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Here we go with MASM means Macro ASseMbler

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Installing MASM32

Target OS versions
The MASM32 SDK is designed to install on Windows operating system versions from WIN2000 upwards. It is not designed to install on earlier Win9x systems or Windows ME.

Installation Safety Design
The MASM32 SDK is targeted at experienced programmers who routinely have multiple development environments setup and configured on their development computer.

Safety for the experienced developer is as follows,
(1) The MASM32 SDK does not write any files to the operating system.
(2) The MASM32 SDK is registry safe and writes nothing to the registry.
(3) The MASM32 SDK does not set up any file extensions and does not steal file extensions from existing applications.
(4) The MASM32 SDK does not depend on any other application for its operation.
(5) The MASM32 SDK is free of DLL Hell.
(6) The MASM32 SDK is DEP safe, it runs on a fully DEP enabled computer.

At the end of the installation a choice is offered to set up an icon on the desktop using a simple VBS script, its contents are displayed and you have the option of using it or not. This safety comes at the price that the programmer who wants to use the advanced capacity of the MASM32 SDK must have sufficient technical knowledge to configure their development computer so that the MASM SDK can be installed.

Assumptions Of The Installation
The installation assumes that,
(a) The computer is correctly configured and is completely free on any trojan/rootkit/viral damage or infection.
(b) The installation is being performed with ADMINISTRATIVE RIGHTS that allow software to be installed.
(c) The installation can write files to disk without the process being obstructed or damaged by computer settings or security software.

Secure Origin Of The MASM32 SDK Content
The MASM32 SDK is built in a completely isolated environment from its own source code and is completely free of any trojan/rootkit/viral transmission code. It has been successfully installed on millions of development computers over a period in excess of 10 years and the only problems that have ever occurred come from false positives in downmarket AV software. The contents of the MASM32 SDK are DEP SAFE and were re-written when Microsoft introduced the new security modification to the Portable Executable specifications to protect computers from a variety of stack based exploits.

Common Problems With Inexperienced Users
(1) The computer is infected with either a trojan/root kit or virus and interferes with the installation by obstructing the file write to disk process.
SOLUTION : Either completely repair the operating system installation ensuring it is free from any infection or damage or do a full installation of the operating system to ensure it is working correctly.

(2) The person installing the MASM32 SDK does not have sufficient Administrative rights to install software.
SOLUTION : Change to the Administrative Profile before attempting to install the software.

(3) Anti-virus and/or similar security software obstructs the installation of the MASM32 SDK.
SOLUTION : Configure, change or remove the AV software so that it does not interfere with the installation of the MASM32 SDK.

This has primarily been a problem for inexperienced users who are using free security software downloaded  from the internet. While reputable AV and security software vendors produce commercial products that properly understand the Microsoft Portable Executable specifications, much of the BETA level freeware available on the internet does not properly do this and regularly delivers FALSE POSITIVES on software that is completely free of any infection. At a design level this follows from the AV product using a naive dictionary approach coupled with poorly designed heuristic scanning methods. The result when using this style of junk is that the AV software often silently removes programs that it does not understand without advising the user and damages the installation of the software.

There is no solution to this problem from the installation end without interfering with the operating system, something that the MASM32 SDK does not do by design. To successfully install the MASM32 SDK you must have the appropriate Administrative rights and the computer must be clean from viral/trojan infections and be configured to allow an installation to write to disk. Unfortunately vendors of this CRAP are protecting their commercial interests by trying to appear as if their software is protecting your computer where in fact their lack of experience exposes end users to risks of large scale unrecoverable damage.

Securing A Development Computer
While most experienced users already know how to protect a development computer, for the small developer or student/person learning a programming environment there are a number of basic guidelines that help to protect a development computer without strangling it with security restrictions designed to protect non-technical users.

(1) Configure the DSL or similar router so that it uses the internal firewall, enables "Stateful Packet Inspection" (SPI) and "Network Address Translation" (NAT) and do not use any form of tunneling that bypasses the router setup. This is particularly important as it cannot be altered by the operating system so even if the operating system is compromised, the router is not and its security features still work.

(2) Unless you know exactly what you are doing with the firewall in  late versions of Windows, disable it completely and install a manual setup firewall that allows you to block specific ports and protocols.

(3) Examine the running services on your computer and TURN OFF those that you don't need. Things like remote management, TAPI, FTP and HTTP servers are rarely ever required by a home developer and by turning OFF services of this type, the vulnerable surface area of the computer is reduced. Do not install the IIS internet server option on a development machine as it is subject to new exploits on a regular basis.

(4) NEVER EVER share the boot partition of a development computer. If you need to transfer data from a development computer to another computer in your LAN, set up a directory on another partition on your development computer that has two (2) subdirectories, "upload" and "download". Set up the directories as shared with read only access to the "download" directory  and normal read/write access for the "upload" directory. You are safer if you share nothing and use a shared drive/directory on another computer in your LAN to write data to when needed.

(5) If you must use your development computer to handle email there are a number of steps that reduce your risk.

(a) Install software that reads the available email on the email server you use WITHOUT downloading it. This allows you to check what is there and delete any of the junk you don't want without it ever being on your computer.

