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

Overview of intel pro-pentium, Overview of Intel Pro-Pentium : The 2 c...

Overview of Intel Pro-Pentium : The 2 chief players in the PC CPU market are Motorola and Intel.  Intel has enjoyed incredible success with its processors since the early 1980

Segment registers-microprocessor, Segment Registers The 8086 addresses ...

Segment Registers The 8086 addresses a segmented memory unlike 8085. The complete 1 megabyte memory, which 8086 is capable to address is divided into 16 logical segments.Thusea

8086 microprocessors, program to find negative and positive integers from g...

program to find negative and positive integers from given signed numbers with output and explanation of every instructions.

Program, move a byte string ,16 bytes long from the offset 0200H to 0300H i...

move a byte string ,16 bytes long from the offset 0200H to 0300H in the segment 7000H..

Hashing, what is double hashing

what is double hashing

Digital and embedded software, hi!im looking for someone who expert in an a...

hi!im looking for someone who expert in an assembly language and help me write the programmed!Thank you

Csc203 assembly language, I need to estimate the value of a definite integr...

I need to estimate the value of a definite integral using Riemann Sums and For our estimation let f(x) = x2 ,a=0, b=10 and n=5. Where a is the lower bound, b is the upper bound and

Al registre, check the al-register for palindromic number

check the al-register for palindromic number

Logical instruction-microprocessor, Logical Instruction : This type of...

Logical Instruction : This type of instructions is utilized for carrying out the bit by bit shift, basic logical operations or rotate. All of the condition code flags are affe

Hi, i have a question.

i have a question.

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