The Solution manual of COAL
Chapter NO 3. exercise
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1. A solution manual to
Assembly language
Programming
and the organization of
the IBM PC
Chapter 3
Organization of the IBM personal Computer
BY:- Ytha Yu & Charles Marut
prepared by :
Warda Aziz
Wardaaziz555@gmail.com
2. Question 1:
What are the main differences between 80286 and 8086 processors?
Answers:
Question 2:
What are differences between registers and memory locations?
Ans:
Question 3:
List one special function for each of the data registers AX, BX, CX, DX
Ans:
AX: input and output operations
BX: works as an address register
CX: used in looping
DX: used in multiplication and division, also in I/0 operation.
Question 4:
Determine the physical address of a memory location given by 0A51:CD90h
Ans:
Address= A51*10+CD90
=172A0h
Question 5:
A memory location has a physical address 4A37Bh. Compute
The offset address if seg number is 40FFh
The segment number if the offset address is 123Bh
Ans:
8086 80286
Invention date 1978 1982
RAM 1 Mb 16Mb
Registers 14 reg 8,15
Functional units 2 4
CPU speed 10 MHz 12.5 MHz
Modes of
operation
Operates in real address mode
only
Operates in real address mode and protected
virtual address mode
Register Memory location
Holds small amount of data Holds a large amount of data
Fast access Slower access of data
Resides inside CPU Resides in RAM
Called by their names Call by their address
3. Question 6:
What is paragraph boundary?
Ans:
A paragraph comprises 16 bytes, so any address divisible by 16 is said to be a paragraph
boundary.
Question 7:
What determines how compatible an IBM PC clone is with an authentic IBM PC?
Ans:
IBM PC compatible computers are those similar to the original/authentic IBM PC; able
to run the same software and support the same expansion cards as those. Such computers
used to be reffered as PC clones or IBM clones.
Compatibility of IBM clones depends on the BIOS routines matching with the authentic
IBM PC.
Question 8:
What is the maximum amount of memory that DOS allocates for loading RUN files?
Assume that DOS occupies up to the byte 0FFFF
Ans:
Usually it depends on the size of RAM used. In 1MB of RAM only 640 kb are
accessible for direct use I.e., for loading and running application programs.
Question 9:
Give the DOS commands to do the following. Suppose that A is the logged drive.
A. Copy FILE1 in the current directory to FILE1A on the disk in drive b
B. Copy all files with an .ASM extension to the disk in drive B
C. Erase all files with the .BAK extension
D. List all file names in the current directory that begin with A
E. Set the date to September 21, 1991
F. Print the file FILE5.ASM on the printer.
Solution:
Copy FILE1 in the current directory to FILE1A on the disk in drive b
COPY A:FILE1 B:FILE1A
Copy all files with an .ASM extension to the disk in drive B
COPY A: *.ASM B:
Erase all files with the .BAK extension
ERASE *.BAK
List all file names in the current directory that begin with A
DIR A:A*.*
Set the date to September 21, 1991
DATE 09-21-9
Print the file FILE5.ASM on the printer.
4. PRINT A:FILE5.ASM
Question 10:
Suppose that (a) the root directory has sub-directories A,B,C (b) A has
sub-directories A1 and A2 (c) A1 has a sub directory A1A.
Give DOS commands to
A) create the preceding directory tree
B) Make A1A the current directory
C) Have DOS display the current directory
D) Remove the preceding directory tree
Solution:
Create the preceding directory tree
C>CD
C>MDROOT
C>MDROOTA
C>MDROOTB
C>MDROOTC
C>MDROOTAA1
C>MDROOTAA2
C>MDROOTAA1A1A
Make A1A the current directory
CDROOTAA1A1A
Have DOS display the current directory
CD
Remove the preceding directory tree
C>ERASEROOTAA1A1A
C>RMROOTAA1
C>RMROOTAA2
C>RMROOTA
C>RMROOTB
C>RMROOTC
C>RMROOT