(b) If your email software automatically downloads email, TURN THE OPTION OFF, only download email when you choose to do so. Malicious software that is never downloaded onto your computer can never do it any harm.

(c) If you must run AV or similar security software, use proven reliable products. Eset's NOD32 and Kaspersky are professional well written products that can be configured properly. Avoid any form of automatic scanning and only operate the software on a needs basis.

Running Security Risks In A Sandbox
If you know what you are doing and properly secure a development computer you can be free of the virus/anti-virus merry go round and be in full control of your development computer but the weakest point in computer security is you the user, no matter what you installation and configuration may happen to be, if you run something that is dangerous you can damage your computer's operating system installation.

If you must do dangerous things like downloading junk from the internet or email that may contain dangerous attachments, do it in a sandbox, install a Virtual Machine on your computer, set up the browser and email programs you want and if the worst happens, you can just shut it down, replace the virtual hard drive file with a backup and no harm is done.

The Final Solution For Computer Security
Obtain a reliable disk imaging program, Norton Ghost, Acronis True Image and similar and learn how to use it correctly. Make a backup image of your BOOT partition and save that file on another partition on your computer. Software like Acronis has the capacity to create a bootable CD so that you can boot the damaged computer from the CD and restore the disk image file saved on another partition. This approach has two (2) major advantages, it cannot be beaten as it completely overwrites the damaged operating system installation and it usually takes less than five (5) minutes to perform.

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Classification of Instruction set of 8086 with sample programs

Click on each instruction to read its function,explanation and its sample Program.

DATA TRANSFER INSTRUCTIONS

GENERAL – PURPOSE BYTE OR WORD TRANSFER INSTRUCTIONS	SIMPLE INPUT AND OUTPUT PORT TRANSFER INSTRUCTION	SPECIAL ADDRESS TRANSFER INSTRUCTION	FLAG TRANSFER INSTRUCTIONS
MOV PUSH POP XCHG XLAT/XLATB	IN OUT	LEA LDS  LES	LAHF  SAHF  PUSHF  POPF

 

ARITHMETIC INSTRUCTIONS

ADDITION INSTRUCTIONS	SUBTRACTION INSTRUCTIONS	MULTIPLICATION INSTRUCTIONS	DIVISION INSTRUCTIONS
ADD  ADC   INC   AAA  DAA	SUB  SBB  DEC   NEG  CMP  AAS  DAS	MUL  IMUL  AAM	DIV  IDIV  AAD  CBW   CWD

BIT MANIPULATION INSTRUCTIONS

LOGICAL INSTRUCTIONS	SHIFT INSTRUCTIONS	ROTATE INSTRUCTIONS
NOT  AND  OR  XOR   TEST	SHL / SAL  SHR  SAR	ROL  ROR   RCL  RCR

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STRING INSTRUCTIONS

STRING INSTRUCTIONS

REP REPE / REPZ REPNE / REPNZ MOVS / MOVSB / MOVSW COMPS / COMPSB / COMPSW SCAS / SCASB / SCASW LODS / LODSB / LODSW STOS / STOSB / STOSW

PROGRAM EXECUTION TRANSFER INSTRUCTIONS

UNCONDITIONAL TRANSFER INSTRUCTIONS	CONDITIONAL TRANSFER INSTRUCTIONS	ITERATION CONTROL INSTRUCTIONS	INTERRUPT INSTRUCTIONS
CALL RET JMP	JA / JNBE JAE / JNB JB / JNAE JBE / JNA JC JE / JZ JG / JNLE JGE / JNL JL / JNGE JLE / JNG JNC JNE / JNZ JNO JNP / JPO JNS JO JP / JPE JS	LOOP LOOPE / LOOPZ LOOPNE / LOOPNZ JCXZ	INT INTO IRET

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PROCESS CONTROL INSTRUCTIONS FLAG SET / CLEAR INSTRUCTIONS EXTERNAL HARDWARE SYNCHRONIZATION INSTRUCTIONS STC CLC CMC STD CLD STI CLI HLT WAIT ESC LOCK NOP

Rotate instructions

ROTATE INSTRUCTIONS

ROL Instruction - Rotate all bits of operand left, MSB to LSB

ROL instruction rotates the bits in the operand specified by oper1 towards left by the count specified in oper2. ROL moves each bit in the operand to next higher bit position. The higher order bit is moved to lower order position. Last bit rotated is copied into carry flag.

Syntax - ROL destination, count.

Example: ( 1 )

ROL AX, 1                  ;Word in AX is moved to left by 1 bit and MSB bit is to LSB, and CF and CF =0 ,BH =10101110
ROL BH, 1                  ;Result: CF ,Of =1 , BH = 01011101

Example : ( 2 )
                                  ;BX = 01011100 11010011 and CL = 8 bits to rotate
ROL BH, CL                ;Rotate BX 8 bits towards left and CF =0, BX =11010011 01011100

• ROR Instruction - Rotate all bits of operand right, LSB to MSB

ROR instruction rotates the bits in the operand oper1 to wards right by count specified in op2. The last bit rotated is copied into CF.

