those who are not familiar with assembler atll, or have a very distant idea about it. Of course if you have knowledge ofome other programming language (Basic, C/C++, Pascal...) that may helpou a lot.

ut even if you are familiar with assembler, it is still a good idea to lookhrough this document in order to study Emu8086syntax.

t is assumed that you have some knowledge about number representationHEX/BIN), if not it is highly recommended to study Num ber i ng Sys tem s

u to r ia l before you proceed.

W hat i s an assem b ly l anguage?

ssembly language is a low level programming language. You need to getome knowledge about computer structure in order to understandnything. The simple computer model as I see it:

he system bus (shown in yellow) connects the various components of aomputer.he CPU is the heart of the computer, most of computations occur insidehe CPU.AM is a place to where the programs are loaded in order to be executed.

ns ide th e CPU

le:///D|/Heep/Assem/SW/Emu8086v3.07/Help/asm_tutorial_01.html (1 of 4)01/05/2006 12:17:14

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ENERAL PURPOSE REGI STERS

086 CPU has 8 general purpose registers, each register has its own name:

q AX - the accumulator register (divided into AH / AL).q BX - the base address register (divided into BH / BL).q CX - the count register (divided into CH / CL).q DX - the data register (divided into DH / DL).q SI - source index register.q DI - destination index register.q BP - base pointer.q SP - stack pointer.

espite the name of a register, it's the programmer who determines thesage for each general purpose register. The main purpose of a register iso keep a number (variable). The size of the above registers is 16 bit, it's

omething like: 0 0 1 1 0 0 0 0 0 0 1 1 1 0 0 1 b (in binary form), or 1 2 3 4 5 inecimal (human) form.

general purpose registers (AX, BX, CX, DX) are made of two separate 8it registers, for example if AX= 0 0 1 1 0 0 0 0 0 0 1 1 1 0 0 1 b , thenH=0 0 1 1 0 0 0 0 b and AL=0 0 1 1 1 0 0 1 b . Therefore, when you modify any ofhe 8 bit registers 16 bit register is also updated, and vice-versa. The sames for other 3 registers, "H" is for high and "L" is for low part.


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ecause registers are located inside the CPU, they are much faster thanmemory. Accessing a memory location requires the use of a system bus,o it takes much longer. Accessing data in a register usually takes no time.herefore, you should try to keep variables in the registers. Register setsre very small and most registers have special purposes which limit theirse as variables, but they are still an excellent place to store temporaryata of calculations.

EGMENT REGI STERS

q CS - points at the segment containing the current program.q DS - generally points at segment where variables are defined.q ES - extra segment register, it's up to a coder to define its usage.q SS - points at the segment containing the stack.

lthough it is possible to store any data in the segment registers, this is

ever a good idea. The segment registers have a very special purpose -ointing at accessible blocks of memory.

egment registers work together with general purpose register to accessny memory value. For example if we would like to access memory at thehysical address 1 2 3 4 5 h (hexadecimal), we should set the D S = 1 2 3 0 h nd SI = 0 0 4 5h . This is good, since this way we can access much more

memory than with a single register that is limited to 16 bit values.PU makes a calculation of physical address by multiplying the segment

egister by 10h and adding general purpose register to it (1230h * 10h +5h = 12345h):

he address formed with 2 registers is called an e f fect i v e addr ess.y default BX, SI and DI registers work with DS segment register;P and SP work with SS segment register.

ther general purpose registers cannot form an effective address!lso, although BX can form an effective address, BH and BL cannot!

PECI AL PURPOSE REGI STERS

q IP - the instruction pointer.q Flags Register - determines the current state of the processor.


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P register always works together with CS segment register and it points to currently executingnstruction.lags Register is modified automatically by CPU after mathematical operations, this allows toetermine the type of the result, and to determine conditions to transfer control to other parts ofhe program.enerally you cannot access these registers directly.

>>> Next Part >>>


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Mem or y Access

o access memory we can use these four registers: BX, SI , DI ,P.ombining these registers inside [ ] symbols, we can get

ifferent memory locations. These combinations are supportedaddressing modes):

[BX + SI][BX + DI][BP + SI][BP + DI]

[SI][DI]d16 (variable offset only)[BX]

[BX + SI] + d8[BX + DI] + d8[BP + SI] + d8[BP + DI] + d8

[SI] + d8[DI] + d8[BP] + d8[BX] + d8

[BX + SI] + d16[BX + DI] + d16[BP + SI] + d16[BP + DI] + d16

[SI] + d16[DI] + d16[BP] + d16[BX] + d16

8 - stays for 8 bit displacement.

