EE353Chap4-1.0

Chapter 4 : Instruction Set for 8086

1EE353: Introduction to MicroprocessorDr. Ridha Jemal

4.1. Data Transfer Instructions

4.2. Arithmetic Instructions

4.3. Logic Instructions

4.4. Comparison Instruction

4.5. Jump Instructions

By Dr. Ridha JemalElectrical Engineering Department

College of Engineering

King Saud University

Data Transfer Instruction

2Dr. Ridha Jemal EE353: Introduction to Microprocessor 1431-1432

• Mov Instruction: This instruction is well studied

• Push and Pop Instructions:

The data is transferred according tot the LIFO (Last In First out)

The Push instruction transfers two bytes to the top of the stack.

Ex: PUSH CX ; [ss:SP-1] CH

; [ss:SP-1] CL

; SP SP-2

PUSH reg16 PUSH BX 16-bit registerPUSH seg PUSH DS Segment register PUSH mem PUSH [DI+2] memory PUSHA PUSHA Save all 16-bit registers PUSHF PUSHF Save flags

PUSHA instruction pushes all the internal registers onto the stack in thefollowing order: AX, CX, DX, BX, SP, BP, SI, DI. The value of the SP is thatbefore the PUSHA instruction.The PUSHF (push flag) instruction copies the contents of the flag register tothe stack.



• Push Instruction:



The operation of PUSHA instruction showing the locationsand stack data (push flag) instruction copies the contents ofthe flag register to the stack.



• PoP Instruction:

It performs the reversion operation of the PUSHinstruction.

• Ex1: POP BX

BL [ss:SP]

BH [ss:SP+1]

SP SP+2

• Ex2:MOV AX, 7C3DH

PUSH AX

POP BX

Int 3





• XCHG Instruction:

It performs a register-to-register or register-to-memory swap.

operation of the PUSH instruction.

• Ex: XCHG AX,BX ; AX BX

XCHG AL,AH ; AL AH

XCHG [SI],DX ;[DS:SI] DX

• LAHF and SAHF Instruction:

LAHF ; AH Flags Low

LAHF ; Flags Low AH





• Ex: XMOVCHG AX,BX; AX BX

XCHG AL,AH ; AL AH

XCHG [SI],DX ;[DS:SI] DX

• LAHF and SAHF Instruction:

LAHF ; AH Flags Low

LAHF ; Flags Low AH



• LDS and LES Instruction:

LDS: Load memory double word into word register and DS.

LES: Load memory double word into word register and ES.

LDS and LES load a 16-bit register with offset address retrieved from amemory location then load either DS or ES with a segment addressretrieved from memory.

LDS BX,word ptr [SI]

BL [SI], BH [SI+1]

DS [SI+3:SI+2] in the data segment

LES BX,word ptr [SI]

BL [SI], BH [SI+1]

ES [SI+3:SI+2] in the extra segment





• This instruction transfers the 32-bit number, addressed by DI in the data segment, into the BX and DS registers.

• LDS and LES instructions obtain a new far address from memory.

• – offset address appears first, followed by the segment address

• This format is used for storing all 32-bit memory addresses.

• A far address can be stored in memory by the assembler.





• STRING DATA TRANSFERS

o Instructions for moving large blocks of data or strings.

o Five string data transfer instructions: LODS, STOS, MOVS, INS, and OUTS.

o Each allows data transfers as a single byte, word, or double word.

o Before the string instructions are presented, the operation of the D flag-bit (direction), DI, and SI must be understood as they apply to the stringinstructions.

• The Direction Flag

o The direction flag (D, located in the flag register) selects the auto-increment or the auto-decrement operation for the DI and SI registersduring string operations. It is used only with the string instructions

o The CLD instruction clears the D flag and the STD instruction sets it . CLDinstruction selects the auto-increment mode and STD selects the auto-decrement mode .



DI and SI

• During execution of string instruction, memory accesses occur through DI and SI registers.

– DI offset address accesses data in the extra segment for all string instructions that use it

– SI offset address accesses data by default in the data segment

Setting the Direction Flag

Two instructions used:

STD: set direction flag so that the pointers are auto decremented.

CLD: clear direction flag so that the pointers are auto incremented.