Syntax - ROR destination, count

Example:( 1 )

ROR BL, 1                  ;Rotate all bits in BL towards right by 1 bit position, LSB bit is moved to MSB and CF has last rotated bit.

Example:( 2 )
                                  ;CF =0, BX = 00111011 01110101
ROR BX, 1                 ;Rotate all bits of BX of 1 bit position towards right and CF =1, BX = 10011101 10111010

Example ( 3 )

                                 ;CF = 0, AL = 10110011,
MOVE CL, 04H           ; Load CL
ROR AL, CL               ;Rotate all bits of AL towards right by 4 bits, CF = 0 ,AL = 00111011

• RCL Instruction - Rotate operand around to the left through CF

RCL instruction rotates the bits in the operand specified by oper1 towards left by the count specified in oper2.The operation is circular, the MSB of operand is rotated into a carry flag and the bit in the CF is rotated around into the LSB of operand.

Syntax - RCL destination, source.

Example(1):

CLC                         ;put 0 in CF
RCL AX, 1                ;save higher-order bit of AX in CF
RCL DX, 1                ;save higher-order bit of DX in CF
ADC AX, 0                ; set lower order bit if needed.

Example (2):

RCL DX, 1                ;Word in DX of 1 bit is moved to left, and MSB of word is given to CF and CF to LSB CF=0, BH = 10110011
 

RCL BH, 1                ;Result : BH =01100110 CF = 1, OF = 1 because MSB changed CF =1,AX =00011111 10101001
 

MOV CL, 2               ;Load CL for rotating 2 bit position
RCL AX, CL              ;Result: CF =0, OF undefined AX = 01111110 10100110

• RCR Instruction - Rotate operand around to the right through CF

RCR Instruction - RCR instruction rotates the bits in the operand specified by operand1 towards right by the count specified in operand2.

Syntax - RCR destination, count

Example:(1)

RCR BX, 1               ;Word in BX is rotated by 1 bit towards right and CF will contain MSB bit and LSB contain CF bit .

Example:(2)

                               ;CF = 1, BL = 00111000
RCR BL, 1               ;Result: BL = 10011100, CF =0 OF = 1 because MSB is changed to 1.

shift instrcuctions

SHIFT INSTRUCTIONS

 SAL/SHL Instruction - Shift operand bits left, put zero in LSB(s)

SAL instruction shifts the bits in the operand specified by op1 to its left by the count specified in op2. As a bit is shifted out of LSB position a 0 is kept in LSB position. CF will contain MSB bit.

Syntax - SAL/AHL destination, count

Example:

                                            ;CF = 0, BX = 11100101 11010011
SAL BX, 1                               ;Shift BX register contents by 1 bit position towards left
                                            ;CF = 1, BX = 11001011 1010011

• SHR Instruction - Shift operand bits right, put zero in MSB

SHR instruction shifts the bits in op1 to right by the number of times specified by op2 .

Example:( 1 )

SHR BP, 1                             ; Shift word in BP by 1 bit position to right and 0 is kept to MSB

Example:( 2 )

MOV CL, 03H                        ;Load desired number of shifts into CL
SHR BYTE PYR[BX]                ;Shift bytes in DS at offset BX and rotate 3 bits to right and keep 3 0’s in MSB

Example:( 3 )

                                           ;SI = 10010011 10101101 , CF = 0
SHR SI, 1                             ; Result: SI = 01001001 11010110 and CF = 1, OF = 1, SF = 0, ZF = 0

• SAR Instruction - Shift operand bits right, new MAB = old MSB

SAR instruction shifts the bits in the operand specified by op1 towards right by count specified in op2.As bit is shifted out a copy of old MSB is taken in MSB. MSB position and LSB is shifted to CF.

Syntax - SAR destination, count.

Example: ( 1 )
                                         ; AL = 00011101 = +29 decimal, CF = 0
SAR AL, 1                           ;Shift signed byte in AL towards right ( divide by 2 )
                                         ;AL = 00001110 = + 14 decimal, CF = 1

Example: ( 2 )
                                      ;BH = 11110011 = - 13 decimal, CF = 1
SAR BH, 1                        ;Shifted signed byte in BH to right and BH = 11111001 = - 7 decimal, CF = 1.

logical instructions

LOGICAL INSTRUCTIONS

 NOT Instruction - Invert each bit of operand

NOT perform the bitwise complement of operand and stores the result back into operand itself.

Syntax–NOT destination

Example :

NOT BX                            ;Complement contents of BX register. - DX =F038h
NOT DX                           ;after the instruction DX = 0FC7h

• AND Instruction - AND corresponding bits of two operands

This Performs a bitwise Logical AND of two operands. The result of the operation is stored in the op1 and used to set the flags. AND op1, op2 To perform a bitwise AND of the two operands, each bit of the result is set to 1 if and only if the corresponding bit in both of the operands is 1, otherwise the bit in the result I cleared to 0 .

