phobia

LINUX COMMANDS

2008-07-03T01:37:00.000-07:00

Linux command line

du - Shows the disk space a file or directory is using in your disk

This is a very useful command you may use to find which file or directory is filling you disk.

So if you need to check the file and directory size in your home directory you can enter

to switch to your home directory

du -S * | sort -n

How to find files, using command line (locate)

Continuing reviewing some of my old posts, I found this one Examples of find command, now I want to show you a faster and easier way to find files.

We are going to use the commands updatedb and locate

How to find which service is listening on a given port

It is really important to know which ports are open in your PC, this is not only useful for Linux, but also for other operating systems, Linux has a lot of tools to check which ports are open, the most common is nmap which is a command line tool, but also exist a Graphical frontEnd for it if you prefer that way.

So to scan you own PC and find open ports you can enter:

sudo nmap -T Aggressive -A -v 127.0.0.1 -p 1-65000

wget - Resume downloads, limit the speed and much

wget is a command line tool used to download files, or complete webpages, it is a great utility with lots of options, as you can see if you read the wget man page

Some months ago, I have written about how to download files with wget, now I want to add some other tips to those already explained that day.

The /etc/default/rcS file

There is some behavior of your Linux Operating System which is easy to change, but not too common to know how.

The things you can change are:

Frequency to erase /tmp/ directory
Use UTC or local time
How Verbose are the boot messages of your Linux
If a disk error should be always repaired while booting automatically

There are more than those, but I will touch only those, for the rest, you can enter:

man /etc/default/rcS

vnstat - Command Line Tool to check how much bandwidth you use

It is always useful to know how much bandwidth you are using, also to check if your provider is honoring the contract you signed with them.

At my office I have installed Cacti and as I have a layer 3 SNMP capable switch, I can graph the bandwidth that all my office's PCs uses, and also the main pipe to the Internet (My ADSL connection), but at home, I do not have such an expensive switch (and I do not need it), and my ADSL modem is not SNMP capable, so Cacti is not a solution, but with Linux Operating System there is always a solution, and the one for this kind of problem is vnstat.

command line scientific calculator

I was looking for a good and light calculator for Linux, and found some command line ones, I liked this one wcalc it is really light, and accept almost all mathematical expressions, and I say "almost" because I do not want to say that are all and make a mistake.

The use of it is really simple, just enter wcalc followed with the math expression in the command line, press ENTER and you will the result.

To install it on Debian run:

sudo aptitude install wcalc

Trash can or Recycle bin in Linux Desktop (managed from console)

Linux Desktops, at least Gnome and KDE has a trash can, where your deleted files go, (only when deleted from a Desktop utility).

Now if you want to manage it from the console, you can, first we need to know that the trash can is only another folder in the File system structure and it is located at:

$HOME/.Trash

so you can send files to Trash just moving them to there, as an example, lets suppose you have a file in your home called balance.ods and want to move it to the trash can.

mv $HOME/balance.ods $HOME/.Trash/

How to assign a password to root user in Ubuntu

Ubuntu by default has no root password, and you can also configure Debian in that way if you choose the advanced installation option, and decide not to use the root password.

But what to do if later you need the root user? well you just need to assign a password to the root user and then you will be able to switch to root user.

sudo passwd

Type a good strong password twice, and you are done.

su -

Type the root password and voila! you are logged as root, you know you have to be carefull on what you do being root right?

Terminator - multiple terminals in one window

I really like to work with the console, and to connect to my servers using ssh, to administrate them, so sometimes I found myself with lots of terminal windows open, I have explored some solutions in the past, that you can see on this post about Terminal emulators, now I have found a new one which I really like, it is called terminator.

For Debian users it is available only on Sid, and as I use Lenny I had to install it from the sources.

Reading compressed Files

I did not know this, but if you need to show a compressed text file on the screen, you do not actually need to uncompress it.

You can use zcat to send the file to the standard output, uncompressed, but the original file remains untouched.

The syntax of the command is:

zcat file.gz

or you can also use,

gunzip -c file.gz

mplayer - command line movie and music player - music over ssh

Mplayer is one of most known movie players in Linux, and also in Mac OS, and Windows.

To install it in Debian / Ubuntu run:

sudo aptitude install mplayer

Mplayer has lots of options, and we will explore some of them here, the files format it can play according to its official site are:

MPEG/VOB, AVI, Ogg/OGM, VIVO, ASF/WMA/WMV, QT/MOV/MP4, RealMedia, Matroska, NUT, NuppelVideo, FLI, YUV4MPEG, FILM, RoQ, PVA files, supported by many native, XAnim, and Win32 DLL codecs. You can watch VideoCD, SVCD, DVD, 3ivx, DivX 3/4/5 and even WMV movies..

for - Control flow statement in shell scripts

One real important part of programming is to control the flow of the program, you need to be able to repeat some part of the code, jump to another part, etc, depending on the estate of some variables, there are a lot of commands to achieve that, and one of them is for

It can be used directly at the command line like this example using basename

for i in *.jpg; do echo $i $(basename $i .jpg); done

basename - removing directories and suffix from file names

If you need to remove the directory or any suffix from a file name you may use basename, this is really useful when you are creating shell scripts.

In the man page, there are this examples

basename /usr/bin/sort

Output "sort".

basename include/stdio.h .h

Output "stdio".

you can enter

info basename

Friendly interactive shell (Fish)

I think Friendly Interactive Shell (a.k.a. fish) is a real good tool for newbies on the console at Linux, it has a lot of features, which you can by typing help.

Well first lets see how to install the application.

Fedora / Centos

sudo yum install fish

Debian / Ubuntu

sudo aptitude install fish

Simple shell script

Maybe most of the readers of this blog do already know about shell scripts, and maybe they know even more than me (which is not difficult), but for those who does not know about it, I will write a little about shell scripts.

A shell script is a sequence of commands -For those coming from DOS is like a .bat file- this command will execute in the sequence they are entered, unless loops, if, do, for, or any other commands like those are used.

Writing the script

Lets go with the classic example, of "Hello World"

Changing screen resolution on the fly

If you want to change your screen resolution, while still running Linux but you are not using Gnome or KDE, which gives good options to change screen resolution, you will have to use the xrandr command

iftop - Check and analyze your traffic

Introduction

iftop is a real good utility both for your own PC/Laptop and for your server and specially good for the latter and even better if this server is acting as the main router in your premises.

sysstat - great utilities package to monitor performance

sysstat is a package of monitoring tools, these are the tools included in the package.

iostat

Reports CPU statistics and input/output statistics for devices, partitions and network filesystems.

mpstat

Reports individual or combined processor related statistics.

pidstat

Reports statistics for Linux tasks (processes) : I/O, CPU, memory, etc.

sar

Collects, reports and saves system activity information (CPU, memory, disks, interrupts, network interfaces, TTY, kernel tables,etc.)

sadc

source

A New Concept Online Advertising

2008-02-10T23:12:00.000-08:00

A New Concept in Online Advertising

If you have not heard of 'word sites' or 'word link advertising' or 'word cloud' before, then get ready to witness a new phenomenon in online advertising.

If you have a business website that is for trading forex, commodities, stocks or options, etc., then you might be interested in a new advertising method that will send massive new traffic to your business sites and get you tons of valuable one-way back links to improve your page rank, all hands-free!

The concept is very simple. You purchase a word relating to your business, product, service or organization from a 'word site' or 'word cloud' then hyperlink that word to your own website and ... presto! You have just instantly "branded" your website with a single, descriptive word.

Now you can create an instant, recognizable brand for your product, service, business, or organization and increase your website exposure with the hottest thing on the net right now...with word link advertising!

Cynthia Macy, who runs 1000 Trading Words, is VERY proud to introduce to you a new concept in online advertising that is proving to be very effective in creating exposure, traffic and 'branding' for your trading website or online presence.

Now you can have YOUR word hyperlink to YOUR website and get the benefit of your word on a page that links to other hi-page rank sites...this will get you a higher page rank too!

If you haven't picked up a keyword linking to your site on a word cloud site yet, you are missing out on the HOTTEST thing to hit the web since pixel ads!

Don't know what a word cloud is? It's a page of words where each word hyperlinks to a different website. The advantage is three-fold:

1. someone will click on your word name out of interest = a new customer!

2. some of the word names will be a higher page rank than yours = your word name,

by association on the word cloud page, will take on a higher page rank too!

3. as your website increases in page rank, it will be seen more often in the

search engines and the result will be increased traffic, increased exposure and sales!

The potential for massive new traffic as the page rank climbs is huge! This means ever-increasing traffic for your site, for at least the next two years!

With the official launch still several days away, advertisers can take advantage of the current generous "pre-launch" discounts being offered to new word link advertisers. Currently, prices are 1/2 price and a soon-to-be-offered free drawing will give potential advertisers the chance to win the word of their choice for free! Upcoming details about the free drawing will soon be posted on the website.

Cynthia Macy is co-author of 3 popular forex training ebooks. Please visit her Word Cloud page at: http://www.1000tradingwords.com

and her other trading websites:

http://www.daytrade-forex.com

http://www.successtrading2000.com

http://www.professionalforextrading.info

http://www.successtrading2000.com/forex

company: Day Trade Forex, LLC

contact person: Cynthia Macy or Erol Bortucene
e-mail: daytradeforex@daytradeforex.com
phone: 928-639-1330
site: http://www.daytradeforex.com/products2.htm

Pay Per Click

2008-02-10T22:42:00.000-08:00

Pay Per Click

Pay Per Click (Cost Per Click) advertising is what takes your website to the next level. More brand recognition, more traffic, more revenue.
At DailyServer we’ll handle your PPC advertising in four-step manner.
1) We figure out the best keywords and PPC programs to use for your specific site. We have relationships with all of the major providers: Google AdWords, Yahoo! Search Marketing, Microsoft adCenter, Ask, LookSmart, Miva, Kanoodle, Yandex, Baidu, and more.
2) Then we work with you to set a budget that will get you the results you’re looking for.
3) Next, DailyServer implements your PPC program with whichever provider(s) you’ve chosen.
4) Lastly, we finely tune your PPC program on an ongoing basis, making sure you’re getting the site traffic you want and that your money is being wisely spent with the best ROI.