• STOS Instruction:

Stores AL or AX at the extra segment memory location addressed by the DIregister; if DF = 0, increment DI, else decrement DI.

STOSB ES:[DI] AL

IF DF=0, DI DI+1

IF DF=1, DI DI-1

STOSW ES:[DI] AL

ES:[DI+1] AH

IF DF=0, DI DI+2

IF DF=1, DI DI-2



• EXAMPLE For STOSB Instruction

ORG 100

LEA DI, Table1 ; Point to the offset of Table1

MOV AL,12h ; Put 12H in the AL

MOV CX,5 ; number of values to be stored in Table 1

REP STOSB

RET

Table 1 db 6 dup (?)

After the execution, Table1 contains 5 values of 12H

12h 12h 12h 12h 12h 00


• Repeat Prefix

Preceding the string instructions STOS or MOVS with REP causes these instructions to be repeated a number of times equal to the contents of CXregister.

REP STOSB ; STOSB ; CX CX-1 (repeat until CX = 0)


• LODS Instruction:

Loads AL or AX with the data stored at the data segment memory locationaddressed by the SI register; if DF = 0, increment SI, else decrement SI

LODSB AL DS:[SI]

IF DF=0, SI SI+1

IF DF=1, SI SI-1

LODSW AL DS:[SI]

AH DS:[SI+1]

IF DF=0, SI SI+2

IF DF=1, SI SI-2



• EXAMPLE For LODSB Instruction

ORG 100

LEA SI, Table2 ; Point to the offset of Table2

MOV CX,5 ; number of values to be loaded from Table

2

REP LODSB

RET

Table2 db ‘1’,’2’,’3’,’4’,’5’,’6’,’7’

After the execution, Table1 contains AL contains 35h

which is the ASCII code of 5




• MOVS Instruction:

Transfers data from a memory location to another.

This is the only memory-to- memory transfer allowed.

The MOVS instruction transfer a byte or a word from the data segment locationaddressed by the SI to the extra segment location addressed by the DI. Thepointers then increment or decrement according to DF.

MOVSB ES:[DI] DS:[SI]

IF DF=0, SI SI+1

DI DI+1

IF DF=1, SI SI-1

DI DI-1

MOVSW ES:[DI] DS:[SI]

ES:[DI+1] DS:[SI+1]

IF DF=0, SI SI+2

DI DI+2

IF DF=1, SI SI-2

DI DI+2



• XLAT Translate Instruction

Operation: AL AL+BX+DS*16

Affected Flags : None

o Converts the contents of the AL register into a number stored in a memorytable.

o Performs the direct table lookup technique often used to convert one codeto another .

• An XLAT instruction first adds the contents of AL to BX to form a memoryaddress within the data segment.

o Copies the contents of this address into AL.

o Only instruction adding an 8-bit to a 16-bit number.



• Example for XLAT Instruction

Assume DS=0300H, BX=0100h and AL=0DH

ODH represents the ASCII code of the character CR.

The execution of XLAT replaces the contents of AL by the content of the memorylocation with the physical address:

PA = DS*16+BX +AL

= 03000+0100+0DH=0310DH

Thus

AL [0310DH].

Assuming this memory locations contains 52H (EBCDIC code for CR) this value isplaced in AL.

Thus = 52H

Performs the direct table lookup technique often used to convert one code toanother .

.



The operation of the XLAT instruction at the point just before 6DH is loaded into AL



• INS OUTS Instructions

• Assume that one needs a table for the values of x2, where x is between 0 and 9.

First the table is generated and stored in memory:

square DB 0,1,4,9,16,25,36,49,64,81

MOV BX,OFFSET square ; Load the offset address

MOV AL,05 ; AL=05 will retrieve 6th element

XLAT ; Pull out of table the element

; Put in AL

After execution of this program, the AL register will have 25 (19H)




An IN instruction transfers data from an external I/O device to AL or AX. An OUT instruction transfers data from AL or AX to an external I/O device.

The I/O device address is called port address.

In direct addressing mode the address is a single byte.

