8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
Ex: Write an ALP to find the sum of the following series:
Sum= 11 + 22 + 33 + 44 + 55 + 66
Ans.
MOV BX , 0000H
MOV SI , 0001H
XOR DI , DI
NEXT: CALL SUBR
ADD BX , AX
ADC DI , 0000H
INC SI
CMP SI , 0007H
JNZ NEXT
HLT
; Subroutine finds the power
SUBR: MOV CX ,SI
MOV AX , 0001H
AGAIN: MUL SI
DEC CX
JNZ AGAIN
RET
49
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
Input / Output (IN / OUT) Instructions
 The IN instruction copies data from a port to the AL or AX register. If an 8bit is read, the data will go to AL. If a 16-bit is read, the data will go to AX.
The IN instruction does not change any flag. The general form is :
IN AL / AX , (input port)
 The OUT instruction copies a byte from AL or a word from AX to the
specified port. The OUT instruction does not affect any flag. The general
form is :
OUT (output port) , AL / AX
The IN / OUT instructions has two possible formats:
 Fixed (Direct) Port : the 8-bit address of a port is specified directly in the
instruction. With this form 256 possible ports can be addressed.
IN AL , 98H
; Input a byte from port 0C8H to AL
OUT 34H , AX
; Output a word from AX to 34H port
 Variable (Indirect) Port : the port address is loaded into the DX register
before the IN / OUT instructions is executed. Since DX is a 16-bit
register, the port address can be any number between 0000H and FFFFH.
Therefore, up to 65,536 ports are addressable in this mode.
MOV DX , 8F78H
; Initialize DX with a port address
IN AX , DX
; Input a word from the port 0FF78H to AX
OUT DX , AL
; Output a byte from AL to the port 0FF78H
Note : The variable-port in IN / OUT instructions has advantage that the port
address can be computed or dynamically determined in the program. Suppose,
for example, that an 8086-based computer needs to input data from 10 terminals
(ports), each having its own port address. Instead of having a separate procedure
to input data from each port, you can write one generalized input procedure and
simply pass the address of the desired port to the procedure in DX.
58
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
Ex: Write a piece of code to send 55H to output port device which its
address is F300H.
MOV AL , 55H
MOV DX , F300H
OUT DX , AL
Ex: Write a piece of code to input the contents of the byte-wide input port
at A000H of the I/O address space into BL.
MOV DX, 0A000H
IN AL, DX
MOV BL, AL
Ex: Write a piece of code to read data from two byte-wide input ports at
addresses AAH and A9H and output the data as a word to the word-wide
output port at address B000H.
IN AL, 0AAH
MOV AH, AL
IN AL, 0A9H
MOV DX, 0B000H
OUT DX, AX
Ex: An array of size 1 Kbyte is stored at addresses starting at 72000H, write
an ALP to send even parity bytes to port address 378H and odd parity bytes
to port 478H.
MOV AX , 7000H
MOV DS , AX
MOV CX , 400H
XOR BX , BX
AGAIN: MOV AL , [BX + 2000H]
OR AL , 00H
JP EVEN
MOV DX , 478H
OUT DX , AL
JMP NEXT
EVEN:
MOV DX , 378H
OUT DX , AL
NEXT:
INC BX
DEC CX
JNZ AGAIN
HLT
51
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
Miscellaneous Instructions
 HLT (HALT PROCESSING)
The HLT instruction causes the 8086 to stop fetching and executing
instructions. The 8086 will enter a halt state. The different ways to get the
processor out of the halt state are with an interrupt signal on the INTR
pin, an interrupt signal on the NMI pin, or a reset signal on the RESET
input.
 NOP (PERFORM NO OPERATION)
This instruction simply takes three clock cycles (3T) and increments the
instruction pointer to point to the next instruction. The NOP instruction
can be used to increase the delay of a delay loop. A NOP can also be used
to hold a place in a program for an instruction that will be added later.
NOP does not affect any flag.
 ESC (ESCAPE)
This instruction is used to pass instructions to a coprocessor, such as the
8087 Math coprocessor, which shares the address and data bus with 8086.
Instructions for the coprocessor are represented by a 6-bit code embedded
in the ESC instruction. As the 8086 fetches instruction bytes, the
coprocessor also fetches these bytes from the data bus and puts them in its
queue. However, the coprocessor treats all the normal 8086 instructions as
NOPs. When 8086 fetches an ESC instruction, the coprocessor decodes
the instruction and carries out the action specified by the 6-bit code
specified in the instruction. In most cases, the 8086 treats the ESC
instruction as a NOP.
 INT – INT TYPE
The term TYPE in the instruction format refers to a number between 0
and 255, which identify the interrupt. When an 8086 executes an INT
instruction, it will
1. Decrement the stack pointer by 2 and push the flags on to the stack.
2. Decrement the stack pointer by 2 and push the content of CS onto the
stack.
3. Decrement the stack pointer by 2 and push the offset of the next
instruction after the INT number instruction on the stack.
52
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
4. Get a new value for IP from an absolute memory address of 4 times the
type specified in the instruction. For an INT 8 instruction, for example,
the new IP will be read from address 00020H.
5. Get a new for value for CS from an absolute memory address of 4 times
the type specified in the instruction plus 2, for an INT 8 instruction, for
example, the new value of CS will be read from address 00022H.
