Memory segmentation-microprocessor, Assembly Language

Assignment Help:

Memory Segmentation :

The  memory in an 8086/8088  based system is organized as segmented memory. In this scheme, the whole physically available memory can be divided into a number of logical segments. Each segment is64K bytes in size and is addressed by 1 of the segment registers. The 16-bit contents of the segment register in fact point to the beginning location of a specific segment. To address a particular memory location within a segment, we need an offset address. The offset address is also 16-bit long so that the maximum offset value can be FFFFH, and thus the maximum size of segment is 64K locations.

To emphasize this segmented memory concept, we will take an example of a housing colony containing  say, 100 houses. The easy method  of  numbering  the  houses  will  be  just  to  assign  the numbers from 1 to 100 to each house sequentially. Imagine, now, if 1 wants to find out house number 67, then he will begin from house number 1 and go on till he search the house, numbered 67. Consider another case where the 100 houses are arranged in the 10 x 10 (rows x columns) pattern. In this case, to search house number 67, 1 will directly go to the 6th row and then to the 7th column. In the second scheme, the efforts required for searching the similar house will be too less.  This second scheme in our example is analogous  to the segmented memory scheme, where  the  addresses are specified in  terms of segment addresses analogous tooffset androws addresses analogous to columns.

The CPU 8086 is able to address 1Mbytes of physical memory. The complete1Mbytes memory may be divided into 16 segments, particular of 64Kbytes size. The addresses of the segments can be assigned as 0000H to F000H respectively. The offset address values are from 0000H to FFFFH so that the physical addresses range from 00000H to FFFFFH. In the above case, the segments are called non-overlapping segments. The non-overlapping segments are revealed in given figure (a).However, in some cases, the segments can be overlapping. Imagine a segment begins at a specific address and its maximum size may be 64Kbytes. But, if another segment begins before these 64Kbytes locations of the first segment, the 2 segments are said to be overlapping segments. The region of memory from the start of the second segment to the possible end of the first segment is known as overlapped segment area. Figure tells the phenomenon more clearly. The locations in the overlapped area can be addressed by the similar physical address generated from 2 different sets of segment and offset addresses. The major advantages of the segmented memory scheme are as follows:

1) Allows the memory capacity to be 1Mbytes although the actual addresses to be handled are of 16-bit size.

2) Let the placing of code, data and stack portions of the same program in different parts (segments) of memory, for data and code protection both.

3) Permits a program and/or its data to be put into different areas of memory eachtime program is executed, for instance provision for relocation may be done.

Inoverlapped Area Locations Physical Address = IF+ Cs = IF + CS + denoted the process of physical address formation.

 

912_memory segmentation.jpg

        Fig: Non-overlapping Segments                     Fig: overlapping segment


Related Discussions:- Memory segmentation-microprocessor

Write an 8086 assembly program, Project Description: Write an 80x86 asse...

Project Description: Write an 80x86 assembly program that performs the following functions: Reads a set of integers from a file into an array. The data file name is to be

Program that will compute and output tuition cost, Write an assembly progra...

Write an assembly program that will compute and output tuition cost: a. accept keyboard input of the number of credit hours taken b. accept keyboard input of the type of classes 1

Flowchart for the sequence of 8251-microprocessor, Flowchart for the sequen...

Flowchart for the sequence of 8251 Whether the control, mode or sync character register is selected depends on the accessing sequence.  A flowchart of the sequencing is given i

Shl/sal-logical instruction-microprocessor, SHL/SAL : Shift logical/Arithm...

SHL/SAL : Shift logical/Arithmetic Left: These instructions shift the operand byte or word bit by bit to the left and insert 0 in the newly introduced least significant bits. In c

Input output memory mapped-microprocessor, Memory Mapped I/O Memory I/O...

Memory Mapped I/O Memory I/O devices are mapped into the system memory map with ROM and RAM. To access a hardware  device, simply write or  read  to  those  'special'  addresse

Program to find the largest number in an array, Write a MC68H12 assembly la...

Write a MC68H12 assembly language program to find the largest number in an array of ten 8-bit numbers. The array is stored in memory locations starting at address $1100. Use branch

Estimation of a definite integral, Can any one assist me with this program....

Can any one assist me with this program. I am not efficient with assembly language and I need assistance badly. I am not asking anyone to do my work I just need help step by step

Write an assembly program to find out largest number, Program :  Write an...

Program :  Write an assembly program to find out the largest number from a given unordered array of 8-bit numbers that stored in the locations starting from a known address. S

Program to perform a conversion on characters in a text file, Write an asse...

Write an assembly language program to perform a rot13 1 conversion on characters in a text file.  The program should read charatcers from the input file, perform the rot13 convers

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd