GNIT Students E- Learning Point

C program for pyramid

2019-10-07T19:44:00.001+05:30

Logic to print pyramid star pattern

    *
   ***
  *****
 *******
*********

Before you read further have a close look at the above pattern. The pattern consists of N (for this case 5) rows. Each row contain exactly 2 * N - 1 stars. In addition to stars the pattern has leading spaces. Each row contain N - i spaces (where i is current row number). To count total spaces per row point your mouse over the above pattern.

Step by step descriptive logic to print Pyramid star pattern.

Input number of rows to print from user. Store it in a variable say rows.
To iterate through rows, run an outer loop from 1 to rows. The loop structure should look like for(i=1; i<=rows; i++).
To print spaces, run an inner loop from i to rows - 1. The loop structure should look like for(j=i; j<rows; j++). Inside this loop print single space.
Note: Iterating from 1 to N - i or i to rows - 1 both are equal.
To print star, run another inner loop from 1 to 2 * i - 1. The loop structure should look like for(j=1; j<=(2*i - 1); j++). Inside this loop print star.
After printing stars for current row, move to next line i.e. print new line.

Program to print pyramid star pattern series

/** * C program to print equilateral triangle or pyramid star pattern */ #include <stdio.h> int main(){ int i, j, rows; /* Input number of rows to print */ printf("Enter number of rows : "); scanf("%d", &rows); /* Iterate through rows */ for(i=1; i<=rows; i++) { /* Print leading spaces */ for(j=i; j<rows; j++) { printf(" "); } /* Print star */ for(j=1; j<=(2*i-1); j++) { printf("*"); } /* Move to next line */ printf("\n"); } return 0;}

Knowledge bank

2019-09-24T16:02:00.003+05:30

CHARGE YOUR SMARTPHONE BY MAKING THIS COOL WIND TURBINE USING A COMPUTER FAN

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2019-09-24T15:51:00.002+05:30

FOR DIRECTLY DOWNLOADING THE TEXT BOOKS :-CLICK HERE

NOTES:

2-1 SEM NOTES 3-1 SEM NOTES

ALL SUBJECTS NOTES ALL SUBJECTS NOTES
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***In case of failure while you are opening the material through drive link, kindly please open them using the zip Extractor since those are RAR files. You can download the file directly from it.**

***FOR EXTRA INFORMATION REGARDING THE SUBJECTS YOU CAN VISIT ELSDM***

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Guide to Computer Forensics and Investigations Fourth Edition - ppt video online download

2018-05-19T11:44:00.001+05:30

Guide to Computer Forensics and Investigations Fourth Edition - ppt video online download

Guide to Computer Forensics and Investigations, Second Edition Chapter 12 Network Forensics. - ppt download

2018-05-19T11:38:00.001+05:30

Guide to Computer Forensics and Investigations, Second Edition Chapter 12 Network Forensics. - ppt download

Guide to Computer Forensics and Investigations Fourth Edition - ppt video online download

2018-05-19T11:29:00.001+05:30

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Guide to Computer Forensics and Investigations Third Edition Chapter 7 Current Computer Forensics Tools. - ppt download

2018-05-19T11:25:00.001+05:30

Guide to Computer Forensics and Investigations Third Edition Chapter 7 Current Computer Forensics Tools. - ppt download

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2018-05-19T11:22:00.001+05:30

Guide to Computer Forensics and Investigations, Second Edition - ppt video online download

Chapter 5 Processing Crime and Incident Scenes Guide to Computer Forensics and Investigations Fourth Edition. - ppt download

2018-05-19T11:20:00.001+05:30

Chapter 5 Processing Crime and Incident Scenes Guide to Computer Forensics and Investigations Fourth Edition. - ppt download

Ransomeware

2018-05-16T13:35:00.000+05:30

BY JUAN MARTINEZ

JUNE 28, 2017, 1:59 A.M.