In indirect addressing mode the address is two bytes in DX

• Ex: IN AL,2EH ; AL port 2EH

IN AX,26H ; AL port 26H

; AH port 27H

IN AL,DX ; AL port DX

IN AX,DX ; AL port DX

; AH port DX+1




An IN instruction transfers data from an external I/O device to AL or AX. An OUT instruction transfers data from AL or AX to an external I/O device.

The I/O device address is called port address.

In direct addressing mode the address is a single byte.

In indirect addressing mode the address is two bytes in DX

• Ex: OUT 2EH,AL ; Port 2EH AL

OUT 2EH,AX ; Port 2EH AL

; Port 2EH AH

OUT DX,AL ; Port DX AL

OUT DX,AX ; Port DX AL

; Port DX+1 AH

Arithmetic Instructions : Basic Instructions (+, -, *, /)


• ADD Instruction: ADD Dest, Source

Dest Source + Dest; Flag bits are affected

Ex: MOV AX,32FAh

MOV BX,1F02h

ADD AL,BL

ADD AL,5

• INC (Increment by 1) Instruction: INC Dest

Ex: MOV AX,67f2h

INC AL

INC AH

Increment instruction (INC) adds 1 to a register or a memory location. The INC canadd 1 to any register or memory location, except a segment register. With indirectmemory increments, the size of the data must be described by using the BYTE PTR orWORD PTR directives. The reason is that the assembler program cannot determine if,for example, the INC [DI] instruction is a byte-, or word -sized increment.

Arithmetic Instructions : Basic Instructions (+, -, *, /)


• ADC (ADD with carry) Instruction: ADC Dest, Source

Ex: MOV AX,32C5h

MOV CX,1CA2h

ADD AL,2

ADC AH,AL

ADD AX,CX

Increment instructions affect the flag bits except the carry flag bit. Carry doesn'tchange because we often use increments in programs that depend upon the contentsof the carry flag.

Note that increment is used to point to the next memory element in a byte-sizedarray of data only.If word-sized data are addressed. it is better to use an ADD DI,2 instruction to modifythe DI pointer in place of two INC DI instructions.

An addition-with-carry instruction (ADC) adds the bit in the carry flag (C) to theoperand data. This instruction mainly appears in software that adds numbers that arewider than 16 bits in the 8086.

Arithmetic Instructions (contd.)


• SUB Instruction: SUB Dest, Source

o Dest Source - Dest; Flag bits is affected

Ex: MOV AL,A7h

MOV BL,3Fh

SUB AL,BL

SUB AL,5

Many forms of subtraction (SUB) appear in the instruction set.

The only types of subtraction not allowed are memory-to-memory and segmentregister subtractions. Like other arithmetic instructions, the subtract instructionaffects the flag bits.



• DEC (decrement by 1) Instruction: DEC Dest

Ex: SUB AX,AX

DEC AX

DEC BYTE PTR [1000h]

Or

DEC BYTE PTR [1000h]

DEC instruction subtracts a 1 from a register or the contents of a memorylocation. The decrement indirect memory data instructions require BYTE PTR, orWORD PTR.

Arithmetic Instructions


• SBB Instruction (Subtract with borrow):o SBB Dest, Source

o Dest Source - Dest;

o Flag bits is affected

Ex: MOV AL,A7h

MOV BL,3Fh

SUB AL,BL

SUB AL,5

A subtraction-with-borrow (SBB) instruction functions as a regular sbtraction,except that the carry flag (C), which holds the borrow, also subtracts from thedifference. The most common use for this instruction is for subtractions that arewider than 16 bits in the 8086.Wide subtractions require that borrows propagate through the subtraction, just aswide additions propagate the carry. Like the SUB instruction, SBB affects the flags.Notice that the immediate subtract from memory instruction in this table requiresa BYTE PTR , or WORD PTR directive.



• MUL Instruction (Multiplication of unsigned number ):

o MUL Data AX AL*Data or DXAX AX*Data

o The source operand can be a memory location or a register.

o When we multiply an n-bit number with an m-bit number, the product will be

at most (n+m) bits.

Ex: MOV AL,05h

MOV BL,04h

MUL BL ; ALAL*BL = 14h



• IMUL Instruction (Multiplication of signed number ):o IMUL Data AX AL*Data or DXAX AX*Data

o When we multiply an n-bit number with an m-bit number, the product will be

at most (n+m) bits.