PPC Management Fees

The beauty of Points Per Click advertising is that you pay as much or as little as you want. We help you set your budget, by way of the keywords you use and the PPC programs you subscribe to, and then DailyServer takes a small percentage every time your links are clicked and your PPC account is debited.
Our fee percentage is extremely competitive and we only make money when your site gets visitors.
Contact us for a quote and to find out exactly what our PPC management program can do for you.

source

MICROPROCESSOR 8085

2007-12-09T20:23:00.000-08:00

DEPAPRTMENT OF ELECTRONICS AND COMMUNICATION

LAB MANUAL

FOR

ADVANCED MICROPROCESSOR LAB (ECL57)

V SEMESTER

PREPARED BY

Lab Incharge K.P. Shashikala

K.N. Puspalatha

G.Ephratha

S.Ramesh

T.Sanjivi Nathan

DAYANANDA SAGAR COLLEGE OF ENGINEERING

SHAVIGE MALLESHWARA HILLS, K.S.LAYOUT

BANGALORE - 560078

ADVANCED MICROPROCESSOR LAB MANUAL

V SEMESTER VTU

CONTENTS

ASSEMBLING AND EXECUTING THE PROGRAM

PROGRAMS

1. PROGRAMS INVOLVING DATA TRANSFER INSTRUCTIONS

i) Byte and word data transfer in different addressing modes 5

ii) Block move Without overlapping 6

iii) Block move With overlapping 7

iv) Block interchange. 9

2. PROGRAM INVOLVING ARITHMETIC AND LOGIC OPERATIONS LIKE

ADDITION AND SUBTRACTION OF MULTIPRECISION NUMBERS

i ) 16 Bit Addition 11

ii) 32 Bit addition 12

iii) 32 Bit addition using DD directive 13

iv) 16 Bit Subtraction 14

v) 32 Bit Subtraction 15

Multiplication and Division of signed and unsigned Hexadecimal numbers

vi)16 Bit multiplication for unsigned numbers 16

vii)16 Bit multiplication for signed numbers 17

viii)8 Bit Division for Unsigned numbers 18

ix)8 Bit Division for Signed numbers 19

x)16 Bit Division for Unsigned numbers 20

xi)16 Bit Division for Signed numbers 21

3.Code Conversion

i) ASCII adjustment instructions 22

ii) Binary to BCD code conversion 23

iii) BCD to Binary code conversion 24

4. Arithmetic programs to find square, cube, LCM, GCD and factorial

i)Program to find square and cube of a number 25

ii)Program to find LCM of a given number 26

iii)Program to find GCD of two numbers 27

iv)Program to find factorial of a given number 28

5.Program involving bit manipulation instruction

i)If given data is positive or negative 29

ii)If given data is odd or even 30

iii) Logical ones and zeros in a given data 31

iv)2 out of 5 code 32

v) Bit wise palindrome 32

vi)Nibble wise palindrome 34

6. Programs involving branch/loop instructions

Programs on arrays

i)Addition of n numbers 35

ii)Program to subtract n numbers 36

Programs to find largest and smallest number

iii)Program to find largest number among the series 37

iv)Program to find the largest number using dos display interrupts 38

v)Program to find the smallest number among the series 40

vi)Program to sort the numbers in ascending/descending order. 42

Programs On String Manipulation Like String Transfer, String Reversing, Searching

For A Character In A String And Palindrome

vii) Program for string transfer 43

viii)Program to reverse a string 44

ix)Program to search for a character in a string 45

x) Program to check for palindrome 46

7. 1. Program to use DOS interrupt INT 21H function called for reading a character from

keyboard, buffered keyboard input, display of character and string on console.48

2.Creation of a new file 49

3.Writing to a file 50

4.Read system date 51

5.Set System Date 54

6.Read system time 57

7.Set system time 59

8.Interfacing Experiments

1)Matrix keyboard interfacing 61

2)Seven segment display interface 63

3)Logical controller interface 64

4)Stepper motor interface 65

QUESTION BANK 66

VIVA QUESTIONS 68

ASSEMBLING AND EXECUTING THE PROGRAM

Writing an ALP

Assembly level programs generally abbreviated as ALP are written in text editor EDIT.

Type EDIT in front of the command prompt to open an untitled text file.

EDIT

After typing the program save the file with appropriate file name with an extension .ASM

Ex:

Add.ASM

Assembling an ALP

To assumble an ALP we needed executable file calledMASM.EXE. Only if this file is in

current working directory we can assemble the program. The command is

MASM

If the program is free from all syntactical errors, this command will give the OBJECT file.In

case of errors it list out the number of errors, warnings and kind of error.

Note:No object file is created until all errors are rectified.

Linking

After successful assembling of the program we have to link it to get Executable file.

The command is

LINK

This command results in which can be executed in front of the command

prompt.

Executing the Program

Open the program in debugger by the command(note only exe files can be open)by the

command.

This will open the program in debugger screen where in you can view the assemble code

with the CS and IP values at the left most side and the machine code. Register content

,memory content also be viewed using VIEW option of the debugger.

Execute option in the menu in the menu can be used to execute the program either in

single steps(F8) or burst execution(F5).

1. Program involving Data transfer instructions

i)Byte and word data transfer in different addressing modes

DATA SEGMENT

DATA1 DB 23H

DATA2 DW 1234H

DATA3 DB 0H

DATA4 DW 0H

DATA5 DW 2345H,6789H

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Initialize DS to point to start of the memory

MOV DS,AX ;set aside for storing of data

MOV AL,25X ;copy 25H into 8 bit AL register

MOV AX,2345H ;copy 2345H into 16 bit AX register

MOV BX,AX ;copy the content of AX into BX register(16 bit)

MOV CL,AL ;copy the content of AL into CL register

MOV AL,DATA1 ;copies the byte contents of data segment memory

;location DATA1 into 8 bit AL

MOV AX,DATA2 ;copies the word contents of data segment memory

;location DATA2 into 16 bit AX

MOV DATA3,AL ;copies the AL content into the byte contents of data

;segment memory location DATA3

MOV DATA4,AX ;copies the AX content into the word contents of

;data segment memory location DATA4

MOV BX,OFFSET DATA5 ;The 16 bit offset address of DS memeory location

; DATA5 is copied into BX

MOV AX,[BX] ; copies the word content of data segment

;memory location addressed by BX into

;AX(register indirect addressing)

MOV DI,02H ;address element

MOV AX,[BX+DI} ; copies the word content of data segment

;memory location addressed by BX+DI into

;AX(base plus indirect addressing)

MOV AX,[BX+0002H] ; copies the word content of data segment

;memory location addressed by BX+0002H into

;(16 bit)

MOV AL,[DI+2] ;register relative addressing

MOV AX,[BX+DI+0002H] ;copies the word content of data segment

;memory location addressed by BX+DI+0002H

;into AX(16 bit)

MOV AH,4CH ; Exit to DOS with function call 4CH

INT 21H

CODE ENDS ; Assembler stop reading

END START

ii)Block move (with and with out overlapping)

Without overlapping

DATA SEGMENT

X DB 01H,02H,03H,04H,05H ;Initialize Data Segments Memory Locations

Y DB 05 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START:MOV AX,DATA ; Initialize DS to point to start of the memory

MOV DS,AX ; set aside for storing of data

MOV CX,05H ; Load counter

LEA SI,X+04 ; SI pointer pointed to top of the memory block

LEA DI,X+04+03 ; 03 is displacement of over lapping, DI pointed to

;the top of the destination block

Before execution

Y,DI

X, SI

After execution

With Overlapping

DATA SEGMENT

X DB 01H,02H,03H,04H,05H ; Initialize Data Segments Memory Locations

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START:MOV AX,DATA ; Initialize DS to point to start of the memory

MOV DS,AX ; set aside for storing of data

MOV CX,05H ; Load counter

LEA SI,X+04 ; SI pointer pointed to top of the memory block

LEA DI,X+04+03 ; 03 is displacement of over lapping, DI pointed to

;the top of the destination block

UP: MOV BL,[SI] ; Move the SI content to BL register

MOV [DI],BL ; Move the BL register to content of DI

DEC SI ; Update SI and DI

DEC DI

DEC CX ; Decrement the counter till it becomes zero

JNZ UP

MOV AH,4CH

INT 21H

CODE ENDS

END START

DS Before execution

DS After execution

iii) Block Interchange

DATA SEGMENT

X DB 01H,02H,03H,04H,05H

Y DB 11H,12H,13H,14H,15H

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START:MOV AX,DATA

MOV DS,AX

MOV CX,05H ; Load the counter

LEA SI,X ; SI pointed to the source location x

LEA DI,Y ; DI pointed to the destination location y

UP: MOV BL,[SI] ; Move the SI content to BL register

MOV AL,[DI] ; Move the DI content to AL register

MOV [SI],AL ; Move AL register content to content of SI

MOV [DI],BL ; Move BL register content to content of DI

INC SI ; Update SI and DI

INC DI

DEC CX ; Decrement the counter till it becomes zero

JNZ UP

MOV AH,4CH

INT 21H

CODE ENDS

END START

DS Before execution

X, SI

Y,DI

DS After execution

2) Program involving Arithmetic and logic operations like addition and subtraction of

multi precision numbers

i ) 16 Bit Addition

DATA SEGMENT

NUM DW 1234H, 0F234H

SUM DW 2 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS: CODE, DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,NUM ; First number loaded into AX

MOV BX,0H ; For carry BX register is cleared

ADD AX,NUM+2 ; Second number added with AX

JNC DOWN ; Check for carry

INC BX ; If carry generated increment the BX

DOWN: MOV SUM,AX ; Storing the sum value

MOV SUM+2,BX ; Storing the carry value

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT : 1234H, F234H

OUTPUT : 10468H

ii) 32 Bit addition

DATA SEGMENT

NUM1 DW 0FFFFH,0FFFFH

NUM2 DW 1111H,1111H

SUM DW 4 DUP(0)

dATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,NUM1 ;Move LSB of NUM1 to AX

ADD AX,NUM2 ;Add LSB of NUM2 to AX

MOV SUM,AX ;Store the LSB in SUM

MOV AX,NUM1+2 ; Move MSB of NUM1 to AX

ADC AX,NUM2+2 ; Add MSB of NUM2 to AX

JNC DOWN ; Check for carry

MOV SUM+4,01H ; Store the carry in SUM+4

DOWN: MOV SUM+2,AX ; Store the MSB in SUM+2

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: 0FFFFFFFFH, 011111111H

OUTPUT: 0111111110H

iii) 32 Bit addition using DD directive

DATA SEGMENT

NUM1 DW 12345678H

NUM2 DW 12345678H

SUM DW 3 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START:MOV AX,DATA

MOV DS,AX

LEA SI,NUM1 ; SI pointed to the of LSB of NUM1

LEA DI,NUM2 ; DI pointed to the of LSB of NUM2

MOV AX,[SI] ; Move the content of SI to AX

ADD AX,[DI] ; Add DI content to AX

MOV CX,[SI+2] ; Move the SI to point MSB of NUM1 and move that

;content to CX

ADC CX,[DI+2] ; Move the DI to point MSB of NUM2 and add

;with carry to CX

JNC DOWN ; Check for carry

MOV SUM+4,01H ; Store the carry in SUM+4

DOWN:MOV SUM,AX ; Store the LSB in SUM

MOV SUM+2,CX ; Store the MSB in SUM+2

MOV AH,4CH

INT 21H

CODE ENDS

END START

INOUT: 12345678H,12345678H

OUTPUT:2468ACF0H

iv) 16 Bit Subtraction

DATA SEGMENT

NUM DW 4567H,2345H

DIF DW 1 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

CLC ; Clearing Carry

LEA SI,NUM ; SI pointed to the NUM

MOV AX,[SI] ; Move NUM1 to AX

SBB AX,[SI+2] ; Move the SI to Num2 and subtract with AX(Takes

;care for both smaller as well as larger

;Number subtraction)

MOV DIF,AX ;Store the result

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: 4567H,2345H

OUTPUT:2222

v) 32 Bit Subtraction

DATA SEGMENT

NUM1 DW 2345H,6762H

NUM2 DW 1111H,1111H

DIF DW 2 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

LEA SI,NUM1 ; SI pointed to the LSB of NUM1

LEA DI,NUM2 ; DI pointed to the LSB of NUM2

MOV AX,[SI] ; Move the content of SI to AX

MOV BX,[DI] ; Move the content of DI to BX

SUB AX,BX ; Subtract from BX to AX

MOV DIF,AX ; Store the LSB result in DIF

INC SI ;Update SI to point the MSB of NUM1(if

;ADD SI,02 instruction its affect carry flag)