Read this to learn more about the WannaCry ransomware crisis and what you need to stay safe from future attacks

Earlier this month, a strain of ransomware infected more than 300,000 Windows PCs around the world. The awesomely namedWannaCry strain demanded that infected businesses and individuals pay $300 in order to unlock each machine—as well as the data stored on their devices. Some people paid the ransom, while others were lucky enough to wait it out and be rescued by a hero who accidentally stopped the attack by registering the unregistered domain on which the ransomware lived.

Now that the attack has been thwarted, it appears that new WannaCry variants are emerging, and a massive, unrelated ransomware attack hit Eastern Europe. As ransomware attacks become trickier and more difficult to stop, your company is more likely than ever to be at risk. As a result, we've compiled this list of post-mortem steps on what happened, how you can protect your business and yourself, and what you should do if you fall victim to an attack.

1. Be Defensive

You're going to need to be smarter about which emails you open, which links you click, and which files you download. Phishing attacks are common and they're easy to fall victim to. Unfortunately, WannaCry wasn't your typical phishing attack. Instead, this attack manipulated a Windows vulnerability, one that had already been patched by Microsoft earlier this year.

So, how did it get through? You know those annoying pop-up notifications that software manufacturers send to your computer? They're not just alerting you to new features; they're adding patches to your software that will help to protect against attacks like WannaCry. The same goes for your endpoint protectionsoftware. If your vendor asks you to update, then update. In this case, it appears the attackers were able to penetrate systems that had not recently been updated and, as a result, hospitals were crippled and lives were put in jeopardy (more on this later).

"The global fallout of this attack could have easily been prevented by deploying the security update once it was made available by Microsoft," said Liviu Arsene, Senior E-Threat Analyst at Bitdefender. "The lesson to be learned from this experience is to always apply security patches and updates when they become available, not just for operating systems but for applications as well. Of course, a security solution might prevent the payload—in this case, ransomware—from infecting victims. But more advanced and sophisticated threats could potentially leverage the operating system vulnerability to gain persistency and bypass traditional security mechanisms undetected."

2. Back that Cache Up

The worst thing about an attack of this variety is that it gains access to your data. However, the responsible among us don't need to worry about this very much because they have been using disaster recovery (DR) software to ensure that their information is alive and well in the cloud. If you wind up getting hit with a ransomware attack, then having access to your full trove of data in the cloud means you can simply factory-reset your machine, pull in your backed up data, and start working again.

Image Via: McAfee

3. Don't Pay, Silly

As much as you'd like to retrieve your unfinished screenplay, paying hostage takers seldom works. Instead, contact the FBI and let them know you've become the victim of a cyberattack. If you desperately need your data and you don't have a backup stored elsewhere, then just sit tight and wait. Also, if you don't need your data or if you have backed it up, then just reset your machine and start from scratch.

Whatever you do, don't pay. Here's why: There's a good possibility the hacker won't actually release your data. Now you're out $300 and you're still out of luck. Also, paying could actually expose you to additional risk because you've shown a willingness to give into the hackers' demands. So, in the very best-case scenario, you've paid, gotten your data back, and given a criminal incentive to try to attack you again in the future.

"No one is ever encouraged to give in to ransomware demands," said Arsene. "In fact, if no backups are available from which to restore lost data, companies or individuals should treat the incident as hardware failure and move on. Paying would only fuel cybercriminals with the financial resources to keep developing new threats. And there's no actual guarantee that you'll actually receive the decryption key. You are actually dealing with criminals here."

4. What You Should Do

As I previously mentioned, backing up your data and running a factory-reset on your hardware, will let you walk away from a ransomware attack without having experienced much real damage. Here's a step-by-step procedure for what to do when that ransom note hits your screen: 1) Unplug your computer and unplug your computer from its network. 2) Fully wipe your device and restore it from a backup. 3) Install all security patches and updates and add a security solutionlike Bitdefender to your software mix. 4) Contact the FBI.

5. Businesses Must Get Serious

"There are security layers that companies can deploy to protect infrastructures from zero-day vulnerabilities in both operating systems and applications," said Arsene. Arsene recommends organizations running virtual infrastructures deploy a hypervisor-based memory introspection technology that's capable of securing virtual workloads.