6. Reset both IF and TF. Other flags are not affected.
 WAIT – WAIT FOR SIGNAL OR INTERRUPT SIGNAL
When this instruction is executed, the 8086 enters an idle state and is
doing no processing. The 8086 will stay in this idle state until the 8086
TEST input pin is made low(TEST=0) or until an interrupt signal is
received on the INTR or the NMI interrupt input pins. WAIT does not
affect any flag. The WAIT instruction is used to synchronize the 8086
with external hardware such as the 8087 Math coprocessor.
53
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
Converting Assembly Language Instructions to Machine Code
The general instruction format for machine code is shown below :
BYTE 1 Specification
 OPCODE field (6-bits) : Specifies the operation to be performed such as
MOV , ADD , SUB …..etc.
 Register direction bit (D-bit):
D = 1 : the register operand specified by REG in byte 2 is a destination
operand.
D = 0 : the register operand specified by REG in byte 2 is a source
operand
 Data size bit (W-bit): Specifies whether the operation will be performed
on 8-bit or 16-bit data.
W = 1 : 16-bit data size
W = 0 : 8-bit data size
BYTE 2 Specification: byte 2 has three fields:
 Mode (MOD) field (2-bits) : Indicates whether the operand is in register
or memory.
MOD
00
01
10
11
Explanation
Memory Mode no displacement
Memory Mode 8-bit displacement
Memory Mode 16-bit displacement
Register Mode (no displacement)
 Register (REG) field (3-bit) : Identifies the register for the first operand
 Register/Memory (R/M) field (3-bit): Together with MOD field to
specify the second operand.
54
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
REG
000
001
010
011
100
101
110
111
W=0
AL
CL
DL
BL
AH
CH
DH
BH
W=1
AX
CX
DX
BX
SP
BP
SI
DI
MOD = 11
Effective Address Calculation
Reg. W=0 W=1 R/M
MOD=00
MOD=01
MOD=10
000 AL AX 000 (BX)+(SI)
(BX)+(SI)+D8 (BX)+(SI)+D16
001 CL
CX 001 (BX)+(DI)
(BX)+(DI)+D8 (BX)+(DI)+D16
010 DL DX 010 (BP)+(SI)
(BP)+(SI)+D8 (BP)+(SI)+D16
011 BL
BX 011 (BP)+(DI)
(BP)+(DI)+D8 (BP)+(DI)+D16
100 AH
SP
100 (SI)
(SI)+D8
(SI)+D16
101 CH
BP
101 (DI)
(DI)+D8
(DI)+D16
110 DH
SI
110 Direct Address (BP)+D8
(BP)+D16
111 BH
DI
111 (BX)
(BX)+D8
(BX)+D16
Ex : Encode the following instruction in machine code. Assume that the
OPCODE for MOV instruction is 1000102.
MOV BL , AL
Ans.
OPCODE = 100010 (for MOV), D = 0 (source), W = 0 (8-bit) this leads to :
BYTE 1 = 100010002 = 8816
In byte 2 the source operand, specified by REG is AL , then:
REG = 000, MOD = 11, R/M = 011
Therefore:
BYTE 2 = 110000112 = C316
The machine code for the instruction is:
MOV BL , AL = 88C3H
Ex : Encode the following instruction in machine code. Assume that the
OPCODE for ADD instruction is 0000002.
ADD AX , [SI]
Ans.
OPCODE = 000000 (for ADD), D = 1 (destination), W = 1 (16-bit) this leads to
BYTE 1 = 000000112 = 0316
55
8086/8088MP
INSTRUCTOR: ABDULMUTTALIB A. H. ALDOURI
In byte 2 the destination operand, specified by REG is AX , then:
REG = 000, MOD = 00, R/M = 100
Therefore:
BYTE 2 = 000001002 = 0416
The machine code for the instruction is:
ADD AX, [SI] = 030416
Ex: Encode the following instruction in machine code. Assume that the
OPCODE for XOR instruction is 0011002.
XOR CL , [1234H]
Ans.
OPCODE = 001100 (for XOR), D = 1 (destination), W = 0 (8-bit) this leads to:
BYTE 1 = 001100102 = 3216
In byte 2 the destination operand, specified by REG is CL , then:
REG = 001 , MOD = 00, R/M = 110
Therefore:
BYTE 2 = 000011102 = 0E16
BYTE 3 = 3416 and BYTE 4 = 1216
The machine code for the instruction is:
XOR CL , [1234H] = 320E341216
Ex: Encode the following instruction in machine code. Assume that the
OPCODE for ADD instruction is 0000002.
ADD [BX+DI+1234H] , AX
Ans.
OPCODE = 000000 (for ADD) , D = 0 (source), W = 1 (16-bit) This leads to :
BYTE 1 = 000000012 = 0116
In byte 2 the destination operand, specified by REG is AX, then :
REG = 000 , MOD = 10 , R/M = 001
Therefore :
BYTE 2 = 100000012 = 8116
BYTE 3 = 3416 and BYTE 4 = 1216
The machine code for the instruction is:
ADD [BX+DI+1234H] , AX = 0181341216
Ex: Encode the following instruction in machine code
MOV [BP+DI+1234H] , 0ABCDH
Ans.
This example does not follow the general format. The OPCODE of MOV
(immediate to memory) is 1100011W, and W = 1 for word-size data, then:
56