INC SI

INC DI ;Update DI to point the MSB of NUM2

INC DI

MOV AX,[SI] ; Move the content of SI to AX

MOV BX,[DI] ; Move the content of DI to BX

SBB AX,BX ; Subtract with borrow from BX to AX

MOV DIF+2,AX ; Store the MSB result in DIF+2

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: 23456762,-11111111

OUTPUT:12345651

INPUT:11111111,-23451234

OUTPUT:EDCBFEDD

Multiplication and Division of signed and unsigned Hexadecimal numbers

vi)16 Bit multiplication for unsigned numbers

DATA SEGMENT

NUM DW 1234H,1234H

PROD DW 2 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

LEA SI,NUM ; SI pointed to the Multiplicand

MOV AX,[SI] ; Multiplicand has to be in AX register

MOV BX,[SI+2] ; SI+2 pointed to the Multiplier and move it to BX

MUL BX ;Perform the multiplication

MOV PROD,AX ;32 bit product stored in DX-AX registers

MOV PROD+2,DX

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: Multiplicand- 1234H,

Multiplier - 1234H

OUTPUT: DX-01 4B

AX-54 90

vii)16 Bit multiplication for signed numbers

DATA SEGMENT

NUM DW -2,1

PROD DW 2 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

LEA SI,NUM ; SI pointed to the Multiplicand

MOV AX,[SI] ; Multiplicand has to be in AX register

MOV BX,[SI+2] ; SI+2 pointed to the Multiplier and move it to BX

IMUL BX ; Perform the sign multiplication using sign

;Multiplication operator (IMUL)

MOV PROD,AX ; 32 bit product stored in DX-AX registers

MOV PROD+2,DX

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: Multiplicand- -2,

Multiplier - 1

OUTPUT: DX – FF FF

AX – FF FE ; Result is in two complement form.

viii)8 Bit Division for Unsigned numbers

DATA SEGMENT

NUM1 DB 72H,

NUM2 DB 02H

QUO DB 1 DUP(0)

REM DB 1 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AL,NUM1 ;Move the Dividend to AL

MOV AH,0H ; Zero extended for 16 bit/8 bit division

DIV NUM2 ; Perform the Division operation

MOV QUO,AL ; Store the quotient to AL

MOV REM,AH ;Store the reminder to AH

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: Dividend - 72H,

Divisor - 02 H,

0UTPUT: AL - 39H (quotient);

AX - 00H (reminder);

INPUT: Dividend - 55H,

Divisor - 04 H,

0UTPUT: AL - 15H (quotient);

AX - 01H (reminder);

ix)8 Bit Division for Signed numbers

DATA SEGMENT

NUM1 DB -10

NUM2 DB 02

QUO DB 1 DUP(0)

REM DB 1 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AL,NUM1 ;Move the Dividend to AL

CBW

IDIV NUM2 ; Perform the Sign Division operation using IDIV operator

MOV QUO,AL ; Store the quotient to AL

MOV REM,AH ;Store the reminder to AH

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: Dividend - -10

Divisor - 02

0UTPUT: AL – FBH (quotient) ; Result is in two complement form

INPUT: Dividend - -10

Divisor - 03

0UTPUT: AL – FDH (quotient);

AX – FF H (reminder) ; Result is in two complement form

x)16 Bit Division for Unsigned numbers

DATA SEGMENT

NUM1 DW 4567H,2345H

NUM2 DW 4111H

QUO DW 2 DUP(0)

REM DW 1 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,NUM1 ;Move the lower bit of Dividend to AX

MOV DX,NUM1+2 ; Move the higher bit of Dividend to DX

DIV NUM2 ; Perform the Division operation

MOV QUO,AX ; Store the quotient to AX

MOV REM,DX ; Store the reminder to DX

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: Dividend - 23454567,

Divisor - 4111,

0UTPUT: AX – 8AC5H (quotient);

DX – 0952H (reminder);

xi)16 Bit Division for Signed numbers

DATA SEGMENT

NUM1 DW 4567H,2345H

NUM2 DW 4111H

QUO DW 2 DUP(0)

REM DW 1 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,NUM1 ; Move the lower bit of Dividend to AX

MOV DX,NUM1+2 ; Move the higher bit of Dividend to DX

CWD

IDIV NUM2 ; Perform the sign Division operation using IDIV

MOV QUO,AX ; Store the quotient to AX

MOV REM,DX ; Store the reminder to DX

MOV AH,4CH

INT 21H

CODE ENDS

END START

INPUT: Dividend - -44444444,

Divisor - 2222,

0UTPUT: AX – FE (quotient);

DX – FF (reminder) ; Result is in two complement form.

3.Code Conversion

i)ASCII adjustment instructions

CODE SEGMENT

ASSUME CS:CODE

START: MOV AX,31H ;Load ASCII 1

ADD AX,39H ;Load ASCII 9

AAA ;ASCII Adjust, AX=0100 UNPACKED BCD

ADD AX,3030H ;Answer in ASCII

MOV BL,9 ;Load divisor

MOV AX,0702H ;Load dividend, AAD instruction requires

Ax register to contain a two digit unpacked

;BCD number before executing

AAD ;AAD appears before division

DIV BL ;Contents of adjusted AX register is devised

;by an unpacked BCD number to generate

;a single digit result in AL with any

;remainder in AH

MOV AL,5 ;Load multiplicand

MOV CL,5 ;Load multiplier

MUL CL ;AX=0019H

AAM ;AX=0205(Unpacked BCD)

ADD AX,3030H ;AX=3235H

MOV AX,38H ;Load ASCII 8

SUB AX,31H ;Load ASCII 1

AAS ;AX=0007

AX,3030H ;AX=3037H

MOV AH,4CH

INT 21H

CODE ENDS

END START

ii)Binary to BCD code conversion

DATA SEGMENT

BIN DW 01A9H

BCD DB 2 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV AX,BIN ;Move the Binary Data to AX.

MOV CL,64H ;100 in decimal

DIV CL ;Perform the division by 100.

MOV BCD+1,AL ;Store the quotient in BCD+1.

MOV AL,AH ;Move the Reminder value to AL.

MOV AH,00H ;Initialize the AH.

MOV CL,0AH ;10 in decimal.

DIV CL ;Perform the division by 10.

MOV CL,04

ROR AL,CL ;Perform the Right side rotation 4 times.

ADD AL,AH ;Adding the Reminder in LSB.

MOV AH,4CH

INT 21H

CODE ENDS

END START

Input: binary--------01A9

Output: bcd------------425

iii)BCD to Binary code conversion

DATA SEGMENT

BCD DW 27H

BIN DW ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV AX,BCD ;Move the BCD Data to AX.

AND AX,07H ;Perform the AND operation between

;07H and input BCD

MOV BX,AX ;Move data AX to BX

MOV AX,BCD ;Move the BCD Data to AX.

AND AX,0F0H ;Perform the AND with 0F0H for shifting operation.

MOV CX,0AH ;10 in decimal.

MUL CX ;Perform the multiplication by 10.

ADD AX,BX ;Perform the addition operation to get the LSB.

MOV BIN,AX ;Move the result to binary.

MOV AH,4CH

INT 21H

CODE ENDS

END START

Input: BCD-------------27

Output:------------------1B

4. Arithmetic programs to find square, cube, LCM, GCD and factorial

i) Program to find square and cube of a number

DATA SEGMENT

X DW 04H

SQUARE DW ?

CUBE DW ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV AX,X ;Move the X number Data to AX.

MOV BX,X ;Move the X number Data to BX.

MUL BX ;Perform the multiplication by BX.

MOV SQUARE,AX ;Store value in SQUARE.

MUL BX ;Perform the multiplication by BX.

MOV CUBE,AX ;Store value in CUBE.

MOV AH,4CH

INT 21H

CODE ENDS

END START

Input: x ----------- 4h

Output: Square ---------10h

Cube -----------40h

ii)Program to find LCM of a given number

DATA SEGMENT

NUM DW 05,04

LCM DW 2 DUP(0)

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV DX,0H ;Initialize the DX.

MOV AX,NUM ;Move the first number to AX.

MOV BX,NUM+2 ;Move the second number to BX.

UP: PUSH AX ;Store the quotient/first number in AX.

PUSH DX ;Store the remainder value in DX.

DIV BX ;Divide the first number by second number.

CMP DX,0 ;Compare the remainder.

JE EXIT ;If remainder is zero, go to EXIT label.

;If remainder is non-zero,

POP DX ;Retrieve the remainder.

POP AX ;Retrieve the quotient.

ADD AX,NUM ;Add first number with AX.

JNC DOWN ;If no carry jump to DOWN label.

INC DX ;Increment DX.

DOWN: JMP UP ;Jump to Up label.

EXIT: POP LCM+2 ;If remainder is zero, store the value at LCM+2.

POP LCM

MOV AH,4CH

INT 21H

CODE ENDS

END START

Input: 0A, 04

Output: 02

iii)Program to find GCD of two numbers

DATA SEGMENT

NUM1 DW 000AH

NUM2 DW 0004H

GCD DW ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV AX,NUM1 ;Move the first number to AX.

MOV BX,NUM2 ;Move the second number to BX.

UP: CMP AX,BX ;Compare the two numbers.

JE EXIT ;If equal, go to EXIT label.

JB EXCG ;If first number is below than second,

;go to EXCG label.

UP1:MOV DX,0H ;Initialize the DX.

DIV BX ;Divide the first number by second number.

CMP DX,0 ;Compare remainder is zero or not.

JE EXIT ;If zero, jump to EXIT label.

MOV AX,DX ;If non-zero, move remainder to AX.

JMP UP ;Jump to UP label.

EXCG:XCHG AX,BX ;Exchange the remainder and quotient.

JMP UP1 ;Jump to UP1.

EXIT:MOV GCD,BX ;Store the result in GCD.

MOV AH,4CH

INT 21H

CODE ENDS

END START

Input: 0A,04

Output: 02

iv)Program to find factorial of a given number

DATA SEGMENT

X DW 06H

FACT DW ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,01H ;Set the value of AX as 01H.

MOV CX,X ;Move the i/p number to CX.

UP: MUL CX ;Perform the Loop multiplication operation.

LOOP UP

MOV FACT,AX ;Store the FACT value.

MOV AH,4CH

INT 21H

CODE ENDS

END START

Input: 06

Output: 2D0H

5.Program involving bit manipulation instruction

i)If given data is positive or negative

DATA SEGMENT

NUM DB 12H

MES1 DB 10,13,'DATA IS POSITIVE $'

MES2 DB 10,13,'DATA IS NEGATIVE $'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AL,NUM ;Move the Number to AL.

ROL AL,1 ;Perform the rotate left side for 1 bit position.

JC NEGA ;Check for the negative number.

MOV DX,OFFSET MES1 ;Declare it positive.

JMP EXIT ;Exit program.

NEGA: MOV DX,OFFSET MES2;Declare it negative.

