Addressing modes of 8086

 The way for which an operand is specified for an instruction in the accumulator, in a general purpose register or in memory location, is called addressing mode.

The 8086 microprocessors have 8 addressing modes. Two addressing modes have been provided for instructions which operate on register or immediate data.

These two addressing modes are:

Register Addressing: In register addressing, the operand is placed in one of the 16-bit or 8-bit general purpose registers.

Example

  • MOV AX, CX
  • ADD AL, BL
  • ADD CX, DX

Immediate Addressing: In immediate addressing, the operand is specified in the instruction itself.

Example

  • MOV AL, 35H
  • MOV BX, 0301H
  • MOV [0401], 3598H
  • ADD AX, 4836H

The remaining 6 addressing modes specify the location of an operand which is placed in a memory.

These 6 addressing modes are:

Direct Addressing: In direct addressing mode, the operand?s offset is given in the instruction as an 8-bit or 16-bit displacement element.

Example

  • ADD AL, [0301]

The instruction adds the content of the offset address 0301 to AL. the operand is placed at the given offset (0301) within the data segment DS.

Register Indirect Addressing: The operand's offset is placed in any one of the registers BX, BP, SI or DI as specified in the instruction.

Example

  • MOV AX, [BX]

It moves the contents of memory locations addressed by the register BX to the register AX.

Based Addressing: The operand's offset is the sum of an 8-bit or 16-bit displacement and the contents of the base register BX or BP. BX is used as base register for data segment, and the BP is used as a base register for stack segment.

Effective address (Offset) = [BX + 8-bit or 16-bit displacement].

Example

  • MOV AL, [BX+05]; an example of 8-bit displacement.
  • MOV AL, [BX + 1346H]; example of 16-bit displacement.

Indexed Addressing: The offset of an operand is the sum of the content of an index register SI or DI and an 8-bit or 16-bit displacement.

Offset (Effective Address) = [SI or DI + 8-bit or 16-bit displacement]

Example

  • MOV AX, [SI + 05]; 8-bit displacement.
  • MOV AX, [SI + 1528H]; 16-bit displacement.

Based Indexed Addressing: The offset of operand is the sum of the content of a base register BX or BP and an index register SI or DI.

Effective Address (Offset) = [BX or BP] + [SI or DI]

Here, BX is used for a base register for data segment, and BP is used as a base register for stack segment.

Example

  • ADD AX, [BX + SI]
  • MOV CX, [BX + SI]

Based Indexed with Displacement: In this mode of addressing, the operand's offset is given by:

Effective Address (Offset) = [BX or BP] + [SI or DI] + 8-bit or 16-bit displacement

Example

  • MOV AX, [BX + SI + 05]; 8-bit displacement
  • MOV AX, [BX + SI + 1235H]; 16-bit displacement

Comments

Post a Comment

Popular posts from this blog

8086 pins configuration

Image
8086 pins configuration The description of the pins of 8086 is as follows: AD0-AD15 (Address Data Bus):  Bidirectional address/data lines. These are low order address bus. They are multiplexed with data. When these lines are used to transmit memory address, the symbol A is used instead of AD, for example, A0- A15. A16 - A19 (Output):  High order address lines. These are multiplexed with status signals. A16/S3, A17/S4:  A16 and A17 are multiplexed with segment identifier signals S3 and S4. A18/S5:  A18 is multiplexed with interrupt status S5. A19/S6:  A19 is multiplexed with status signal S6. BHE /S7 (Output):  Bus High Enable/Status. During T1, it is low. It enables the data onto the most significant half of data bus, D8-D15. 8-bit device connected to upper half of the data bus use BHE signal. It is multiplexed with status signal S7. S7 signal is available during T3 and T4. RD  (Read):  For read operation. It is an output signal. It is active when...
Read more

Architecture of 8086

Image
A   Microprocessor   is an Integrated Circuit with all the functions of a CPU. However, it cannot be used stand-alone since unlike a microcontroller it   has no memory or peripherals .  8086 does not have RAM or ROM inside it. However, it has  internal registers  for storing intermediate and final results and interfaces with memory located outside it through the System Bus.  In the case of 8086, it is a 16-bit  Integer processor  in a 40-pin, Dual Inline Packaged IC.  The size of the internal registers(present within the chip) indicates how much information the processor can operate on at a time ( in this case 16-bit registers ) and how it moves data around internally within the chip, sometimes also referred to as the internal data bus.  8086 provides the programmer with 14 internal registers, each of 16 bits or 2 bytes wide.  The main advantage of the 8086 microprocessor is that it supports Pipelining. 8086 Architecture M...
Read more

Assignment Questions (Subject:- Microprocessor and Microcontroller)

  Microprocessor and Microcontroller Assignment Questions Instructions: ·          Solve any 8 questions in each section. ·          Make appropriate assumptions wherever necessary. Unit 1 1.       What is microprocessor? 2.       What is the difference between the microprocessor and microcontroller? 3.       Explain the register organization of 8086 4.       How many data lines and address lines are available in 8086? 5.       Explain briefly about the internal hardware architecture of 8086 microprocessor with a neat diagram 6.       What is the maximum memory size that can be addressed by 8086? 7.       Calculate the physical address for fetching the next instruction to be executed, in 8086? 8.  ...
Read more