Example :

AND BH, CL                 ;AND byte in CL with byte in BH ;result in BH
AND BX,00FFh             ;AND word in BX with immediate 00FFH. Mask upper byte,                         leave lower unchanged

AND CX,[SI]                ; AND word at offset [SI] in data segment with word in CX register . Result in CX register and BX = 10110011 01011110

AND BX,00FFh              ;Mask out upper 8 bits of BX. ;Result BX = 00000000 01011110 and CF =0 , OF = 0, PF = 0, SF = 0 ,ZF = 0.

• OR Instruction - Logically OR corresponding of two operands

OR Instruction - OR instruction perform the bit wise logical OR of two operands .Each bit of the result is cleared to 0 if and only if both corresponding bits in each operand are 0, other wise the bit in the result is set to 1.

Syntax–- OR destination, source.

Examples :

OR AH, CL                         ;CL is OR’ed with AH, result in AH.
                                       ;CX = 00111110 10100101
OR CX,FF00h                     ;OR CX with immediate FF00h result in CX = 11111111 10100101
                                       ;Upper byte are all 1’s lower bytes are unchanged.

• XOR Instruction - Exclusive XOR destination, source

XOR Instruction - XOR performs a bit wise logical XOR of the operands specified by op1 and op2. The result of the operand is stored in op1 and is used to set the flag.

Syntax–- XOR destination, source.

Example : ( Numerical )
                                       ; BX = 00111101 01101001 and CX = 00000000 11111111
XOR BX, CX                       ;Exclusive OR CX with BX and Result BX = 00111101 10010110

• TEST Instruction – AND operand to update flags

TEST Instruction - This instruction ANDs the contents of a source byte or word with the contents of specified destination word. Flags are updated but neither operand is changed . TEST instruction is often used to set flags before a condition jump instruction

Examples:

TEST AL, BH                     ;AND BH with AL. no result is stored . Update PF, SF, ZF TEST CX, 0001H ;AND CX with immediate number
                                      ;no result is stored, Update PF,SF

Example :

                                      ;AL = 01010001
TEST Al, 80H                    ;AND immediate 80H with AL to test f MSB of AL is 1 or 0
                                     ;ZF = 1 if MSB of AL = 0 and AL = 01010001 (unchanged),PF = 0 , SF = 0,
                                     ;ZF = 1 because ANDing produced is 00

DIVISION INSTRUCTIONS

DIVISION INSTRUCTIONS

• DIV Instruction - Unsigned divide-Div source

When a double word is divided by a word, the most significant word of the double word must be in DX and the least significant word of the double word must be in AX. After the division AX will contain the 16 –bit result (quotient ) and DX will contain a 16 bit remainder. Again , if an attempt is made to divide by zero or quotient is too large to fit in AX ( greater than FFFFH ) the 8086 will do a type of 0 interrupt .

Example:

DIV CX                                   ; (Quotient) AX= (DX:AX)/CX 
                                             ; (Reminder) DX=(DX:AX)%CX

For DIV the dividend must always be in AX or DX and AX, but the source of the divisor can be a register or a memory location specified by one of the 24 addressing modes.
If you want to divide a byte by a byte, you must first put the dividend byte in AL and fill AH with all 0’s . The SUB AH,AH instruction is a quick way to do. If you want to divide a word by a word, put the dividend word in AX and fill DX with all 0’s. The SUB DX,DX instruction does this quickly.

Example:

                                           ; AX = 37D7H = 14, 295 decimal and BH = 97H = 151 decimal
DIV BH                                 ;AX / BH
                                           ; AX = Quotient = 5EH = 94 decimal and AH = Remainder = 65H = 101 

 decimal.

• IDIV Instruction - Divide by signed byte or word IDIV source

This instruction is used to divide a signed word by a signed byte or to divide a signed double word by a signed word.  If source is a byte value, AX is divided by register and the quotient is stored in AL and the remainder in AH.  If source is a word value, DX:AX is divided by register,and the quotient is stored in AL and the remainder in DX.

Example:

IDIV BL                                 ;Signed word in AX is divided by signed byte in BL

• AAD Instruction - ASCII adjust before Division

ADD converts unpacked BCD digits in the AH and AL register into a single binary number in the AX register in preparation for a division operation. Before executing AAD, place the Most significant BCD digit in the AH register and Last significant in the AL register. When AAD is executed, the two BCD digits are combined into a single binary number by setting AL=(AH*10)+AL and clearing AH to 0.

Example:

MOV AX,0205h                    ;The unpacked BCD number 25
AAD                                    ;After AAD , AH=0 and AL=19h (25).

After the division AL will then contain the unpacked BCD quotient and AH will contain the unpacked BCD remainder.

Example:

                                         ;AX=0607 unpacked BCD for 67 decimal CH=09H.
AAD                                   ;Adjust to binary before division AX=0043 = 43H =67 decimal.
DIV CH                               ;Divide AX by unpacked BCD in CH, AL = quotient = 07 unpacked BCD, AH = remainder = 04 unpacked BCD

• CBW Instruction - Convert signed Byte to signed word

CBW converts the signed value in the AL register into an equivalent 16 bit signed value in the AX register by duplicating the sign bit to the left. This instruction copies the sign of a byte in AL to all the bits in AH. AH is then said to be the sign extension of AL.