EXIT: MOV AH,09H

INT 21H

MOV AH,4CH

INT 21H

CODE ENDS

END START

Output: Data is positive

Positive Numbers: 00-7F

Negative numbers: 80-FF

ii)If given data is odd or even

DATA SEGMENT

X DW 27H

MSG1 DB 19,13,'NUMBER IS EVEN$'

MSG2 DB 10,13,'NUMBER IS ODD$'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,X

TEST AX,01H ;Test for Even/Odd number.

JNZ EXIT ;If it is Even go to Exit label.

;(alternate logic)

;MOV BL,2

;DIV BL

;CMP AH,0H

;JNZ EXIT

LEA DX,MSG1 ;Declare it is Even number.

MOV AH,09H

INT 21H

JMP LAST

EXIT: LEA DX,MSG2 ;Declare it is Odd number.

MOV AH,09H

INT 21H

LAST: MOV AH,4CH

INT 21H

CODE ENDS

END START

Output: Number is ODD

iii) Logical ones and zeros in a given data

DATA SEGMENT

X DB 0AAH

ONE DB ?

ZERO DB ?

DATA ENDS

CODE SEGMENT

ASSUME CS: CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AH,X

MOV BL,8 ;Initialize BL to 8.

MOV CL,1 ;Initialize CL to 1.

UP: ROR AH,CL ;Perform the single bit rotate operation

;with respect to right.

JNC DOWN ;If no carry go to DOWN label.

INC ONE ;Increment one.

JMP DOWN1 ;Jump to DOWN1.

DOWN: INC ZERO ;Increment ZERO.

DOWN1: DEC BL ;Decrement the BL.

JNZ UP ;If no zero go to UP label.

MOV AH,4CH

INT 21H

CODE ENDS

END START

Output: Ones--------04

Zeros--------04

iv) 2 out of 5 code

DATA SEGMENT

X DW 82H

MES DB 10,13,'VALID 2 OUT OF CODE $'

MES1 DB 10,13,'NOT A VALID 2 OUT OF CODE $'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV AX,X ;Move the Data word to AX.

MOV BX,0H ;Initialize the BX.

AND AX,0E0H ;Perform the AND operation of first 3 bit.

JNZ DISP ;If no zero jump to DISP label.

MOV CL,05 ;If zero, Initialize the counter for check the last 5 bit.

MOV AX,X ;Move the Data word to AX.

UP: ROR AX,1 ;Rotate right side one time.

JNC DOWN ;If no carry jump to DOWN label.

INC BX ;Increment the BX.

DOWN:DEC C ;Decrement the counter.

JNC UP ;If no carry jump to UP label.

CMP BX,02H ;Compare the BX with 2.

JNZ DISP ;If no zero jump to DISP label.

LEA DX,MES ;Declared as 2 out of 5 code .

MOV AH,09H

INT 21H

JMP EXIT

DISP: LEA DX,MES1 ;Declared as not valid code .

MOV AH,09H

INT 21H

EXIT:MOV AH,4CH

INT 21H

CODE ENDS

END START

Output: Not a valid 2 out of 5 code.

v) Bit wise palindrome

DATA SEGMENT

X DW 0FFFFH

MSG1 DB 10,13,'NUMBER IS PALINDROME$'

MSG2 DB 10,13,'NUMBER IS NOT PALINDROME$'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;Load the Data to AX.

MOV DS,AX ;Move the Data AX to DS.

MOV AX,X ;Move DW to AX.

MOV CL,10H ;Initialize the counter 10.

UP: ROR AX,1 ;Rotate right one time.

RCL DX,1 ;Rotate left with carry one time.

LOOP UP ;Loop the process.

CMP AX,DX ;Compare AX and DX.

JNZ DOWN ;If no zero go to DOWN label.

LEA DX,MSG1 ;Declare as a PALINDROME.

MOV AH,09H

INT 21H

JMP EXIT ;Jump to EXIT label.

DOWN: LEA DX,MSG2 ; Declare as not a PALINDROME

MOV AH,09H

INT 21H

EXIT:MOV AH,4CH

INT 21H

CODE ENDS

END START

Output: Number is Palindrome

vi) Nibble wise palindrome

DATA SEGMENT

X DW 2662H

TEMP DW 0H

MES DB 10,13,'THE WORD IS NIBBLEWISE PALINDROME $'

MES1 DB 10,13,'THE WORD IS NOT NIBBLEWISE PALINDROME $'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AX,X

MOV BX,X

MOV CL,04 ;Initialize counter.

AND AX,0000FH ;Perform the and operation between

;last nibble of AX and 000FH.

AND BX,0F000H ;Perform the and operation between

;last nibble of BX and 000FH.

ROL BX,CL ;Rotate left side 4 times the BX.

CMP AX,BX ;Compare AX with BX.

JNZ TER ;If no zero go to TER label.

MOV AX,X ;Move the DW to AX.

ROR AX,CL ;Rotate right side 4 times the AX.

MOV BX,AX ;Move AX to BX.

AND AX,000FH ; Perform the and operation with last nibble.

ROR BX,CL ;Rotate right side 4 times the BX.

AND BX,000FH ;Perform the and operation with last nibble of BX.

CMP AX,BX ;Compare AX with BX.

JNZ TER ;If no zero go to TER label.

MOV AH,09H

LEA DX,MES ;Declared as a PALINDROME.

INT 21H

JMP LAST

TER:MOV AH,09H

LEA DX,MES1 ;Declared as a non palindrome.

INT 21H

LAST:MOV AH,4CH

INT 21H

CODE ENDS

END START

Output: The word is nibble wise palindrome

6. PROGRAMS INVOLVING BRANCH/LOOP INSTRUCTIONS / PROGRAMS ON

ARRAYS

i)ADDITION OF n NUMBERS

DATA SEGMENT ;start of data segment

ARR DW 0010H,0020H,0030H,0040H,0050H

LEN EQU ($-ARR)/2

SUM DW ?

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

LEA SI,ARR ;SI points to the LSB of data ARR

CLC ;clear carry

XOR AX,AX ;clear AX register

MOV CX,LEN ;load CX with the number of data words in

ARR

UP: ADC AX,[SI] ;add the number pointed by SI to A

INC SI ;point to the next data word

INC SI

DEC CX ;decrement Cx

JNZ UP ;and check if all numbers are added

if no then add

MOV SUM,AX ;store the addition result in user

defined memory location sum

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: 00F0

ii)PROGRAM TO SUBTRACT N NUMBERS

DATA SEGMENT ;start of data segment

ARR DW 50H,10H,20H,10H,05H

LEN EQU ($-ARR)/2

DIF DW ?

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

LEA SI,ARR ;SI points to the LSB of data ARR

CLC ;clear carry flag

MOV CX,LEN-1 ;load CX register with the number of

data words in ARR

MOV AX,[SI] ;make a copy of the first number

pointed by SI in AX

UP: SUB AX,[SI+2] ;subtract the next number from the

contents of AX and store the result in AX

INC SI ;point to the next number

INC SI

DEC CX ;decrement CX

JNZ UP ;and check if all subtraction of all

numbers is complete if no then subtract

MOV DIF,AX ;store the difference in user defined

memory location DIFF

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: 0005

PROGRAMS TO FIND LARGEST AND SMALLEST NUMBER

iii)PROGRAM TO FIND LARGEST NUMBER AMONG THE SERIES

DATA SEGMENT ;start of data segment

X DW 0010H,52H,30H,40H,50H

LAR DW ?

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV CX,05H ;load CX register with number of datawords

in X

LEA SI,X ;initialize SI to point to the first number

MOV AX,[SI] ;make a copy of the number pointed by SI in

DEC CX ;set count value in CX for comparison

UP: CMP AX,[SI+2] ;compare two adjacent numbers(one is in

AX and the other is pointed by SI+2)

JA CONTINUE ;if contents of AX is greater than the next

number in array retain the contents of AX

MOV AX,[SI+2] ;if not make a copy of the larger number in

CONTINUE:ADD SI,2 ;point to the next number

DEC CX ;decrement CX to check if all numbers are

compared

JNZ UP ;if no continue to compare

MOV LAR,AX ;if yes make a copy of AX(largest number)

in user defined memory location LAR

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

iv)PROGRAM TO FIND THE LARGEST NUMBER USING DOS DISPLAY

INTERRUPTS

DATA SEGMENT ;start of data segment

X DW 0010H,0052H,0030H,0040H,0050H

MES DB 10,13,'LARGEST NUMBER AMONG THE SERIES IS $'

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV CX,05H ;load CX register with

number of datawords in array X

LEA SI,X ;SI points to start of dataword

array X

MOV AX,[SI] ;make a copy of the first

number in AX

DEC CX ;initialize CX with count

value for comparison

UP: CMP AX,[SI+2] ;compare the contents of AX

and the number pointed by SI+2

JA CONTINUE ;if AX is greater than the next

number in array then retain the

contents of AX

MOV AX,[SI+2] ;else make a copy of the next

number (larger number)in AX

CONTINUE:ADD SI,2 ;point to next number in array

DEC CX ;decrement CX

JNZ UP ;check if all numbers are

compared if no continue comparison

AAM ;if yes convert largest binary

number in AX to unpacked BCD

ADD AX,3030H ;convert unpacked BCD to

unpacked ASCII equivalent

MOV BX,AX ;make a copy of it in AX

MOV AX,09H ;display the message stored at

user defined memory location MES

LEA DX,MES

INT 21H

MOV DL,BH ;display the largest number

MOV AH,02H

INT 21H

MOV DL,BL

INT 21H

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: LARGEST NUMBER AMONG THE SERIES IS 0052

v)PROGRAM TO FIND THE SMALLEST NUMBER AMONG THE SERIES

DATA SEGMENT ;start of data segment

X DW 0060H,0020H,0030H,0040H,0050H

MES DB 10,13,'SMALLEST NUMBER AMONG THE SERIES IS $'

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV CX,05H ;load CX with number of

datawords in array X

LEA SI,X ;SI points to the first number

in array X

MOV AX,[SI] ;make a copy of the first

number in AX

DEC CX ;intialize CX with count

value for comparison

UP: CMP AX,[SI+2] ;compare the contents of AX

with next number in array pointed

by SI+2

JB CONTINUE ;if AX is smaller than the

next number retain the contents of

MOV AX,[SI+2] ;else make a copy of the smaller

number in AX

CONTINUE:ADD SI,2 ;SI points to the next number

DEC CX ;decrement the count value

JNZ UP ;check if all the numbers are

compared if no continue

comparison

AAM ;if yes convert the smallest

binary number to unpacked BCD

ADD AX,3030H ;convert the unpacked BCD

to unpacked ASCII equivalent

MOV BX,AX ;make a copy of the unpacked

ASCII in BX

MOV AH,09H ;display the message stored at

user defined memory location

MES using DOS interrupts

LEA DX,MES

INT 21H

MOV DL,BH ;display the smallest number

in array X using DOS interrupts

MOV AH,02H

INT 21H

MOV DL,BL

INT 21H

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: SMALLEST NUMBER AMONG THE SERIES IS 0020

vi)PROGRAM TO SORT THE NUMBERS IN ASCENDING/DESCENDING ORDER

DATA SEGMENT ;start of data segment

x DW 42H,34H,26H,17H,09H

LEN EQU 05

ASCD DB 10 DUP(0)

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV BX,LEN-1 ;load BX(counter1) with count

value(number of datawords in array - 1)