Ex: MOV AL,-1 ; MOV AL,FFh

MOV BL,3

IMUL BL ; AX = FFFDh



Examples

An 8-bit unsigned integer, X, is present in the data segment at offset 2200h. Write an

assembly code to calculate R=X4+1. We assume That R can be stored in 32 bits

and should be stored in the data segment starting at offset 7000H.

MOV AL,[2200h] ; AL=X

MUL AL ; AX=X2

MUL AX ; DXAX=X4

ADD AX,1

ADC DX,0 ; DXAX= X3+1

MOV [7000h],AX

MOV [7002h],DX



Examples

Suppose an 8-bit register unsigned number, X, is present in the data segment

of memory at offset 1200h. Write a program to calculate: R=X3+1 . We

assume That R can needs 32 bits at most and should be stored in the data

segment starting at offset 2000H.

MOV AL,[1200h] ; AL=X

MOV BL,AL ; BL=X

MUL AL ; AX=X2

MOV BH,0 ; BL=X

MUL BX ; DXAX=X3

ADD AX,1

ADC DX,0 ; DXAX= X3+1

MOV [2000h],AX

MOV [2002h],DX



• DIV Instruction (Multiplication of unsigned number ):o DIV Data Quotient in AL ; Reminder in AH

Ex: MOV AL,19

MOV CL,3

DIV CL ; AL 6 (quotient); AH 1 (reminder)

• IDIV Instruction (Multiplication of signed number ):o DIV Data Quotient in AL ; Reminder in AH

Ex: MOV AX,16 ; AL = 10h

MOV BL,-3 ; BL = FDh (2’S comp. of 3)

IDIV BL

Example: Suppose an 8-bit number X is present in AL.

Write an assembly code to calculate

MOV AL,X

MOV BL,4

CBW BL ; Convert Byte to word BH is filled with zeros

IDIV BL


Logic Instructions

• AND Instruction : Affected Flag: OF, SF, ZF, PF, CFAND Dest, Source; Dest (Dest Source)

Some flag bits change (S, Z and P. C = 0)

Uses any addressing mode except memory to memory and

segment registers addressing.

Ex: MOV AX,3FCAh

MOV CX,CC09h

AND AL,CL ;AL = 08

AND AL,F1h ;AL = 0

The AND operation is used for masking. We can set a particular bit in aregister to 0 without changing other bits.

x.0 = 0 x x x x x x x x

x.1 = x 1 1 1 1 1 1 0 0

x x x x x x 0 0

• Logical Operation : AND, OR, NOT, XOR


• OR Instruction : OR Dest, Source; Affected Flag: OF, SF, ZF, PF, CF

Dest (Dest Source)

Some flag bits change (S, Z and P. C = 0)

Uses any addressing mode except memory to memory and

segment registers addressing.

Ex: MOV AL,13h

MOV BL,30h

OR AL,BL ;AL = 33h

OR instruction is used to set a particular bit in a register to 1 (Bit Setting).

x+1 = 1 x x x x 0 x x 0

x+0 = x 0 0 0 0 1 0 0 1

x x x x 1 x x 1

Ex: OR AL,BH

OR SI,DI

OR BX,[SI]

OR Data[SI+20h],AL

Logic Instructions


• XOR Instruction : XOR Dest, Source; Affected Flag: OF, SF, ZF, PF, CF

Dest (Dest Source)

XOR is used to complement a particular bits (BitComplementing).

0 0 = 0

0 1 = 1

1 1 = 0

x 1 = x’

x 0 = x

Ex: XOR AL,AL

OR CX,600h

AND CX,FFFCh

XOR CX,1000h ;

Logic Instructions


• NOT Instruction : NOT Dest; Affected Flag: None

NOT is a logical inversion

Ex: NOT CL ;CH is one’s complement

NOT TEMP

NOT BYTE PTR [1000h] ;1’s complement of the byte in 1000h

• NEG Instruction : NEG Dest;

NEG is an arithmetic inversion (2’s complement)

Ex: NEG DL ;DL is 2’s complement

NEG BX

Logic Instructions


• SHL Shift Logical Left Instruction : SHL Target, Count;

o SHL Target, 1;

o SHL Target, CL;

SHL multiplies the number signed or unsigned by 2 oneach shift

The target can be a register or a memory

Ex: MOV BL,10110101B

SHL BL,1

MOV CL,3

SHL BL,CL ;

Shift Instruction

PF, SF and ZF flags are affected by shift instructions but left unchanged by

the rotate instructions.

