Addressing modes of 8086

2.  Select the appropriate addressing mode to
accomplish a given task.
 Detail the difference between addressing
memory data using real mode and protected
mode operation.
 Describe the sequence of events that place data
onto the stack or remove data from the stack.
 Explain how a data structure is placed in
memory and used with software.

3.  An addressing mode means the method by which
an operand can be specified in a register or a
memory location
 8086provide a seven Addressing Modes:
(1) Register Addressing
(2) Immediate Addressing
(3) Direct Addressing
(4) Register Indirect Addressing
(5) Based Indexed Addressing
(6) Register Relative Addressing
(7) Relative Based Indexed Addressing

4. 8-bit 8-bit 8-bit
OP CODE OPERAND
byte 1 to 2 byte
 An opcode is a short of “operation code”
 An opcede is a singe instruction can be executed by the CPU.
In machine language it is a binary or hexadecimal value such
as B7 loaded into the instruction register.
 In assembly language mnemonic form an opcode is a
command such as MOV or ADD or JMP.
 Example:
MOV AX, 1000H ; MOV is the opcode.
; AX (register) is an operand.
 Operands are manipulated by the opcode. In this example,
the operands are the register AX and the value 1000H.

5.  Transfers a copy of a byte or word from the source register
or memory location to the destination register or memory
location.
 Use of registers to hold the data to be manipulated
 Memory is not accessed when this addressing mode is
executed
 Example:
MOV BX, DX
MOV ES, AX
ADD AL, BH
; copy the contents of DX into BX
; copy the contents of AX into ES
; add the contents of BH to
contents of AL
destination registers must have the same size

7.  Transfers the source, an immediate byte or word of data,
into the destination register or memory location
 The source operand is a constant
 The operand comes immediately after the opcode
 For this reason, this addressing mode executes quickly
 Immediate addressing mode can be used to load
information into any of the registers except the segment
registers and flag registers.

8.  Example:
MOV AX, 2550H
MOV CX, 625
MOV BL, 40H
; move 2550H into AX
; load the decimal value 625 into
CX
; load 40H into BL
 The data must first be moved to a general-purpose
register and then to the segment register.
 Example:
MOV AX, 2550H MOV DS, AX
MOV DS, 0123H ; illegal instruction!

10. Moves a byte or word between a memory location and
a register.
The data is in some memory location(s) and the
address of the data in memory comes immediately
after the instruction
This address is the offset address
Example:
MOV AX, [2400] ; move contents of DS:2400H
into AX
The physical address is calculated by combining the
contents of offset location 2400 with DS

11.  Example:
Find the physical address of the memory location and its
contents after the execution of the following, assuming that DS
= 1512H.
AL, 3BH
[3518], AL
MOV
MOV
Sol
ution
:
 First 3BH is copied into AL,
 Then in line two, the contents of AL are moved to logical
address DS:3518 which is 1512:3518.
 Shifting DS left and adding it to the offset gives the physical
address of 18638H (15120H + 3518H = 18638H).
 After the execution of the second instruction, the memory
Prof. Fayleoz Fc. aMa.tEil-oSonusy with address 18638H will contain the
value 3BH.

13. Transfers a byte or word between a register and a
memory location addressed by an index or base
register
The address of the memory location where the
operand resides is held by a register
The registers used for this purpose are SI, DI, and BX
They must be combined with DS in order to generate
the 20-bit physical address.

14. Example:
MOV AX, [BX] ; moves into AX the contents of
the memory location pointed
to by DS:BX, 1000:1234
The physical address is calculated as
1000x10+1234=11234H
The same rules apply when using register SI or DI.
Example:
MOV CL, [SI]
MOV [DI], AH
; move contents of DS:SI into CL
; move contents of AH into DS:DI

15.  Example:
 Assume that DS = 1120, SI = 2498, and AX = 17FE Show
the contents of memory locations after the execution of
MOV [SI], AX ; move contents of AX into DS:SI
 Solution:
The contents of AX are moved into memory locations
with logical address DS:SI and DS:SI + 1;
The physical address starts at DS (shifted left) + SI =
13698. According to the little endian convention,
Low address 13698H contains FE, the low byte,
High address 13699H will contain 17, the high byte.