1 6 - stays for 16 bit displacement.

isplacement can be a immediate value or offset of a variable, orven both. It's up to compiler to calculate a single immediatealue.

isplacement can be inside or outside of[ ] symbols, compilerenerates the same machine code for both ways.

isplacement is a s igned value, so it can be both positive oregative.

enerally the compiler takes care about difference between d8 nd d1 6 , and generates the required machine code.

or example, let's assume that D S = 1 0 0 , BX = 3 0 , SI = 7 0 .he following addressing mode: [ BX + SI ] + 2 5 s calculated by processor to this physical address: 1 00 * 1 6 +0 + 7 0 + 2 5 = 1 72 5 .


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y default DS segment register is used for all modes excepthose with BP register, for these SS segment register is used.

here is an easy way to remember all those possibleombinations using this chart:

ou can form all valid combinations by taking only one item fromach column or skipping the column by not taking anything from. As you see BX and BP never go together. SI and DI alsoon't go together. Here is an example of a valid addressing

mode: [ B X + 5 ] .

he value in segment register (CS, DS, SS, ES) is called asegm en t ",nd the value in purpose register (BX, SI, DI, BP) is called anof fse t ".

When DS contains value 1 2 3 4 h and SI contains the value

8 9 0 h it can be also recorded as 1 2 3 4 : 7 8 9 0 . The physicalddress will be 1234h * 10h + 7890h = 19BD0h.

n order to say the compiler about data type,hese prefixes should be used:

YTE PTR - for byte.W ORD PTR - for word (two bytes).

or example:

YTE PTR [BX] ; byte access.or

WORD PTR [BX] ; word access.


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Emu8086supports shorter prefixes as well:

. - for BYTE PTR. - for W ORD PTR

ometimes compiler can calculate the data type automatically,ut you may not and should not rely on that when one of the

perands is an immediate value.

MOV inst ru ct ion

q Copies the second operand (source) to the first operand (destination).

q The source operand can be an immediate value, general-purpose register ormemory location.

q The destination register can be a general-purpose register, or memorylocation.

q Both operands must be the same size, which can be a byte or a word.

These types of operands are supported:

MOV REG, memoryMOV memory, REGMOV REG, REGMOV memory, immediateMOV REG, immediate

REG: AX, BX, CX, DX, AH, AL, BL, BH, CH, CL, DH, DL, DI, SI, BP, SP.

memory: [BX], [BX+SI+7], variable, etc...

immediate: 5, -24, 3Fh, 10001101b, etc...


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For segment registers only these types ofMOV are supported:

MOV SREG, memoryMOV memory, SREGMOV REG, SREGMOV SREG, REG

SREG: DS, ES, SS, and only as second operand: CS.

REG: AX, BX, CX, DX, AH, AL, BL, BH, CH, CL, DH, DL, DI, SI, BP, SP.

memory: [BX], [BX+SI+7], variable, etc...

he MOV instruction cannot be used to set the value of the CS and IP registers.

Here is a short program that demonstrates the use ofMOV instruction:

#MAKE_COM# ; instruct compiler to make COM file.ORG 100h ; directive required for a COM program.MOV AX, 0B800h ; set AX to hexadecimal value of B800h.MOV DS, AX ; copy value of AX to DS.

MOV CL, 'A' ; set CL to ASCII code of 'A', it is 41h.MOV CH, 01011111b ; set CH to binary value.MOV BX, 15Eh ; set BX to 15Eh.MOV [BX], CX ; copy contents of CX to memory at B800:015ERET ; returns to operating system.

ou can copy & paste the above program toEmu8086code editor, and pressCompile and Emulate] button (or press F5 key on your keyboard).

he Emulator window should open with this program loaded, click [Single Step]utton and watch the register values.

ow to do copy & paste:


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1. Select the above text using mouse, click before the text and drag it downuntil everything is selected.

2. Press Ctrl + C combination to copy.

3. Go toEmu8086source editor and press Ctrl + V combination to paste.

s you may guess, ";" is used for comments, anything after ";" symbol is ignoredy compiler.

ou should see something like that when program finishes:

ctually the above program writes directly to video memory, so you may see thatMOV is a very powerful instruction.

>


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Var iab les

ariable is a memory location. For a programmer it is much easiero have some value be kept in a variable named "va r1 " then at theddress 5A73:235B, especially when you have 10 or more variables.

ur compiler supports two types of variables: BYTE and WORD .