SHL 0

… …Carry

Target

Logic Instructions


Ex: MOV AL,00000011B

SHL AL,1

SHL AL,-1

MOV AL,-1

SHL AL,1

SHL AL,1

Shift Instruction

Logic Instructions


• SHR Shift Logical Right Instruction : SHR Target, Count;o SHR Target, 1;

o SHR Target, CL;

SHL divides the number signed or unsigned by 2 on eachshift


Ex: MOV AL,8

SHR AL,1

MOV BL,16

MOV CL,3

SHR BL,CL ;

Shift Instruction

SHR 0

… …Carry

Target

Logic Instructions


• SAL Shift Arithmetic left Instruction :o SAL Target, 1;

o SAL Target, CL;

SAL divides a signed number by 2 on each shift


Ex: MOV AL,8

SAL AL,1

MOV BL,16

MOV CL,3

SAL BL,CL ;

Shift Instruction

SAL 0

… …Carry

Target

Logic Instructions


• SAR Shift Arithmetic Right Instruction :o SAR Target, 1;

o SAR Target, CL;

SAL divides a signed number by 2 on each shift(The MSBremains the same)


Ex: MOV AL,6

SAR AL,1

MOV BL,-16

MOV CL,3

SAR BL,CL ;

Shift Instruction

SAR

… …Carry

Target

Logic Instructions


• Shifting Example

An 16-bit unsigned number X is present in AX register. Write a program to compute 10.X, assuming that final result can be filled in 16 bits.

Solution #1 using MUL (Delay consuming solution)

Ex: MOV BX,10 ; 2 clock cycles

MUL BX ; DXAX=10X -> 120 clock cycles

Solution #2 using SHIFT an ADD

Ex: MOV AX,1 ; AX=2X -> 2 clock cycles

MOV BX,AX ; BX=2X -> 2 clock cycles

SHL AX,1 ; AX=4X -> 2 clock cycles

SHL AX,1 ; AX=8X -> 2 clock cycles

ADD AX,BX ; AX=10X -> 4 clock cycles

-> 12 clock cycles

Example

Logic Instructions


• ROL/ROR Rotate Left (Right) Instruction :o ROL/ROR Target, 1;

o ROL/ROR Target, CL;

All except immediate and segment registers.

Ex: MOV AL,BCh

ROL AL,1

ROL AL,1

ROR AL,1

ROR AL,1

Rotate Instruction

ROL

… …Carry

Target

ROR

… … Carry

Target

Logic Instructions


• RCL/RCR Rotate Through Carry Left (Right) Instruction :o RCL/RCR Target, 1;

o RCL/RCR Target, CL;


Rotate Instruction

RCL

… …Carry

Target

RCR

… … Carry

Target

Ex: Consider a 64-bit number X=X63X62…X0 located in DXCXBXAX

registers. Write an assembly code to shift X one position left.

Sol1: SHL AX,1

SHL BX,1

SHL CX,1

SHL CX,1

Sol2: SHL AX,1

RCL BX,1

RCL CX,1

RCL CX,1

Logic Instructions


• RCL Shift Arithmetic Right Instruction :o SAR Target, 1;

o SAR Target, CL;


Count must be 1 or CL.

Ex: MOV AL,6

SAR AL,1

MOV BL,-16

MOV CL,3

SAR BL,CL ;

Rotate Instruction

Logic Instructions


Comparison Instruction

• CMP (Compare)Instruction :• CMP Source1 – Source2; Update flags

The comparison instruction (CMP) is a subtraction that changes only the flag bits;the destination operand never changes. A comparison is useful for checking theentire contents of a register or a memory location against another value.

• A CMP is normally followed by a conditional jump instruction, which tests thecondition of the flag bits. The only disallowed forms of compare are memory-to-memory and segment register compares.