Example:

                                        ; AX = 00000000 10011011 = - 155 decimal
CBW                                 ; Convert signed byte in AL to signed word in AX.
                                        ; Result in AX = 11111111 10011011 and  = - 155 decimal

• CWD Instruction - Convert Signed Word to - Signed Double word

CWD converts the 16 bit signed value in the AX register into an equivalent 32 bit signed value in DX: AX register pair by duplicating the sign bit to the left.
The CWD instruction sets all the bits in the DX register to the same sign bit of the AX register. The effect is to create a 32- bit signed result that has same integer value as the original 16 bit operand.

Example:

Assume AX contains C435h. If the CWD instruction is executed, DX will contain FFFFh since bit 15 (MSB) of AX was 1. Both the original value of AX (C435h) and resulting value of DX : AX (FFFFC435h) represents the same signed number.

Example:

                                   ;DX = 00000000 00000000 and AX = 11110000 11000111 = - 3897 decimal
CWD                            ;Convert signed word in AX to signed double word in DX:AX
                                   ;Result DX = 11111111 11111111 and AX = 11110000 11000111 = -3897 decimal.

multiplication instructions

MULTIPLICATION INSTRUCTIONS

• MUL Instruction - Multiply unsigned bytes or words-MUL source

MUL Instruction - This instruction multiplies an unsigned multiplication of the accumulator by the operand specified by op. The size of op may be a register or memory operand .

Syntax: MUL op

Example:

                                             ;AL = 21h (33 decimal),BL = A1h(161 decimal )
MUL BL                                   ;AX =14C1h (5313 decimal) since AH≠0, CF and OF will set to 1.
MUL BH                                  ; AL times BH, result in AX
MUL CX                                  ;AX times CX, result high word in DX,low word in AX.

• IMUL Instruction - Multiply signed number-IMUL source

This instruction performs a signed multiplication.There are two types of syntax for this instruction. They are:

IMUL op                               ;In this form the accumulator is the multiplicand and op is the multiplier. op may be a register or a memory operand.
IMUL op1, op2                      ;In this form op1 is always be a register operand and op2 may be a register or a memory operand.

Example:

IMUL BH                               ;Signed byte in AL times multiplied by signed byte in BH and result in AX .

Example:

                                          ; 69 * 14
                                          ; AL = 01000101 = 69 decimal
                                          ; BL = 00001110 = 14 decimal
IMUL BL                               ;AX = 03C6H = + 966 decimal ,MSB = 0 because positive result , - 28 * 59
                                          ; AL = 11100100 = - 28 decimal ,BL = 00001110 = 14 decimal
IMUL BL                               ;AX = F98Ch = - 1652 decimal, MSB = 1 because negative result

• AAM Instruction - ASCII adjust after Multiplication

AAM Instruction - AAM converts the result of the multiplication of two valid unpacked BCD digits into a valid 2-digit unpacked BCD number and takes AX as an implicit operand. To give a valid result the digits that have been multiplied must be in the range of 0 – 9 and the result should have been placed in the AX register. Because both operands of multiply are required to be 9 or less, the result must be less than 81 and thus is completely contained in AL. AAM unpacks the result by dividing AX by 10, placing the quotient (MSD) in AH and the remainder (LSD) in AL.

Example:

MOV AL, 5
MOV BL, 7
MUL BL                       ;Multiply AL by BL , result in AX
AAM                           ;After AAM, AX =0305h (BCD 35)

subtarction (arithmetic instruction set)

SUBTRACTION INSTRUCTIONS

SUB Instruction - Subtract two numbers

These instructions subtract the source number from destination number destination and put the result in the specified destination. The source and destination must be of same type , means they must be a byte location or a word location.

• SBB Instruction - Subtract with borrow SBB destination, source

SBB instruction subtracts source from destination, and then subtracts 1 from source if CF flag is set and result is stored destination and it is used to set the flag.

ie.,destination = destination -(source + CF)

Example:

SUB CX, BX                                 ;CX – BX . Result in CX
SUBB CH, AL                               ; Subtract contents of AL and contents CF from contents of CH .
;Result in CH
SUBB AX, 3427H                          ;Subtract immediate number from AX

Example:

- Subtracting unsigned number
                                               ; CL = 10011100 = 156 decimal BH = 00110111 = 55 decimal
SUB CL, BH                               ; CL = 01100101 = 101 decimal CF, AF, SF, ZF = 0, OF, PF = 1
- Subtracting signed number
                                               ; CL = 00101110 = + 46 decimal BH = 01001010= + 74 decimal
SUB CL, BH                               ;CL = 11100100 = - 28 decimal,CF = 1, AF, ZF =0,SF = 1 result negative

• DEC Instruction - Decrement destination register or memory DEC destination.

DEC instruction subtracts one from the operand and sets the flag according to the result. DEC instruction is treated as an unsigned binary number.

Example:

                                              ; AX =8000h
DEC AX                                    ;After this instruction AX = 7999h
DEC BL                                    ; Subtract 1 from the contents of BL register

• NEG Instruction - From 2’s complement – NEG destination

NEG performs the two’s complement subtraction of the operand from zero and sets the flags according to the result.