Syntax for a variable declaration:

nameDBvalue

nameDWvalue

DB - stays for Define Byte.DW - stays for Define Word.

name - can be any letter or digit combination, though it should start with a letter.It's possible to declare unnamed variables by not specifying the name (thisvariable will have an address but no name).

value - can be any numeric value in any supported numbering system(hexadecimal, binary, or decimal), or "?" symbol for variables that are not

initialized.

s you probably know from part 2of this tutorial, MOV instruction issed to copy values from source to destination.et's see another example with MOV instruction:


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#MAKE_COM#ORG 100h

MOV AL, var1MOV BX, var2

RET ; stops the program.

VAR1 DB 7var2 DW 1234h

opy the above code to Emu8086source editor, and press F5 keyo compile and load it in the emulator. You should get somethingke:

s you see this looks a lot like our example, except that variablesre replaced with actual memory locations. When compiler makes

machine code, it automatically replaces all variable names with theirf f se ts. By default segment is loaded in DS register (when COM


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les is loaded the value ofDS register is set to the same value asS register - code segment).

n memory list first row is an of fse t , second row is a hexadec ima la lue, third row is decim a l va lue , and last row is an ASCI I haracter value.

ompiler is not case sensitive, so "VAR1 " and "va r1 " refer to theame variable.

he offset ofVAR1 is 0 1 0 8 h , and full address is 0B56 :0108 .

he offset ofva r2 is 0 1 0 9 h , and full address is 0B56 :0109 , thisariable is a WORD so it occupies 2 BYTES. It is assumed that lowyte is stored at lower address, so 3 4 h is located before 1 2 h .

ou can see that there are some other instructions after the RET

nstruction, this happens because disassembler has no idea aboutwhere the data starts, it just processes the values in memory and itnderstands them as valid 8086 instructions (we will learn them

ater).ou can even write the same program using DB directive only:

#MAKE_COM#ORG 100h

DB 0A0hDB 08hDB 01h

DB 8BhDB 1EhDB 09hDB 01h

DB 0C3h

DB 7

DB 34hDB 12h


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opy the above code to Emu8086source editor, and press F5 keyo compile and load it in the emulator. You should get the sameisassembled code, and the same functionality!

s you may guess, the compiler just converts the program source tohe set of bytes, this set is called m ach ine code, processor

nderstands the m ach ine code and executes it.

ORG 10 0 h is a compiler directive (it tells compiler how to handlehe source code). This directive is very important when you work

with variables. It tells compiler that the executable file will be loadedt the of fse t of 100h (256 bytes), so compiler should calculate theorrect address for all variables when it replaces the variable names

with their o f fse ts. Directives are never converted to any realm ach ine code.Why executable file is loaded at of fse t of1 0 0 h? Operating systemeeps some data about the program in the first 256 bytes of the CScode segment), such as command line parameters and etc.hough this is true for COM files only, EXE files are loaded at offsetf0 0 0 0 , and generally use special segment for variables. Maybe

we'll talk more about EXE files later.

A r r a y s

rrays can be seen as chains of variables. A text string is anxample of a byte array, each character is presented as an ASCIIode value (0..255).

ere are some array definition examples:

DB 48h, 65h, 6Ch, 6Ch, 6Fh, 00hDB 'Hello', 0

is an exact copy of the aarray, when compiler sees a string insideuotes it automatically converts it to set of bytes. This chart showspart of the memory where these arrays are declared:


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ou can access the value of any element in array using squarerackets, for example:

MOV AL, a[3]

ou can also use any of the memory index registers BX, SI , DI , BP,or example:

MOV SI, 3MOV AL, a[SI]

f you need to declare a large array you can use DUP operator.he syntax for DUP:

umber DUP ( value(s) )umber - number of duplicate to make (any constant value).alue - expression that DUP will duplicate.

or example:DB 5 DUP(9)

s an alternative way of declaring:DB 9, 9, 9, 9, 9

ne more example:DB 5 DUP(1, 2)

s an alternative way of declaring:DB 1, 2, 1, 2, 1, 2, 1, 2, 1, 2

f course, you can use DW instead ofDB if it's required to keep

alues larger then 255, or smaller then -128. DW cannot be used toeclare strings!

he expansion ofDUP operand should not be over 1020 characters!the expansion of last example is 13 chars), if you need to declareuge array divide declaration it in two lines (you will get a singleuge array in the memory).