Ex1: CMP CL,BL

CMP AX,SP

CMP AX,SP

CMP AX,2000h

CMP [DI],CH

CMP CL,[BP]

Ex2: CMP AL,10h ;compare AL with 10h

JAE label ;if 10h or obove

•


Jump Instructions

Short Jump (relative jump): is a two-byte instruction where the targetaddress within -128 to +127 of the IP value

• JMP Instruction : JMP Disp.

o Unconditional jump (without testing any condition)

o Conditional jump where condition is tested through one of the following flagbits : Z, C, P, O, S

Ex: XOR BX,BX ; BX= 0

Start: MOV AX,1 ; AX= 1

ADD AX,BX ; BX= 0

JMP SHORT Next ; or JMP Next (Opcode=EB)

Next: MOV BX,AX

JMP Start


Jump Instructions

Near Jump :

• Direct: this is a 3-bytes instruction. the first byte is the code and the nexttwo are the signed number displacement value.

• Register indirect jump

JMP DI ; IP DI

o Any non-segment register can be used.

• Memory indirect jump:

JMP WORD PTR [BX]

Far Jump: Target address is outside the present code segment. A far jump instruction replaces the contents of both CS and IP with the four bytes following the opcode

• Direct far JMP

• Memory indirect far jump

JMP DWORD PTR [BX]


Jump Instructions

Opcode

Disp. Low Disp. High

CS Low CS HighIP Low IP High

Disp. EB

E9

EA

Short Jump

Near Jump

Far Jump


Labels can be defined in

separated lines.

Example: calc andstop

Labels can be defined in the

same line

Example: back

Example: Add two numbers in ax and bx

Comments

Unconditional Jump

Jump Instructions


Instruction Description ConditionInstruction inverse

JZ , JE Jump if Zero (Equal). ZF = 1 JNZ, JNE

JC , JB, JNAEJump if Carry (Below, Not Above Equal)

CF = 1 JNC, JNB, JAE

JS Jump if Sign. SF = 1 JNS

JO Jump if Overflow. OF = 1 JNO

JPE, JP Jump if Parity Even. PF = 1 JPO

JNZ , JNE Jump if Not Zero (Not Equal). ZF = 0 JZ, JE

JNC , JNB, JAE

Jump if Not Carry (Not Below, Above Equal)

CF = 0 JC, JB, JNAE

JNS Jump if Not Sign. SF = 0 JS

JNO Jump if Not Overflow. OF = 0 JO

JPO, JNP Jump if Parity Odd (No Parity). PF = 0 JPE, JP

Conditional Jump Checking only one flag

Jump Instructions


Conditional Jump Checking only one flag

Jump Instructions

Instruction Name condition

JZ label Jump if Zero Jump if ZF=1

JNZ label Jump if Not Zero Jump if ZF=0

JE label Jump if Equal Jump if ZF=1

JNE label Jump if Not Equal Jump if ZF=0

JC label Jump if Carry Jump if CF=1

JNC label Jump if Not Carry Jump if CF=0

JS label Jump if Sign Jump if SF=1

JNS label Jump if Not Sign Jump if SF=0

JO label Jump if Overflow Jump if OF=1

JNO label Jump if Not Overflow Jump if OF=0

JP label Jump if Parity Jump if PF=1

JNP label Jump if Not Parity Jump if PF=0


Instruction Description ConditionInstruction

inverse

JE , JZJump if Equal (=).Jump if Zero.

ZF = 1 JNE, JNZ

JNE , JNZJump if Not Equal (<>).Jump if Not Zero.

ZF = 0 JE, JZ

JG , JNLEJump if Greater (>).Jump if Not Less or Equal (not <=).

ZF = 0 et SF = OF JNG, JLE

JL , JNGEJump if Less (<).Jump if Not Greater or Equal (not >=).

SF <> OF JNL, JGE

JGE , JNLJump if Greater or Equal (>=).Jump if Not Less (not <).

SF = OF JNGE, JL

JLE , JNGJump if Less or Equal (<=).Jump if Not Greater (not >).

ZF = 1 ou SF <> OF JNLE, JG

Conditional Jump Checking on Signed numbers

Jump Instructions


Instruction Description Condition Inst.inverse

JE , JZJump if Equal (=).Jump if Zero.