MOV CX,BX ;make a copy of the count value in CX(counter2)

UP1: MOV BX,CX ;load the updated CX in BX

LEA SI,X ;SI points to the first number in the array

UP: MOV AX,[SI] ;make a copy of the number pointed by SI in

MOV DX,[SI+2] ;make a copy of the next number in DX

CMP AX,DX ;compare both the numbers

JB DOWN/JA DOWN ;if AX <> DX retain them as it is

MOV [SI],DX ;if not sort the numbers in ascending order

MOV [SI+2],AX

DOWN: INC SI ;point to the next number

INC SI

DEC BX ;decrement the counter1

JNZ UP ;compare till the larger number is sorted at

the end of the array

DEC CX ;decrement counter2

JNZ UP1 ;compare till the numbers are sorted in

ascending order

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: 09 17 26 34 42

PROGRAMS ON STRING MANIPULATION LIKE STRING TRANSFER, STRING

REVERSING, SEARCHING FOR A CHARACTER IN A STRING AND PALINDROME

vii) PROGRAM FOR STRING TRANSFER

DATA SEGMENT ;start of data segment

STR1 DB 'HOW ARE YOU'

LEN EQU $-STR1

STR2 DB 20 DUP(0)

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA,ES:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV ES,AX ;initialize extra segment for string operations

LEA SI,STR1 ;SI points to starting address of string at

;STR1

LEA DI,STR2 ;DI points to starting address of where the

string has to be transferred

MOV CX,LEN ;load CX with length of the string

CLD ;clear the direction flag for auto increment SI

;and DI

REP MOVSB ;the source string is moved to destination

address till CX=0(after every move CX is

;decremented)

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

viii)PROGRAM TO REVERSE A STRING

DATA SEGMENT ;start of data segment

STR1 DB 'HELLO'

LEN EQU $-STR1

STR2 DB 20 DUP(0)

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA,ES:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV ES,AX ;initialize extra segment for string operations

LEA SI,STR1 ;SI points to the starting address of the string

at STR1

LEA DI,STR2+LEN-1 ;DI points to the address of the last character in

the string(here address of '0')

MOV CX,LEN ;load CX with count value equal to number of

characters in the string

UP: CLD ;clear the direction flag to autoincrement SI

LODSB ;load AX with the character pointed SI

STD ;set the direction flag to autodecrement DI

STOSB ;the contents of AX is stored at the address

pointed by DI

LOOP UP ;decrement CX and continue the transfer till

CX is zero

MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: OLLEH

ix)PROGRAM TO SEARCH FOR A CHARACTER IN A STRING

DATA SEGMENT ;start of data segment

MSG DB 'HELLO'

CNT EQU $-MSG

SRC EQU 'E'

MSG1 DB 10,13,'CHARACTER FOUND$'

MSG2 DB 10,13,'CHARACTER NOT FOUND$'

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA,ES:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV ES,AX ;initialize extra segment

LEA SI,MSG ;SI points to the starting address of

the string

MOV AL,SRC ;the character to be searched in the

string is stored in AL

MOV CL,CNT ;CX is loaded with count value equal

to number of characters in the string

MOV CH,00H

CLD ;clear the direction flag for

auto increment SI and DI

UP: SCASB ;check if the character in AL is the

same as that pointed by index register

JZ DOWN ;if it is same jump to label DOWN

LOOP UP ;if not decrement CX and continue

checking till CX is zero

LEA DX,MSG2 ;display the message at MSG2 that is

CHARACTER NOT FOUND

MOV AH,09H

INT 21H

JMP EXIT ;jump to label EXIT

DOWN: LEA DX,MSG1 ;if the character is found display the

message CHARACTER FOUND

MOV AH,09H

INT 21H

EXIT: MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: CHARACTER FOUND

x)PROGRAM TO CHECK FOR PALINDROME

DATA SEGMENT ;start of data segment

STR1 DB 'LIRIL'

LEN EQU $-STR1

STR2 DB 20 DUP(0)

MES1 DB 10,13,'WORD IS PALINDROME$'

MES2 DB 10,13,'WORD IS NOT PALINDROME$'

DATA ENDS ;end of data segment

CODE SEGMENT ;start of code segment

ASSUME CS:CODE,DS:DATA,ES:DATA

START: MOV AX,DATA ;initialize data segment

MOV DS,AX

MOV ES,AX ;initialize extra segment for string

operations

LEA SI,STR1 ;SI points to starting address of string

LEA DI,STR2+LEN-1 ;DI points the last character in the

string

MOV CX,LEN ;load CX with count value equal to

number of characters in the string

UP: CLD ;clear the direction flag to

auto increment SI

LODSB ;get the character in AL from the

address pointed by SI

STD ;set the direction flag equal to

auto decrement DI

STOSB ;store the character in AL at address

pointed by DI

LOOP UP ;decrement CX and continue with

reversing the string till CX=0

LEA SI,STR1 ;SI points to the starting address of

original string

LEA DI,STR2 ;DI points to the starting address of

the string reversed

CLD

MOV CX,LEN ;set CX as counter for checking if

palindrome

REPE CMPSB ;compare the strings pointed by SI

and DI

CMP CX,0H ;do the comparison till CX=0(if

palindrome)

JNZ NOTPALIN ;if CX is not zero then jump to

display WORD NOT PALINDROME

LEA DX,MES1 ;display the message at MES1 which

MOV AH,09H is WORD IS PALINDROME

INT 21H

JMP EXIT ;jump to end of the program

NOTPALIN: LEA DX,MES2 ;display the message WORD NOT

MOV AH,09H PALINDROME using DOS

INT 21H interrupts

EXIT: MOV AH,4CH ;terminate the process

INT 21H

CODE ENDS ;end of code segment

END START

OUTPUT: WORD IS PALINDROME

7.1.Program to use DOS interrupt INT 21H function called for reading a character from

keyboard, buffered keyboard input, display of character and string on console.

DATA SEGMENT

INKEY DB ?

BUF DB 20 DUP(0)

MES DB 10,13, 'DAYANANDA SAGAR COLLEGE OF ENGINEERING $'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE , DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AH,01H ;DOS function to read a character from keyboard ;with

echo. [AL = 8bit character]

INT 21H

MOV INKEY,AL ;Returns ASCII value of the pressed key.

MOV BUF,10 ;Load how many characters to enter.

MOV AH,0AH ;Dos function to read string of characters from

;keyboard.

LEA DX,BUF

INT 21H

MOV AH,06H ;Dos function to display a character.

MOV DL,'A' ;Load the character to be displayed.

INT 21H

MOV AH,09H ;Dos function to read string of characters from

;keyboard.

LEA DX,MES ;DX = offset address of the message

INT 21H

MOV AH,4CH

INT 21H

CODE ENDS

END START

2.Creation of a new file

DATA SEGMENT

FILENAME DB'HELLO.NEW'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

START: MOV AX,DATA ;initialise data segment

MOV DS,AX

MOV AH,3CH ;dos function call to create

;new file

MOV CX,0 ;CX = file attribute

MOV DX,OFFSET FILENAME ; dx has offset address of

;filename

INT 21H

CODE ENDS

END START

3.Writing to a file

DATA SEGMENT

MES DB 10,13,'ENTER SOME DATA IN THE FILE$'

FILENAME DB 'HELLO.NEW'

BUFFER DB 22 DUP(0)

MES1 DB 10,13, 'ERROR IN FILE HANDLING$'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AH,09H

LEA DX,MES

INT 21H

MOV BUFFER,20

MOV AH,0AH

MOV DX,OFFSET BUFFER

INT 21H

MOV AH,3CH

MOV CX,0

MOV DX,OFFSET FILENAME

INT 21H

MOV BX,AX

MOV AH,40H ;function to write in a file

MOV CX,20

MOV DX, OFFSET BUFFER

INT 21H

JC ERROR

JMP EXIT

ERROR: MOV DX,OFFSET MES1

MOV AH,09H

INT 21H

EXIT: MOV AH,4CH

INT 21H

CODE ENDS

END START

4.Read system date

DATA SEGMENT

YY DB

MM DB

D DB

TDAY DW UN,MON,TUE,WED,THU,FRI,SAT

SUN DB'SUNDAY,$'

MON DB'MONDAY,$'

TUE DB'TUESDAY,$'

WED DB'WEDNESDAY,$'

THU DB'THURSDAY,$'

FRI DB FRIDAY,$'

SAT DB'SATURDAY,$'

TMON DW JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC

JAN DB'JANUARY,$'

FEB DB'FEBRUARY,$'

MAR DB'MARCH,$'

APR DB'APRIL,$'

MAY DB'MAY,$'

JUN DB'JUNE,$'

JUL DB'JULY,$'

AUG DB'AUGUST,$'

SEP DB'SEPTEMBER,$'

OCT DB'OCTOBER,$'

NOV DB'NOVEMBER,$'

DEC DB'DECEMBER,$'

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

DISCHAR MACRO CHAR

PUSH AX

PUSH DX

MOV DL,CHAR

MOV AH,02

INT 21H

POP DX

POP AX

ENDM

START: MOV AX,DATA

MOV DX,AX

CALL PDATE

MOV AH,4CHH

INT 21H

PDATE PROC NEAR

MOV AH,2AH

INT21H

MOV YY,CX

MOV MM,DH

MOV D,DL

MOV AH,0

ROL AX,1

MOV SI,OFFSET TDAY

ADD SI,AX

MOV DX,[SI]

MOV AH,09H

INT21H

MOV AL,D

MOV AH,00H

AAM

OR AH,AH

JZ DIGIT0

ADD AH,30H

DISCHAR AH

DIGIT0: ADD AL,30H

DISCHAR AL

DISCHAR ''

MOV AL,MM

SUB AL,1

MOV AH,0

ROL AX,1

MOV SI,OFFSET TMON

ADD SI,AX

MOV DX,[SI]

MOV AH,09H

INT21H

MOV AX,YY

CMP AX,2000

JB DIS19

SUB AX,2000

DISCHAR'2'

DISCHAR'0'

JMP SKIP

DIS19: SUB AX,1900

DICHAR'1'

DISCHAR'9'

SKIP: AAM

ADD AX,3030H

DISCHAR AH

DISCHAR AL

RET

PDATE ENDP

CODE ENDS

END START

Output: sunday,september 25,2005

5.Set System Date

DATA SEGMENT

MES DB 10,13,'ENTER THE DATE WITH FORMAT: DD:MM:YY $'

MES1 DB 10,13,"DATE:$"

BUFF DB 10 DUP(0)

DB 0

DB 10 DUP(0)

YY DB ?

MM DB ?