                                           ;AX = 2CBh
NEG AX                                 ;after executing NEG result AX =FD35h.

Example:

NEG AL                                 ;Replace number in AL with its 2’s complement

NEG BX                                ;Replace word in BX with its 2’s complement

NEG BYTE PTR[BX]                ; Replace byte at offset BX in DS with its 2’s complement

• CMP Instruction - Compare byte or word -CMP destination, source.

The CMP instruction compares the destination and source ie.,it subtracts the  source from destination.The result is not stored anywhere. It neglects the  results, but sets the flags accordingly.This instruction is usually used before a conditional jump instruction.

Example:

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

• AAS Instruction - ASCII Adjust for Subtraction

AAS converts the result of the subtraction of two valid unpacked BCD digits to a single valid BCD number and takes the AL register as an implicit operand. The two operands of the subtraction must have its lower 4 bit contain number in the range from 0 to 9 .The AAS instruction then adjust AL so that it contain a correct BCD digit.

MOV AX,0901H                     ;BCD 91
SUB AL, 9                            ;Minus 9
AAS                                    ; Give AX =0802 h (BCD 82)

Example:( a )
                                         ;AL =0011 1001 =ASCII 9
                                         ;BL=0011 0101 =ASCII 5
SUB AL, BL                          ;(9 - 5) Result : ;AL = 00000100 = BCD 04,CF = 0
AAS                                    ;Result : AL=00000100 =BCD 04 , CF = 0 NO Borrow required

Example:( b )

                                         ;AL = 0011 0101 =ASCII 5
                                         ;BL = 0011 1001 = ASCII 9
SUB AL, BL                          ;( 5 - 9 ) Result : AL = 1111 1100 = – 4 in 2’s complement CF = 1
AAS                                    ;Results :AL = 0000 0100 =BCD 04, CF = 1 borrow needed

• DAS Instruction - Decimal Adjust after Subtraction

This instruction corrects the result (in AL) of subtraction of two packed BCD values. The flags which modify are AF, CF, PF, SF, ZF

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

- AL = AL – 6

- AF = 1

if AL > 9Fh or CF = 1 then:

- AL = AL - 60h

- CF = 1

Example:

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

Arithmetic instructions

• ADD Instruction - ADD destination, source
These instructions add a number from source to a number from some destination and put the result in the specified destination. The source and destination must be of same type , means they must be a byte location or a word location. If you want to add a byte to a word, you must copy the byte to a word location and fill the upper byte of the word with zeroes before adding.
• ADC Instruction - Add with carry
After performing the addition, the add with carry instruction ADC, adds the status of the carry flag into the result.
EXAMPLE:
ADD AL,74H ;Add immediate number 74H to content of AL
ADC CL,BL ;Add contents of BL plus carry status to contents of CL Results in CL
ADD DX, BX ;Add contents of BX to contents of DX
ADD DX, [SI] ;Add word from memory at offset [SI] in DS to contents of DX
; Addition of Un Signed numbers
ADD CL, BL ;CL = 01110011 =115 decimal + BL = 01001111 = 79 decimal Result in CL = 11000010 = 194 decimal
; Addition of Signed numbers
ADD CL, BL ;CL = 01110011 = + 115 decimal + BL = 01001111 = +79 decimal Result in CL = 11000010 = - 62 decimal
; Incorrect because result is too large to fit in 7 bits.
• INC Instruction - Increment - INC destination
INC instruction adds one to the operand and sets the flag according to the result. INC instruction is treated as an unsigned binary number.
Example:
; AX = 7FFFh
INC AX ;After this instruction AX = 8000h
INC BL ; Add 1 to the contents of BL register
INC CL ; Add 1 to the contents of CX register.
• AAA Instruction - ASCII Adjust after Addition
AAA converts the result of the addition of two valid unpacked BCD digits to a valid 2-digit BCD number and takes the AL register as its implicit operand.
Two operands of the addition must have its lower 4 bits contain a number in the range from 0-9.The AAA instruction then adjust AL so that it contains a correct BCD digit. If the addition produce carry (AF=1), the AH register is incremented and the carry CF and auxiliary carry AF flags are set to 1. If the addition did not produce a decimal carry, CF and AF are cleared to 0 and AH is not altered. In both cases the higher 4 bits of AL are cleared to 0.
AAA will adjust the result of the two ASCII characters that were in the range from 30h (“0”) to 39h(“9”).This is because the lower 4 bits of those character fall in the range of 0-9.The result of addition is not a ASCII character but it is a BCD digit.
Example:
MOV AH,0 ;Clear AH for MSD
MOV AL,6 ;BCD 6 in AL
ADD AL,5 ;Add BCD 5 to digit in AL
AAA ;AH=1, AL=1 representing BCD 11.
• DAA Instruction - Decimal Adjust after Addition
The contents after addition are changed from a binary value to two 4-bit binary coded decimal (BCD) digits. S, Z, AC, P, CY flags are altered to reflect the results of the operation.
If the value of the low-order 4-bits in the accumulator is greater than 9 or if AC flag is set, the instruction adds 6 to the low-order four bits.
If the value of the high-order 4-bits in the accumulator is greater than 9 or if the Carry flag is set, the instruction adds 6 to the high-order four bits.
Example:
MOV AL, 0Fh ; AL = 0Fh (15)
DAA ; AL = 15h
RET

FLAG TRANSFER INSTRUCTIONS

FLAG TRANSFER INSTRUCTIONS

• LAHF Instruction - Load Register AH From Flags

LAHF instruction copies the value of SF, ZF, AF, PF, and CF, into bits of 7, 6, 4, 2, 0 respectively of AH register. This LAHF instruction was provided to make conversion of assembly language programs written for 8080 and 8085 to 8086 easier.