ZF = 1 JNE, JNZ

JNE , JNZJump if Not Equal (<>).Jump if Not Zero.

ZF = 0 JE, JZ

JA , JNBEJump if Above (>).Jump if Not Below or Equal (not <=).

CF = 0 and ZF = 0 JNA, JBE

JB , JNAE, JCJump if Below (<).Jump if Not Above or Equal (not >=).Jump if Carry.

CF = 1 JNB, JAE, JNC

JAE , JNB, JNCJump if Above or Equal (>=).Jump if Not Below (not <).Jump if Not Carry.

CF = 0 JNAE, JB

JBE , JNAJump if Below or Equal (<=).Jump if Not Above (not >).

CF = 1 ou ZF = 1 JNBE, JA

Conditional Jump Checking on Unsigned numbers

Jump Instructions


Example


Example

We need to add two signed numbers N1 and N2 se located at offsets

1100H and 1101H.

The result is stored at the offset 1103H if it is negative and at the offset

1102H if tit is positive and at the offset 1104H if it is null.

1. Write the corresponding assembly code


Example

Start

Result

<0

N1+N2

Result

=0

End

Store Result

in 1103h

Yes

No

Yes

No

Store Result

in 1104h

Store Result

in 1102h

Ex: MOV AL,[1100h] ; Al=N1

ADD AL,[1101h] ; Al=N1+N2

JS Negative

JZ AL,Null

MOV[1102h],AL ;Positive result

JMP End

Negative: MOV[1103h],AL ;Negative result

JMP End

Null: MOV[1104h],AL ;Null result

End: HLT


Control Transfer Instructions

Call and Jump Instructions




data segment

SrcBuff DB 1,2,3,4,5,6,7,8,9,10

DesBuff DW 10 dup (0)

Ends

code segment

PI proc near

mov al,cl

mul cl

add ax,2 ; x2 +2

ret

PI endp

start:

mov ax, data

mov ds, ax

mov bx, 10

mov si,offset SrcBuff

mov di,offset DesBuff

x1: mov cl, [si]

call PI

mov [di],ax

inc si

add di,2

dec bx

jnz x1 ; These 2 instructions can be replaced by loop

mov ax, 4c00h ; exit to operating system.

int 21h

Ends

end start ; set entry point and stop the assembler.

Suppose two arrays X and Y containing 10

elements located in the data segment

X is defined within 8 bits (byte) where Y is

defined using 16 bits (word)



Call and Loop Instructions

data segment

SrcBuff DB 1,2,3,4,5,6,7,8,9,10

DesBuff DW 10 dup (0)

Ends

code segment

PI proc near

mov al,bl

mul bl

add ax,2 ; x2 +2

ret

PI endp

start:

mov ax, data

mov ds, ax

mov cx, 10

mov si,offset SrcBuff

mov di,offset DesBuff

x1: mov bl, [si]

push cx

call PI

pop cx

mov [di],ax

inc si

add di,2

lopp x1


int 21h

Ends


Suppose two arrays X and Y containing 10

elements located in the data segment

X is defined within 8 bits (byte) where Y is

defined using 16 bits (word)




data segment

ends

code segment

start:

PI proc near

mov al,cl

mul cl

add ax,2 ; x2 +2

ret

PI endp

mov bx, 100

mov si,1000h

mov di,3000h

x1: mov cl, [si]

call PI

mov [di],ax

in si

add di,2

dec bx

jnz x1


int 21h

ends



; multi-segment executable file template.

data segment

MSG1 DB 30h,31h,32h,0Dh,0Ah,'aaa',0dh,0Ah,0

MSG2 DB 'This is a sample line.',0

ends

code segment

string proc near

lodsb

cmp al,0

je stop

mov dl,al

mov ah,2

int 21h

jmp string

stop: ret

start:

; set segment registers:

mov ax, data

mov ds, ax

; add your code here

cld

mov si, offset MSG1

call string

mov si, offset MSG2

call string


int 21h

ends


EE353Chap4-1.0

Documents

indirect addressing

called port

data segment

data transfer

div data quotient

direct addressing

ah flags low

flags low