1. Computer Organization and ASSEMBLY LANGUAGE
Lecture 11 & 12
x86 Memory Organisation
Memory Segmentation
Course Instructor: Engr. Aisha Danish

2. The 8086 has a 20-bit Address Bus
With 20 bits, 1,048,576 different combinations are available. Each memory location is assigned a different combination. Each memory location is 1-byte wide. Therefore, the memory space of the 8086 consists of 1,048,576 bytes or 524, bit words.
JNM
1/2002

3. Problem
The 8086 has a 20-bit address bus. Therefore, it can access 1,048,576 bytes of memory. How many bits and how many HEX digits are required to access 1M memory?
2N = 1M (where N is in bits)
N = 20 bits = 20/4 = 5 HEX digits
JNM
1/2002

4. Positional Notation (Hex Digits)
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1/2002

5. Problem Continued
But each register is only 16 bits (4 HEX digits) wide. How are all of the locations accessed?
Segmented Memory – all of memory is divided into 64- kByte segments. (4 HEX digits (16 bits) can access 64k different locations)
216  64k
JNM
1/2002

6. Segmented Memory
Within the 1 MB of memory, the 8086 defines KB memory blocks.
DS: E CS: B300
SS: ES: 5D27
7FFFF
The segment registers point to location 0 of each segment. (The base address)
JNM
1/2002

7. Segmentation
Segments are variable-sized areas of memory used by a program containing either code or data.
Segmentation provides a way to isolate memory segments from each other. This permits multiple programs to run simultaneously without interfering with each other.
A segment selector is a 16-bit value stored in a segment register.
A logical address is a combination of a segment selector and an offset(16-bit for 8086).
JNM
1/2002

8. A program can have more than four or six segments,
Think of segments as windows that can be moved over any area of memory to access data or code
A program can have more than four or six segments,
but only access four or six segments at a time

9. Segmented Memory Memory Organization: Linear vs. Segmented
Linear Addressing: (MC68K, i8085)
The entire memory is regarded as a whole
Specify absolute addresses in instructions
The entire memory space is available all the time
Segmented Addressing: (ix86)
Memory is divided into segments
Specify an address as offset relative to segment base address
Programs use offsets as logical address, independent of where segments are located

10. Figure 2–3  The 8086 (or real mode) memory-addressing scheme, using a segment address plus an offset.
this shows a memory segment beginning at 10000H, ending at location 1FFFFH
64K bytes in length
also shows how an offset address, called a displacement, of F000H selects location 1F000H in the memory

11. 8086 Programming Model: Segment Registers
Segment Group: divide memory into
CS: Code Segment [64KB for each segment]
DS: Data Segment
ES: Extra Segment
SS: Stack Segment
Segment Registers: CS, DS, ES, SS
Save Base addresses to particular segments
SEG(16-bit):OFFSET(16-bit) is used by BIU to calculate (20-bit) physical memory address
[DS:0] = byte (the 1-st byte) in Data Segment 0 = offset to begin-of-DS

12. Memory Banks
Even though the 8086 has a 16-bit data bus, it is byte addressable.
Memory is divided into two 8-bit banks
Even Addresses
Odd Addresses
BE0
BE1
JNM
1/2002

13. 8086 Memory Organization Features of 8086 Memory Organization
Memory Banks (x2):
for concurrently accessing two bytes
Segmented Memory Address Space
for flexible memory management
compatible with 8085 memory management (16-bit addresses)

14. (started at odd-byte address) Aligned words
FIGURE The memory space of the 8086 consists of 1,048,576 bytes or 524, bit words.
Mis-aligned words
(started at odd-byte address)
Aligned words
(started at even-byte address)
John Uffenbeck The 80x86 Family: Design, Programming, and Interfacing, 3e
Copyright ©2002 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

15. 8086 Memory Organization: Memory Banks
2^20 addresses can be arranged as:
8-bit bytes in continuous locations, or
16-bit words (in parallel banks)
Not a single way, but …
Memory organization & interfacing should reflect the access behavior of the processor in the most natural way. Two types of operations:
Byte: I/O, character data (e.g., ASCII)
Words: large 16-bit integers arithmetic operations

16. 8086 Memory Organization: Memory Banks
Even and Odd Memory Banks
16-bit data bus two-byte / two one-byte access
Allows processor to work on bytes or on (16-bit) words
IO operations are normally conducted in bytes
Arithmetic operations may use large 16-bit words
Can handle odd-length instructions
Single byte instructions
Multiple byte (and very long) instructions

17. FIGURE (a) By reading from an even-addressed bank and an odd-addressed bank the 8086 can read two bytes from memory simultaneously. (b) If the 16-bit word begins at an odd address, the 8086 will require two memory read or write cycles.
word 4
word 0
John Uffenbeck The 80x86 Family: Design, Programming, and Interfacing, 3e
Copyright ©2002 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.