• SAHF instruction - Store AH Register into FLAGS

SAHF instruction transfers the bits 0-7 of AH of SF, ZF, AF, PF, and CF, into the Flag register.

• PUSHF Instruction - Push flag register on the stack

This instruction decrements the SP by 2 and copies the word in flag register to the memory location pointed to by SP.

• POPF Instruction - Pop word from top of stack to flag - register.

This instruction copies a word from the two memory location at the top of the stack to flag register and increments the stack pointer by 2.    

SPECIAL ADDRESS TRANSFER INSTRUCTION

SPECIAL ADDRESS TRANSFER INSTRUCTION

• LEA Instruction - Load Effective Address

LEA Instruction - This instruction indicates the offset of the variable or memory location named as the source and put this offset in the indicated 16 – bit register.

Syntax – LEA register, source

Example:

LEA BX, PRICE              ;Load BX with offset of PRICE in DS
LEA BP, SS:STAK          ;Load BP with offset of STACK in SS
LEA CX, [BX][DI]           ;Load CX with EA=BX + DI

• LDS Instruction - Load register and Ds with words from memory

LDS Instruction - This instruction loads a far pointer from the memory address specified by op2 into the DS segment register and the op1 to the register.

Syntax – LDS register, memory address of first word or LDS op1, op2

Example:

LDS BX, [4326]           ; copy the contents of the memory at displacement 4326H in DS to BL, contents of the 4327H to BH. Copy contents of 4328H and 4329H in DS to DS register.

• LES Instruction - Load register and ES with words from memory

This instruction loads a 32- bit pointer from the memory address specified to destination register and Extra Segment. The offset is placed in the destination register and the segment is placed in Extra Segment. Using this instruction, the loading of far pointers may be simplified.

Syntax – LES register, memory address of first word

SIMPLE INPUT AND OUTPUT PORT TRANSFER INSTRUCTION

SIMPLE INPUT AND OUTPUT PORT TRANSFER INSTRUCTION

•  IN Instruction - Copy data from a port IN accumulator, port

This IN instruction will copy data from a port to the AL or AX register. For the Fixed port IN instruction type the 8 – bit port address of a port is specified directly in the instruction.

Example:

IN AL,0C8H                     ;Input a byte from port 0C8H to AL
IN AX, 34H                      ;Input a word from port 34H to AX
A_TO_D EQU 4AH
IN AX, A_TO_D                ;Input a word from port 4AH to AX

For a variable port IN instruction, the port address is loaded in DX register before IN instruction. DX is 16 bit. Port address range from 0000H – FFFFH.

Example:

MOV DX, 0FF78H             ;Initialize DX point to port
IN AL, DX                        ;Input a byte from a 8 bit port 0FF78H to AL
IN AX, DX                        ;Input a word from 16 bit port to 0FF78H to AX.

• OUT Instruction - Output a byte or word to a port – OUT port, accumulator AL or AX.

The OUT instruction copies a byte from AL or a word from AX or a double from the accumulator to I/O port specified by op. Two forms of OUT instruction are available : (a) Port number is specified by an immediate byte constant, ( 0 - 255 ).It is also called as fixed port form. (b) Port number is provided in the DX register ( 0 – 65535 )

Example: (a)

OUT 3BH, AL              ;Copy the contents of the AL to port 3Bh
OUT 2CH,AX               ;Copy the contents of the AX to port 2Ch

Example: (b)

MOV DX, 0FFF8H        ;Load desired port address in DX
OUT DX, AL                ; Copy the contents of AL to FFF8h
OUT DX, AX                ;Copy content of AX to port FFF8H

GENERAL – PURPOSE BYTE OR WORD TRANSFER INSTRUCTIONS

GENERAL – PURPOSE BYTE OR WORD TRANSFER INSTRUCTIONS

•  MOV Instruction - MOV destination, source

The MOV instruction copies a word or a byte of data from a specified source to a specified destination . MOV op1, op2

Example:

MOV CX, 037AH             ; MOV 037AH into the CX.
MOV AX, BX                   ;Copy the contents of register BX to AX
MOV DL,[BX]                  ;Copy byte from memory at BX to DL , BX contains the offset of byte in DS.

• PUSH Instruction - PUSH source

PUSH instruction decrements the stack pointer by 2 and copies a word from a specified source to the location in the stack segment where the stack pointer pointes.