17.  Transfers a byte or word between a register and the memory location
addressed by a base register (BP or BX) plus an index register (DI or
SI).
 Combining based and indexed addressing modes.
 One base register and one index register are used.
 Examples:
MOV [BX+DI], CL ; move contents of CL into DS:BX+DI
 Physical Address = DSx10 + BX+DI
MOV CH, [BX+SI] ; move contents of the DS:BX+SI into CH
 Physical Address = DSx10 + BX+SI
MOV AH, [BP+DI] ; move contents of the SS:BP+SI into AH
 Physical Address = SSx10 + BP+DI
MOV [BP+SI], AL ; move contents of AL into SS:BP+SI
= SSx10 + BP+SI

19.  Moves a byte or word between a register and the memory location
addressed by an index or base register plus a displacement.
 The data in a segment of memory are addressed by adding the
displacement to the contents of a base or an index register (BP, BX, DI,
or SI).
 Examples:
MOV AX, [BX+4] ; move contents of DS:BX+4 into AX
 Physical Address = DSx10 + BX+4
MOV CH, [SI+5] ; move contents of the DS:SI+5 into CH
 Physical Address = DSx10 +SI+5
MOV AH, [DI+1] ; move contents of the DS:DI+1 into AH
 Physical Address = DSx10 + DI+1
MOV [BP+2], AL ; move contents of AL into SS:BP+2
= SSx10 + BP+2

20.  Example:
 Assume that DS = 4500, SS = 2000, BX = 2100, SI = 1486, DI =
 8500, BP= 7814, andAX = 2512. Show the exact physical memory location where
AX is stored in each of the following.All values are in hex.
1 MOV [BX+20], AX
2 MOV [SI+10], AX
3 MOV [DI+4], AX
4 MOV [BP+12], AX
 Solution:
 Physical Address = segment reg. x 10 + (offset reg.) +
displacement
1 DS:BX+20
2 DS:SI+10
3 DS:DI+4
4- SS:BP+12
location 47120 = (12) and 47121 = (25)
location 46496 = (12) and 46497 = (25 )
location 4D504 = (12) and 4D505 = (25)
location 27826 = (12) and 27827 = (25)

22.  The base relative-plus-index addressing mode is similar to the
base-plus-index addressing mode, but adds a displacement
besides using a base register and an index register to form the
memory address.
 This type of addressing mode often addresses a two-dimensional
array of memory data.
 The data in a segment of memory are addressed by adding the
displacement to the contents of a base and an index register (BP,
BX, DI, or SI).
 Examples:
MOV [BX+DI+1], AX ; move contents of AX into
DS:BX+DI+1
 Physical Address = DSx10 + BX+DI+1H
MOV AX, [BX+SI+10]; move contents of the DS:BX+SI+10
into AX
PhysicalAddress = DSx10 + BX+SI+10H

23. MOV AH, [BP+DI+3] ; move contents of the SS:BP+SI+3
into AH
 Physical Address = SSx10 + BP+DI+3H
MOV [BP+SI+6], AL ; move contents of AL into
SS:BP+SI+6
 Physical Address = SSx10 + BP+SI+6
MOV AX, FILE[BX+DI] ; move contents of the
DS:FILE+BX+DI into AX
 Physical Address = DSx10 + BX+DI+FILE
MOV LIST[BP+SI+4], DH ; move contents of DH into
SS:LIST+BP+SI+4
Address = SSx10 +LIST+ BP+SI+4

25. Segment Register CS DS ES SS
Offset Register IP SI, DI, BX SI, DI, BX SP, BP
The following Table provides a summary of the
offset registers that can be used with the four
segment registers of the 8086