D DB ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

START: MOV AX,DATA

MOV DS,AX

CALL DATEP

MOV AH,4CH

INT 21H

DATEP PROC NEAR

MOV AH,09H

LEA DX,MES

INT 21H

MOV AH,09H

LEA DX,MES1

INT 21H

MOV AH,0AH

LEA DX,BUFF

INT 21H

MOV CL,04

MOV DL,0H

LEA SI,BUFF

ADD SI,02

BACK: MOV AL,[SI]

CMP AL,':'

JZ TER

ROL DL,CL

SUB AL,30H

ADD DL,AL

INC SI

JMP BACK

TER: MOV DH,DL

ADD DL,0F0H

ROR DL,CL

MOV AL,10

MUL DL

AND DH,0FH

ADD AL,DH

MOV DH,AL

MOV DL,0

INC SI

BACK1: MOV AL,[SI]

CMP AL,':'

JZ TER1

ROL DL,CL

SUB AL,30H

ADD DL,AL

INC SI

JMP BACK1

TER1: MOV DH,DL

AND DL,0F0H

ROR DL,CL

MOV AL,10

MUL DL

AND DH,0FH

ADD AL,DH

MOV MM,AL

MOV DL,0

INC SI

BACK2: MOV AL,[SI]

CMP AL,13

JZ TER2

ROL DL,CL

SUB AL,30H

ADD DL,AL

INC SI

JMP BACK2

TER2: MOV DH,DL

AND DL,0F0H

ROR DL,CL

MOV AL,10

MUL DL

AND DH,0FH

ADD AL,0DH

MOV YY,AL

MOV AH,2BH

MOV CL,YY

MOV CH,00

ADD CX,2000

MOV DH,MM

MOV DL,0DH

INT 21H

RET

DATEP ENDP

CODE ENDS

END START

6.READ SYSTEM TIME

DATA SEGMENT

HOUR DB ?

MIN DB ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

DISCHAR MACRO CHAR

PUSH AX

PUSH DX

MOV DL,CHAR

MOV AH,02

INT 21H

POP DX

POP AX

ENDM

START: MOV AX,DATA

MOV DS,AX

CALL TIME

MOV AH,4CH

INT21H

TIME PROC NEAR

MOV AH,2CH ;function to read system time

INT21H

MOV HOUR,CH

MOV MIN,CL

CMP CH,12

JB DOWN

SUB CH,12

DOWN: MOV AL,CH

MOV AH,00H

AAM

MOV AX,3030H

DISCHAR AH

DISCHAR AL

DISCHAR’:’

MOV AL,CL

MOV AH,00H

AAM

ADD AX,3030H

DISCHAR AH

DISCHAR AL

DISCHAR’ ’

CMP HOUR,12

JB AM

DISCHAR ‘P’

JMP DOWN1

AM: DISCHAR’A’

DOWN1: DISCHAR’M’

RET

TIME ENDP

CODE ENDS

END START

7.Set system time

DATA SEGMENT

MES DB 10,13,’ENTER TIME WITH THE FORMAT :HOUR FOLLOWED BY MIN

FOLLOWED BY AM OR PM$’

MES1 DB 10,13 ,’TIME:$’

BUF DB 10

DB 0

DB 10 DUP(0)

HOUR DB?

MIN DB?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

START: MOV AX,DATA

MOV DS,AX

CALL TIME

MOV AX,4CH

INT 21H

TIME PROC NEAR

MOV AH,09H

LEA DX,MES

INT 21H

MOV AH,09H

LEA DX,MES1

INT 21H

MOV AH,0AH

LEA DX,BUF

INT 21H

MOV CL,4

MOV DL,00H

LEA SI,BUF

ADD SI,2

UP: MOV AL,[SI]

CMP AL,’:’

JZ DOWN

ROL DL,CL

SUB AL,30H

ADD DL,AL

INC SI

JMP UP

DOWN: MOV DH,DL

AND DL,0F0H

ROR DL,CL

MOV AL,10

MUL DL

AND DH,0FH

ADD AL,DH

MOV HOUR,AL

MOV DL,0

INC SI

UP1: MOV AL,[SI]

CMP AL,’ ’

JZ DOWN1

ROL DL,CL

SUB AL,30H

ADD DL,AL

INC SI

JMP UP1

DOWN1: MOV DH,DL

AND DL,0F0H

ROR DL,CL

MOV AL,10

MUL DL

AND DH,0FH

ADD AL,DH

MOV MIN,AL

INC SI

MOV CH,[SI]

CMP CH,’P’

JNZ SKIP

ADD HOUR,0CH

SKIP: MOV AH,2DH

MOV CH,HOUR

MOV CL,MINUTE

MOV CX,0000H

INT 21H

RET

TIME ENDP

CODE ENDS

END START

8.INTERFACING EXPERIMENTS

1)MATRIX KEYBOARD INTERFACING

DATA SEGMENT

PORTA EQU 120H

PORTC EQU 122H

CWRD EQU 123H

ARRAY DB '0123456789.+-*/%ACK=MMMM'

DATA ENDS

CODE SEGMENT

ASSUME CS: CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX ;initialise data segment

MOV AL,90H ;initialise 8255 porta as i/p and portc as o/p

MOV DX,CWRD

OUT DX,AL

REPEAT: MOV DX,PORTC ;make first row of the keyboard high through pc0

MOV AL,01

OUT DX,AL

MOV DX,PORTA

IN AL,DX ; input contents of porta and check if key is pressed-

CMP AL,00 ; in first row.

JZ NEXT

JMP FIRSTROW

NEXT: MOV DX,PORTC ;if key not found in first row, check if key is in

;second row

MOV AL,02

OUT DX,AL

MOV DX,PORTA

IN AL,DX

CMP AL,00

JNZ SECONDROW

MOV AL,04 ; if key not found then check for key closure in

;third row

MOV DX,PORTC

OUT DX,AL

MOV DX,PORTA

IN AL,DX

CMP AL,00H

JNZ THIRDROW

JMP REPEAT

FIRSTROW: CALL DELAY ;check all the keys one by onein first row

LEA SI,ARRAY

UP: SHR AL,1

JC DISPLAY ;if key found jump to the display subroutine

INC SI

JMP UP

JMP DISPLAY

SECONDROW:CALL DELAY

LEA SI,ARRAY+08H ;second row keys from array +08

UP1:SHR AL,1

JC DISPLAY ;if key found jump to the display subroutine

INC SI

JMP UP1

THIRDROW: CALL DELAY

LEA SI,ARRAY+10H ;third row keys from array +16(dec)

UP2: SHR AL,1

JC DISPLAY ;if key found jump to the display subroutine

INC SI

JMP UP2

JMP DISPLAY

DISPLAY: MOV DL,[SI]

CMP DL,97 ;24 in decimal. 8x3rows = 24keys

JZ EXIT

MOV AH,02H ; display key no in ascii

INT 21H

JMP REPEAT

DELAY: MOV BX,0FFFFH

L1: MOV CX,0FFFH

L2: DEC CX

JNZ L2

DEC BX

JNZ L1

RET

EXIT:MOV AH,4CH

INT 21H

CODE ENDS

END START

2)SEVEN SEGMENT DISPLAY INTERFACE

DATA SEGMENT

PORTA EQU 120H

PORTB EQU 121H

PORTC EQU 122H

CWRD EQU 123H

TABLE DB 8CH,0C7H,86H,89H

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

START: MOV AX,DATA ;intialise data segment

MOV DS,AX

MOV AL,80H ;initialise 8255 portb and portc as o/p

MOV DX,CWRD

OUT DX,AL

MOV BH,04 ; BH = no of digitsto be displayed

LEA SI,TABLE ; SI = starting address of lookup table

NEXTDIGIT:MOV CL,08 ; CL = no of segments = 08

MOV AL,[SI]

NEXTBIT: ROL AL,01

MOV CH,AL ;save al

MOV DX,PORTB ;one bit is sent out on portb

OUT DX,AL

MOV AL,01

MOV DX,PORTC ;one clock pulse sent on pc0

OUT DX,AL

DEC AL

MOV DX,PORTC

OUT DX,AL

MOV AL,CH ; get the sevensegment code back in al

DEC CL ;send all 8 bits,thus one digit is displayed

JNZ NEXTBIT

DEC BH

INC SI ;display all the four digits

JNZ NEXTDIGIT

MOV AH,4CH ;exit to dos

INT 21H

CODE ENDS

END START

3) LOGICAL CONTROLLER INTERFACE

DATA SEGMENT

PA EQU 120H ;INITIALIZE THE ADDRESS OF PORT A OF 8255

PB EQU 121H ;INITIALIZE THE ADDRESS OF PORT B OF 8255

PC EQU 122H ;INITIALIZE THE ADDRESS OF PORT C OF 8255

CR EQU 123H ;INITIALIZE THE ADDRESS OF CONTROL WORD

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

START: MOV AX, DATA

MOV DS, AX

MOV AX, 082H ;load the control word

MOV DX, CR

OUT DX,AX

REPEAT: MOV DX, PB ;input the data (from dip switch)from port b

IN AL,DX

AND AL, 03H

CMP AL,00H

JZ DISPLAY

CMP AL,03H ;check if input is 11

JZ DISPLAY

MOV AL,0FFH ;display 11 if input is 01 else 10

MOV DX, PA ;output to porta

OUT DX,AL

JMP REPEAT

DISPLAY: MOV AL,00H ;display oo if input is 00 else 11

MOV DX,PA ;output to porta

OUT DX, AL

JMP REPEAT

CODE ENDS

END START

4)STEPPER MOTOR INTERFACE

DATA SEGMENT

PORTA EQU 120H

PORTB EQU 121H

PORTC EQU 122H

CWRD EQU 123H

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA

MOV DS,AX

MOV AL,80H ;initialise 8255 ,porta as o/p port

MOV DX,CWRD

OUT DX,AL

MOV DX,PORTA

MOV AL,88H ;load initial bit pattern

OUT DX,AL ;output on porta

UP: CALL DELAY

ROL AL,01H ;rotate left to get exitation sequence of 11,22,44,88

OUT DX,AL

JMP UP

DELAY: MOV CX,0FFFFH ;delay can be adjusted to get different speeds

UP2: MOV BX,0FFH

UP1: DEC BX

JNZ UP1

DEC CX

JNZ UP2

RET

MOV AH,4CH

INT 21H

CODE ENDS

END START

VISVESWARAIAH TECHNOLOGICAL UNIVERSITY,BELGAUM

Branch: ELECTRONICS AND COMMUNICATION Semester: V

Subject Code: ECL57 Duration Of Exam:3hrs

Subject Title: Advanced Microprocessor Lab Max Exam Marks: 50

QUESTION BANK

PART – A

1A) Write an ALP to show the byte and word transfers in different addressing modes.

1B) Write an ALP to transfer a given block of data word from source memory to

Destination memory without overlap.

1C) Write an ALP to transfer a given block of data word from source memory to

destination memory with overlap.

1D) Write an ALP to interchange two blocks of data.

2A) Write an ALP to add / subtract two16 bit numbers.

2B) Write an ALP to add / subtract two32 bit numbers.

2C) Write an ALP to add / subtract two32 bit numbers using DD Directive.

2D) Write an ALP to multiply two 16 bit unsigned / signed numbers and display.

2E) Write an ALP to divide two 8 bit numbers (signed and unsigned)

2F). Write an ALP to divide two 16 bit numbers(signed and unsigned)

3A)Write an ALP to to add/substract/ multiply/divide two ASCII numbers.

3B) Write an ALP to convert 16 bit binary No to BCD.

3C) Write an ALP to convert BCD No to binary.

3D) Write an ALP to find square and cube of an 8 bit number .

3E) Write an ALP to find LCM of a 16 bit No.

3F) Write an ALP to find the GCD of two 16 bit unsigned integers.

3G) Write an ALP to find the factorial of an 8 bit number.

4A) Write an ALP to check if the number is positive or negative.

4B) Write an ALP to check if the given number is even or odd.