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Get t ing t he Addr ess o f a Var iab le

here is LEA (Load Effective Address) instruction and alternativeOFFSET operator. Both OFFSET and LEA can be used to get theffset address of the variable.EA is more powerful because it also allows you to get the address

f an indexed variables. Getting the address of the variable can beery useful in some situations, for example when you need to passarameters to a procedure.

eminder :n order to tell the compiler about data type,ese prefixes should be used:

YTE PTR - for byte.ORD PTR - for word (two bytes).

or example:

YTE PTR [BX] ; byte access.orORD PTR [BX] ; word access.

mu8086supports shorter prefixes as well:

- for BYTE PTR

. - for W ORD PTR

ometimes compiler can calculate the data type automatically, but you may not andhould not rely on that when one of the operands is an immediate value.

ere is first example:


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ORG 100h

MOV AL, VAR1 ; check value of VAR1 by moving it to AL.

LEA BX, VAR1 ; get address of VAR1 in BX.

MOV BYTE PTR [BX], 44h ; modify the contents of VAR1.


RET

VAR1 DB 22h

END

ere is another example, that uses OFFSET instead ofLEA:

ORG 100h


MOV BX, OFFSET VAR1 ; get address of VAR1 in BX.

MOV BYTE PTR [BX], 44h ; modify the contents of VAR1.


RET

VAR1 DB 22h

END

oth examples have the same functionality.

hese lines:


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EA BX, VAR1MOV BX, OFFSET VAR1re even compiled into the same machine code: MOV BX, numum is a 16 bit value of the variable offset.

lease note that only these registers can be used inside squarerackets (as memory pointers): BX, SI , D I , BP!

see previous part of the tutorial).

Cons tan ts

onstants are just like variables, but they exist only until yourrogram is compiled (assembled). After definition of a constant itsalue cannot be changed. To define constants EQU directive is used:

name EQU < any expression >

or example:

k EQU 5

MOV AX, k

he above example is functionally identical to code:

MOV AX, 5

ou can view variables while your program executes by selectingVar iab les" from the "View " menu of emulator.


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o view arrays you should click on a variable and set Elementsroperty to array size. In assembly language there are not strict

ata types, so any variable can be presented as an array.

ariable can be viewed in any numbering system:

q HEX - hexadecimal (base 16).q BI N - binary (base 2).q OCT - octal (base 8).q SI GNED - signed decimal (base 10).q UNSI GNED - unsigned decimal (base 10).q

CHAR - ASCII char code (there are 256 symbols, somesymbols are invisible).

ou can edit a variable's value when your program is running,mply double click it, or select it and click Edit button.

t is possible to enter numbers in any system, hexadecimal numbershould have "h " suffix, binary "b " suffix, octal "o" suffix, decimalumbers require no suffix. String can be entered this way:

he l l o w o r l d ' , 0this string is zero terminated).

rrays may be entered this way:, 2 , 3 , 4 , 5the array can be array of bytes or words, it depends whether BYTEr WORD is selected for edited variable).

xpressions are automatically converted, for example:


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when this expression is entered:+ 2will be converted to 7 etc...

< < < Pr ev io us Par t < < < > > > Nex t Par t > > >


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n t e r r u p t s

nterrupts can be seen as a number of functions. These functionsmake the programming much easier, instead of writing a code torint a character you can simply call the interrupt and it will do

verything for you. There are also interrupt functions that workwith disk drive and other hardware. We call such functions

o f t w a r e i n t e r r u p t s.

nterrupts are also triggered by different hardware, these arealled h a r d w a r e i n t e r r u p t s. Currently we are interested ino f t w a r e i n t e r r u p t s only.

o make a s of t w a r e in t e r r u p t there is an I NT instruction, itas very simple syntax:

I NT va lue

Where va lue can be a number between 0 to 255 (or 0 to 0FFh),enerally we will use hexadecimal numbers.ou may think that there are only 256 functions, but that is notorrect. Each interrupt may have sub-functions.

o specify a sub-function AH register should be set before callingnterrupt.ach interrupt may have up to 256 sub-functions (so we get 256256 = 65536 functions). In general AH register is used, but

ometimes other registers maybe in use. Generally otheregisters are used to pass parameters and data to sub-function.

he following example uses I N T 1 0 h sub-function 0 Eh to type aHello!" message. This functions displays a character on the

creen, advancing the cursor and scrolling the screen asecessary.


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086 Assembler

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