Example:

PUSH BX                     ;Decrement SP by 2 and copy BX to stack
PUSH DS                     ;Decrement SP by 2 and copy DS to stack
PUSH TABLE[BX]        ;Decrement SP by 2 and copy word from memory in DS at EA = TABLE + [BX] to stack .

• POP Instruction - POP destination

POP instruction copies the word at the current top of the stack to the operand specified by op then increments the stack pointer to point to the next stack.

Example:

POP DX                   ;Copy a word from top of the stack to DX and increments SP by 2.
POP DS                   ; Copy a word from top of the stack to DS and increments SP by 2.
POP TABLE [BX]     ;Copy a word from top of stack to memory in DS with EA = TABLE + [BX].

• XCHG Instruction - Exchange XCHG destination, source

The Exchange instruction exchanges the contents of the register with the contents of another register (or) the contents of the register with the contents of the memory location. Direct memory to memory exchange are not supported.

Syntax: XCHG op1, op2 - The both operands must be the same size and one of the operand must always be a register .

Example:

XCHG AX, DX                      ;Exchange word in AX with word in DX
XCHG BL, CH                      ;Exchange byte in BL with byte in CH
XCHG AL, Money [BX]        ;Exchange byte in AL with byte in memory at EA.

• XLAT/XLATB Instruction - Translate a byte in AL

XLAT exchanges the byte in AL register from the user table index to the table entry, addressed by BX.  It transfers 16 bit information at a time. The no-operands form (XLATB) provides a "short form" of the XLAT instructions.

Example:MOV AL, [BX+AL]

Instruction set of 8086

1. DATA TRANSFER INSTRUCTIONS
2. ARITHMETIC INSTRUCTIONS
3. BIT MANIPULATION INSTRUCTIONS
4. STRING INSTRUCTIONS
5. PROGRAM EXECUTION TRANSFER INSTRUCTIONS
6. PROCESS CONTROL INSTRUCTIONS

1.DATA TRANSFER INSTRUCTIONS

The DATA TRANSFER INSTRUCTIONS are those, which transfers the DATA from any one source to any one destination.The datas may be of any type. They are again classified into four groups.They are:

GENERAL – PURPOSE BYTE OR WORD TRANSFER INSTRUCTIONS	SIMPLE INPUT AND OUTPUT PORT TRANSFER INSTRUCTION	SPECIAL ADDRESS TRANSFER INSTRUCTION	FLAG TRANSFER INSTRUCTIONS
MOV PUSH POP XCHG XLAT	IN OUT	LEA LDS LES	LAHF SAHF PUSHF POPF

2.ARITHMETIC INSTRUCTIONS

These instructions are those which are useful to perform Arithmetic calculations, such as addition, subtraction, multiplication and division.They are again classified into four groups.They are:

ADDITION INSTRUCTIONS	SUBTRACTION INSTRUCTIONS	MULTIPLICATION INSTRUCTIONS	DIVISION INSTRUCTIONS
ADD ADC INC AAA DAA	SUB SBB DEC NEG CMP AAS DAS	MUL IMUL AAM	DIV IDIV AAD CBW CWD

3.BIT MANIPULATION INSTRUCTIONS

These instructions are used to perform Bit wise operations.

LOGICAL INSTRUCTIONS	SHIFT INSTRUCTIONS	ROTATE INSTRUCTIONS
NOT AND OR XOR TEST	SHL / SAL SHR SAR	ROL ROR RCL RCR

4. STRING INSTRUCTIONS

The string instructions function easily on blocks of memory.They are user friendly instructions, which help for easy program writing and execution. They can speed up the manipulating code.They are useful in array handling, tables and records.

STRING INSTRUCTIONS

REP REPE / REPZ REPNE / REPNZ MOVS / MOVSB / MOVSW COMPS / COMPSB / COMPSW SCAS / SCASB / SCASW LODS / LODSB / LODSW STOS / STOSB / STOSW

5.PROGRAM EXECUTION TRANSFER INSTRUCTIONS

These instructions transfer the program control from one address to other address. ( Not in a sequence). They are again classified into four groups.They are:

UNCONDITIONAL TRANSFER INSTRUCTIONS	CONDITIONAL TRANSFER INSTRUCTIONS	ITERATION CONTROL INSTRUCTIONS	INTERRUPT INSTRUCTIONS
CALL RET JMP	JA / JNBE JAE / JNB JB / JNAE JBE / JNA JC JE / JZ JG / JNLE JGE / JNL JL / JNGE JLE / JNG JNC JNE / JNZ JNO JNP / JPO JNS JO JP / JPE JS	LOOP LOOPE / LOOPZ LOOPNE / LOOPNZ JCXZ	INT INTO IRET

6.PROCESS CONTROL INSTRUCTIONS

These instructions are used to change the process of the Microprocessor. They change the process with the stored information. They are again classified into Two groups.They are:

FLAG SET / CLEAR INSTRUCTIONS	EXTERNAL HARDWARE SYNCHRONIZATION INSTRUCTIONS
STC CLC CMC STD CLD STI CLI	HLT WAIT ESC LOCK NOP