4C) Write an ALP to check number of ones and zeroes n the given data.

4D) Write an ALP to check if the given byte is 2 out of 5 code or not (i.e., the code is

first 3 MSB must be 0 and the last 5 LSB should have two 1s).

4E) Write and Alp to check if the given 16 bit data is a palindrome (bitwise).

4F. Write and Alp to check if the given 16 bit data is a palindrome (nibble-wise).

5A) Write an Alp to add / substract ‘N’ 16 bit numbers and display the result.

5B)Write an ALP to find the largest of ‘N’ 16 bit numbers and display the result.

5C) Write an ALP to find the smallest of ‘N’ 16 bit numbers and display the result.

5D) Write an ALP to sort the given set of 16 bit unsigned integers in ascending order

using bubble sort algorithm.

6A) Write an ALP to transfer a given source string to destination using string

instructions.

6B) Write an ALP to reverse a string.

6C) Write an ALP to search for a character in a string.

6D) Write an ALP to check if the given string is a palindrome or not.

7A) Write an ALP to read a character from a keyboard with and without echo.

7B) Write an ALP to read a string of characters from the keyboard and display.

7C) Write an ALP to create a new file.

7D) Write an ALP to read the contents of a file.

7E) Write an ALP to write a new file.

8A) Write an ALP to read the system date.

8B)Write an ALP to set the system date.

8C) Write an ALP to read the system time.

8D) Write an ALP to set the system time.

PART - B

1A) Write an ALP to scan the keyboard for key closure and store the code of the key

pressed in memory location.

1B) Write an ALP to implement a rolling display of set characters using a display

interface.

1C) Interface a logic controller via 8255 using I/O cards and perform the following

Operations: Read all the 8 inputs from the logic controller, Complement

/XOR/AND/OR/NAND and display at the output.

1D) Write an ALP to control the speed of a stepper motor & to drive the stepper motor

interface to rotate the motor in clockwise and anticlockwise directions.

VIVA QUESTIONS IN ADVANCED MICROPROCESSOR

1. List all the modern microprocessor

2. Name some 16 bit Processor (8086, 80286, 80386L, EX)

3. Name some 32 bit processors (80386DX, 80486, PENTIUM OVERDRIVE)

4. Name some 64 bit processor (Pentium, Pentium pro, Pentium II, Xeon, Pentium III, and

Pentium IV)

5. List the address bus width and the memory size of all the processor.

Processor address bus memory size

8086 20 1M

8088 20 1M

80186 20 1M

80286 24 16M

80386 24 16M

80386DX 32 4G

80386EX 26 64M

80486 32 4G

PENTIUM 64 4G

PENTIUM O 32 4G

PENTIUM P 32 4G

PENTIUM 2,3,4 36 64G

6. The memory map of any IBM COMPATIBLE PC consists of three main parts, name them

[transient memory area, system area, Extended memory system]

7. The first I MB of the memory area is called as …………… (Real memory area)

8. What does the TPA hold (interrupt vectors, bios, DOS, IO.SYS, MSDOS, DEVICE

DRIVERS, command.com)

9. The system area contain programs in …………memory(ROM)

10. What are the main two parts of 8086 internal architecture.(BIU,EU)

11. Name the registers in BIU (CS, DS, ES, SS, IP)

12. Name the registers in EU.( AX, BX, CX, DX, SP, BP, SI, DI)

13. Name the flag registers in 8086. (O, D, I, T, S, Z, A, P, C)

14. How is the real memory semented?

15. What is the advantage of segmentation.

16. Name the default segment and offset register combinations.

17. What is the relocatable program.

18. Name the three main addressing modes in 8086.

19. Name the data addressing modes. And the program addressing modes. Give examples

20. Explain MOV AL, ‘A’, MOV AX, NUMBER, MOV [BP], DL, MOV CH,[1000],

MOV[BX+SI],SP, MOV ARRAY[SI],BL, MOV DH,[BX+DI+10H]

21. Name the programme memory addressing modes. (Direct, relative, indirect)

22. What is an intersegment and intrasegment jump.

23. Differentiate near and short jumps (+_32k and +127to_128 bytes)

24. Differentiate near and far jumps.

25. Differentiate push and pop instructions.

26. Explain PUSH word ptr [BX], POP F.

27. JMP TABLE[BX]

28. Explain the following : ASSUME,DB,DD,DW,DQ,END

29. Give the opcode format for 8086 instructions.

(op(1-2b),(mode,reg,rem),(displacement-0-2b))

30. Explain LES BX, LEA AX, DATA, LDS DI,LIST

31. Explain how the string instructions are executed.

32. List some string instructions

33. Explain the significance of REP Prefix.

34. Explain XCHG, LAHF, SAHF, XLAT

35. What are the two types of I/O addressing modes. ( fixed port ,variable port)

36. What do you mean by segment override prefix.

37. Explain the following directives. NEAR ,FAR,BYTE PTR,ORG,OFFSET,ORG

38. Differentiate END, ENDP, ENDM

39.Differntiare PROC AND

40. What are the two basic formats used by assemblers. Where are they used.

(Models, full segment definition)

41. Explain ADD BYTE PTR (.model tiny (64kb), .model small(128 kb), .model huge.

42. Explain ADD BYTE PTR [DI], 3, SBB BYTE PTR [DI],5, CMP[DI], CH

IMUL BYTE PTR [BX], IDIV SI, CWD, CBW.

43. DAA, (ONLY ON AL), AAA, AAD, AAM, AAS.

44. Name the logical instructions. How can we invert number .(XOR WITH 1s)

45. Differentiate TEST and CMP, and NOT& NEG, SAR & SHR, RCL & ROL, SCAS &

CMPS, REPE SCASB &REPNE &SCASB

46. Which are the flags affected. JA(Z=0 C=0), JB(C=0), JG (Z=0 S=0), JLE( Z=1 S<>0)

47. LOOP, LOOPNE, LOOPE LOOPZ

48. Differentiate NEAR & FAR CALL, NEAR RET & FAR RET

49. Explain, maskable, non maskable, vectored, non vectored, software & Hardware Interrupts.

50. What are interrupt vectors. (4 byte no. stored in the first 1024 bytes of memory. There are

256 interrupt vectors. Each vector contains value of CS & IP, 32 vectors are reserved for

present and future. 32 to 255 are available for users.

51. Name the interrupt instructions. ( INT, INT0, INT3)

52. Give significance of INT0, INT3.

53. Give the significance of IRET instruction how is it different from RET.

(Like far RET retrieves 6 bytes from stack, two for IP, two for CS and two for flags.)

54. Explain the operation of real mode interrupt.

55. Explain the protected mode interrupt.

56. Explain how the interrupt flag bit IF and TF are used during an interrupt

57. Name the hardware and soft ware interrupt of 8086, explain about them. (NMI, INTR are

hardware interrupts. INT, INT0, INT3, BOYND, are the software interrupts)

58. How can you expand the interrupt structure. ( using 74LS 244 7 more interrupts can

accommodated. Daisy chained interrupt is better as it requires only one interrupt vector.)

59. Give a general description of 8259 interrupt controller.

61. Explain the above pins of 8086 TEST, READY, RESET, BHE/S7, MN/MX, ALE, DT/R,

DEN, HOLD, HLDA, SO, RO/GT1, LOCK, QS1-QS0.

62. Name the maximum mode pins.

63. Name the minimum mode pins.

64. Name the function of 8284

65 How does the RESET function.

66. What is the clock frequency of the 8086.

67. How are the address and data buses are separated.

68. What is the function of the following 74LS373, 245, 244

69. Differentiate between minimum mode and maximum mode of operation.

70. What are the two methods of interfacing memory. ( linear and absolute decoding)

71. What do you understand by linear and absolute decoding.

72. What is the maximum memory capacity of 8086

73. Name the difference between 8086,8088.

74, Name the difference between 8085 and 8086.

75. Name the types of memory used in microprocessor based system.

76. What is the function of the 8288 controller

77. What are the various signals in a RAM and ROM memories.

78. Name the following. 8255, 8155, 8259, 8253, 8257, 8251

79. Give the format of control word register.

80. Explain the PPI you know.

81. Explain the modes of 8255.

82. Explain the basic function of 8279.

83. How are the delays obtained in a microprocessor based system.

84. What is an arithmetic coprocessor, What are its functions. (multiply, devide, ad, subtract,

square root, calculate partial tangent, partial arctangent and logarithms)

85. What are the data types used. ( 16,32, 64 bit signed integers, 18 bit BCD data, 32, 64 and

80 bit floating point nos.)

86. What are the advantages of the 8087 coprocessor. (many times faster than the

microprocessor)

87. How can we use the DOS function calls.

88. What is the function of int21 interrupts.

89. Explain PUBLIC and EXTERN statements.

90. What do you mean by modular programming, how is it accomplished in 8086.

91. what are libraries.

92. Differentiate between MACRO and PROCEDURE.

93. What are the conditional statements used in a MACRO. (REPEAT, WHILE)

94. What are the different methods of reading the keyboard using DOS function calls.

95. How can we use XLAT instruction for look up tables.

96. What are the two methods of interfacing I/O ( memory mapped I/O and I/O mapped I/O)

Intel® Core™2 Duo

2007-12-07T00:37:00.000-08:00

Making the performance mold, and breaking the performance barrier.

In the world of small, incremental performance increases, Intel has delivered something completely different. The Intel® Core™2 Duo desktop processor has broken the dual-core performance barrier with up to 40 percent faster¹ performance and designed to be more power efficient than previous desktop processors, making it a true revolution in system responsiveness for both single- and multi-threaded applications.

Intel® Centrino® Duo processor technology has just gotten a mobility upgrade with the new Intel Core 2 Duo mobile processor. Its advanced capabilities provide up to 2x more performance when multi-tasking and designed to be more power efficient² so you get the benefit of a powerful dual-core PC plus all the benefits of mobility.³

¹Performance measured Intel® Core™2 Duo desktop processors compared to Intel® Pentium® D Processor 805 on SPECint_base2000 and SPECint_rate_base2000(2 copies.) Actual performance may vary. See http://www.intel.com/performance for more information.

²Performance compared to prior generation Pentium® M Intel processors. Actual performance may vary. See http://www.intel.com/performance for processor and benchmark details.

³System performance, battery life, high-definition quality and functionality, and wireless performance and functionality will vary depending on your specific operating system, hardware and software configurations. References to enhanced performance including wireless as measured by SYSmark* 2004 SE, PCMark*05, 3DMark*05, 3DMark*06, SPEC* CPU2000* and Adjacent Channel Interference (ACI)* refer to comparisons with previous generation Intel® Centrino® processor technology platforms. References to improved battery life as measured by MobileMark* 2005, if applicable, refer to previous generation Intel Centrino processor technology platforms. Wireless connectivity and some features may require you to purchase additional software, services or external hardware. Availability of public wireless LAN access points is limited, wireless functionality may vary by country and some hotspots may not support Linux-based Intel Centrino processor technology systems. See www.intel.com/products/centrino/ for more information. SPEC, SPECint, SPECfp, SPECrate, SPECweb, SPECjbb are trademarks of the Standard Performance Evaluation Corporation. See: http://www.spec.org for more information on the benchmarks.

Features

Delivering the best overall performance. Period.

With Intel Core 2 Duo processors powering your desktop and laptop PCs you'll get the latest arsenal of performance-rich technologies, including up to 4MB of shared L2 cache, up to 1333 MHz Front Side Bus for desktop, and up to 800 MHz Front Side Bus for laptop, you've got the future of computing now, and only from Intel:

Intel® Wide Dynamic Execution, enabling delivery of more instructions per clock cycle to improve execution time and energy efficiency
Intel® Intelligent Power Capability, designed to deliver more energy-efficient performance
Intel® Smart Memory Access, improving system performance by optimizing the use of the available data bandwidth
Intel® Advanced Smart Cache, providing a higher-performance, more efficient cache subsystem. Optimized for multi-core and dual-core processors
Intel® Advanced Digital Media Boost, accelerating a broad range of applications, including video, speech and image, photo processing, encryption, financial, engineering and scientific applications

Big, big performance. More energy efficient.¹ Now available in smaller packages.

The Intel Core 2 Duo processor-based desktop PC was designed from the ground up for energy efficiency, letting you enjoy higher performing, ultra-quiet, sleek, and low-power desktop PC designs.

Multitask with reckless abandon.

Do more at the same time, like playing your favorite music, running virus scan in the background, and all while you edit video or pictures. The powerful Intel Core 2 Duo desktop processor provides you with the speed you need to perform any and all tasks imaginable.

Not all features and benefits are included with each system based on this Intel product. Please check system specifications with manufacturer or retailer before making your purchase decision.

¹Performance based on SPECint*_rate_base2000 (2 copies) and energy efficiency based on Thermal Design Power (TDP), comparing Intel® Core™2 Duo E6700 to Intel® Pentium® D Processor 960. Actual performance may vary. See www.intel.com/performance for more information.

For more legal information please go to www.intel.com/sites/corporate/tradmarx.htm

ABOUT SUN CLOCK

2007-10-23T21:49:00.000-07:00

Not until somewhat recently (that is, in terms of human history) did people find a need for knowing the time of day. As best we know, 5000 to 6000 years ago great civilizations in the Middle East and North Africa initiated clock making as opposed to calendar making. With their attendant bureaucracies and formal religions, these cultures found a need to organize their time more efficiently.

Sun Clocks

After the Sumerian culture was lost without passing on its knowledge, the Egyptians were the next to formally divide their day into parts something like our hours. Obelisks (slender, tapering, four-sided monuments) were built as early as 3500 B.C. Their moving shadows formed a kind of sundial, enabling citizens to partition the day into two parts by indicating noon. They also showed the year's longest and shortest days when the shadow at noon was the shortest or longest of the year. Later, markers added around the base of the monument would indicate further time subdivisions.

Another Egyptian shadow clock or sundial, possibly the first portable timepiece, came into use around 1500 B.C. to measure the passage of "hours." This device divided a sunlit day into 10 parts plus two "twilight hours" in the morning and evening. When the long stem with 5 variably spaced marks was oriented east and west in the morning, an elevated crossbar on the east end cast a moving shadow over the marks. At noon, the device was turned in the opposite direction to measure the afternoon "hours."

The merkhet, the oldest known astronomical tool, was an Egyptian development of around 600 B.C. Two merkhets were used to establish a north-south line by lining them up with the Pole Star. They could then be used to mark off nighttime hours by determining when certain other stars crossed the meridian.

In the quest for more year-round accuracy, sundials evolved from flat horizontal or vertical plates to forms that were more elaborate. One version was the hemispherical dial, a bowl-shaped depression cut into a block of stone, carrying a central vertical gnomon (pointer) and scribed with sets of hour lines for different seasons. The hemicycle, said to have been invented about 300 B.C., removed the useless half of the hemisphere to give an appearance of a half-bowl cut into the edge of a squared block. By 30 B.C., Vitruvius could describe 13 different sundial styles in use in Greece, Asia Minor, and Italy.

Elements of a Clock

Having described a variety of ways devised over the past few millennia to mark the passage of time, it is instructive to define in broad terms what constitutes a clock. All clocks must have two basic components

regular, constant or repetitive process or action to mark off equal increments of time. Early examples of such processes included movement of the sun across the sky, candles marked in increments, oil lamps with marked reservoirs, sand glasses ("hourglasses"), and in the Orient, small stone or metal mazes filled with incense that would burn at a certain pace.
A means of keeping track of the increments of time and displaying the result. Our means of keeping track of time passage include the position of clock hands and a digital time display.

The history of timekeeping is the story of the search for ever more consistent actions or processes to regulate the rate of a clock.

Water Clocks

Water clocks were among the earliest timekeepers that did not depend on the observation of celestial bodies. One of the oldest was found in the tomb of Amenhotep I, buried around 1500 B.C. Later named clepsydras ("water thief") by the Greeks, who began using them about 325 B.C., these were stone vessels with sloping sides that allowed water to drip at

a nearly constant rate from a small hole near the bottom. Other clepsydras were cylindrical or bowl-shaped containers designed to slowly fill with water coming in at a constant rate. Markings on the inside surfaces measured the passage of "hours" as the water level reached them. These clocks were used to determine hours at night, but may have be

en used in daylight as well. Another version consisted of a metal bowl with a hole in the bottom; when placed in a container of water the bowl would fill and sink in a certain time. These were still in us

e in North Africa this century.

More elaborate and impressive mechanized water clocks were developed between 100 B.C. and 500 A.D. by Greek and Roman horologists and astronomers. The added complexity was aimed at making the

flow more constant by regulating the pressure, and at providing fancier displays of the passage

of time. Some water clocks rang bells and gongs; others opened doors and windows to show little figures of people, or moved pointers, dials, and astrological models of the universe.

A Greek astronomer, Andronikos, supervised the construction of the Tower of the Winds in Athens in the 1^st century B.C. This octagonal structure showed scholars and marketplace shoppers both sundials and mechanical hour indicators. It featured a 24-hour mechanized clepsydra and

indicators for the eight winds from which the tower got its name, and it displayed the seasons of the year and astrological dates and periods. The Romans also developed mechanized clepsydras, though their complexity accomplished little improvement over simpler methods for determining the passage of time.

In the Far East, mechanized astronomical/astrological clock making developed from 200 to 1300 A.D. Third-century Chinese clepsydras drove various mechanisms that illustrated astronomical phenomena. One of the most elaborate clock towers was built by Su Sung and his associates in 1088 A.D. Su Sung's mechanism incorporated a water-driven escapement invented about 725 A.D. The Su Sung clock tower, over 30 feet tall, possessed a bronze power-driven armillary sphere for observations, an automatically rotating celestial globe, and five front panels with doors that permitted the viewing of changing manikins which rang bells or gongs, and held tablets indicating the hour or other special times of the day.

Since the rate of flow of water is very difficult to control accurately, a clock based on that flow could never achieve excellent accuracy. People were naturally led to other approaches.

souce

Windows 7

2007-09-22T06:12:00.001-07:00

Typically when Microsoft ships a new OS (like Windows Vista), we immediately start talking about the next version-which begs two questions: 1) is Microsoft working on a new version of Windows, and if so, 2) why aren't you talking about it?
I thought I would spend a minute giving you an update on where we are. First, yes, we are working on a new version of Windows. As you likely know, it's called Windows 7.We are always looking for new ways to deliver great experiences for our customers. This is especially true of Windows - where we're constantly examining trends in hardware, software and services to ensure that we continue to drive the innovation that has both made Windows the world's most popular operating system and has provided a foundation on which our partners built great products and businesses. When we shipped Windows 2000, we were already working on Windows XP and we started working on Windows Vista even before we released Windows XP. So naturally, we've been thinking about the investments we made in Windows Vista and how we can build on these for the next version of Windows.
What is a little different today is when and how we are talking about the next version of Windows. So, why the change in approach? We know that when we talk about our plans for the next release of Windows, people take action. As a result, we can significantly impact our partners and our customers if we broadly share information that later changes. With Windows 7, we're trying to more carefully plan how we share information with our customers and partners. This means sharing the right level of information at the right time depending on the needs of the audience. For instance, several months ago we began privately sharing our preliminary plans for Windows 7 with software and hardware partners who build on the Windows platform. This gave them an opportunity to give us feedback and gave us the opportunity to incorporate their input into our plans. As the product becomes more complete, we will have the opportunity to share our plans more broadly. Steven Sinofsky, Windows and Windows Live Engineering SVP, talks more about this in his interview with CNET's Ina Fried, published today: http://news.cnet.com/8301-13860_3-9951638-56.html.
We know that this is a change in our approach, but we are confident that it will help us not only to build even better products, but also to be more predictable in the delivery of our products. We also know that this change has led to some confusion, so we would like to share information today that will hopefully clear up some of this.
Before we talk about what's ahead, we should take a look at where we are today with Windows Vista. From a quality perspective, both Windows Vista SP1 and the ecosystem have delivered measurable progress in the six dimensions of quality we track -- device compatibility, application compatibility, reliability, performance, battery life and security. The business results speak for themselves. As of March 31, we had sold more than 140 million Windows Vista licenses, and analyst firm forecasts indicate that Windows Vista adoption among businesses is on a similar pace as Windows XP in similar timeframes. Millions of enterprise users are already running Windows Vista, and we invite you to read their stories published in more than 100 case studies. The benefits they are experiencing range from energy conservation, lower TCO for mobile users, and greater security. Our job is not done, but we've worked hard with our ecosystem to improve the quality of Windows Vista and we're pleased with the customer response.
Another question we often get asked is whether Windows 7 is a major release. The answer is "yes" -- it's hard to describe any product that is used by millions of people and worked on by thousands of engineers as anything else. That said, the long-term architectural investments we introduced in Windows Vista and then refined for Windows Vista SP1 and Windows Server 2008 will carry forward in Windows 7. Windows Vista established a very solid foundation, particularly on subsystems such as graphics, audio, and storage. Windows Server 2008 was built on that foundation and Windows 7 will be as well. Contrary to some speculation, Microsoft is not creating a new kernel for Windows 7. Rather, we are refining the kernel architecture and componentization model introduced in Windows Vista. While these changes will increase our engineering agility, they will not impact the user experience or reduce application or hardware compatibility. In fact, one of our design goals for Windows 7 is that it will run on the recommended hardware we specified for Windows Vista and that the applications and devices that work with Windows Vista will be compatible with Windows 7.
We are well into the development process of Windows 7, and we're happy to report that we're still on track to ship approximately three years after the general availability of Windows Vista. As always, we will be releasing early builds of Windows 7 prior to its general availability as a means to gain feedback, but we're not yet ready to discuss timing and specific plans for any Beta releases. In the meantime, customers can confidently continue with their Windows Vista deployment plans.

sourse

about me

2007-09-22T05:57:00.000-07:00

I'm ranjeeth

This blogg is informe about
Top 10 billiners
C-language and
Computer tips

phobia

LINUX COMMANDS

Linux command line

Introduction

A New Concept Online Advertising

A New Concept in Online Advertising

Pay Per Click

MICROPROCESSOR 8085

Intel® Core™2 Duo

Delivering the best overall performance. Period.

Big, big performance. More energy efficient.1 Now available in smaller packages.

Multitask with reckless abandon.

ABOUT SUN CLOCK

Elements of a Clock

Water Clocks

Windows 7

about me

Big, big performance. More energy efficient.¹ Now available in smaller packages.