"This new security layer that sits below the operating system can detect zero-day vulnerabilities, like the SMB v1 vulnerability leveraged by WannaCry, and prevent attackers from ever exploiting it, even if the system is unpatched or the vulnerability is completely unknown," Arsene explained. "This complementary security layer, coupled with traditional in-guest security solutions and constant software patching, increases the cost of attack for cybercriminals while giving organizations more visibility into advanced attacks."

Computer Organization Lab

2018-03-13T13:57:00.000+05:30

Dear Students attach the outputs in the lab record.

Click to Download

MASM

2018-02-20T15:34:00.001+05:30

MASM 8086 Assembler in Ubuntu or Windows 7(x64) using DOSBox

Run MASM 8086 Assembler in Ubuntu or Windows 7(x64) using DOSBox

Here’s how to run 16 bit DOS executables like the MASM assembler or Turbo C compiler in Ubuntu (GNU/Linux) or in 64 bit editions of Windows 7 using DOSBox, a DOS environment emulator. DOSBox is available for Linux as well as Windows.

DOSBox installation

For Ubuntu users (using repository)
Open the terminal and type in the following commands to download and install DOSbox in Ubuntu
sudo apt-get update
sudo apt-get install dosbox
You will find it installed under Applications_Menu->Games->DOSBox Emulator

For other GNU/Linux users
Download DOSbox from below link.
Open terminal and cd to the directory containing the downloaded tar.gz file. Type in the following commands to build and install :
tar -xzvf dosbox-0.74.tar.gz
cd dosbox-0.74
./configure
make
Check the src subdir for the binary.

For Windows users
Download DOSbox from below.
Run the downloaded .exe file and install it like any other software.

Now that you’ve installed DOSBox, you’ll be able to run any 16bit or 32bit DOS executable inside it.

Download the 8086 MASM Assembler from below. The zip file contains the following files :
masm.exe, tasm.exe, link.exe, bin2hex.exe, exe2bin.exe, td.exe, edit.com and debug.exe

Windows users extract the .zip file into C:\ so that the path C:\8086 contains all the above mentioned files. GNU/Linux users can extract it and place it in say /home/imran/8086

Launch DOSBox and type the following commands :
For Linux users :
mount c /home/imran/8086
c:
For Windows :
mount c c:\8086
c:

DOSBox running in Ubuntu

Now the contents of the folder /home/imran/8086 or c:\8086 is mounted as c: drive inside the DOS emulator. You can assemble programs inside DOSBox as you do in your Microprocessor Lab under Windows XP; i.e your usual sequence of commands -
edit file.asm
masm file.asm
link file
debug file.exe

When you are done, type exit to quit DOSBox.

P.S : For GNU/Linux users, there’s an alternative assembler known as the NASM. NASM is considered to be one of the most popular assemblers for GNU/Linux.

Click to Downloads :
DOSBox for Linux : dosbox-0.74.tar.gz
DOSBox for Windows : DOSBox0.74-win32-installer.exe
8086 Assembler : 8086_Assembler.zip

MASM Programs

2018-02-17T11:03:00.001+05:30

8086 ASSEMBLY PROGRAM FOR ADDITION OF TWO 8 BIT NUMBERS

data segment

a db 09h

b db 02h

c dw ?

data ends

code segment

assume cs:code,ds:data

start:

mov ax,data

mov ds,ax

mov al,a

mov bl,b

add al,bl

mov c,ax

int 3

code ends

end start

C:\TASM>masm an8add.asm

Microsoft (R) Macro Assembler Version 5.00

Object filename [an8add.OBJ]:

Source listing [NUL.LST]:

Cross-reference [NUL.CRF]:

50402 + 450254 Bytes symbol space free

0 Warning Errors

0 Severe Errors

C:\TASM>link an8add.obj

Microsoft (R) Overlay Linker Version 3.60

Run File [AN8ADD.EXE]:

List File [NUL.MAP]:

Libraries [.LIB]:

LINK : warning L4021: no stack segment

C:\TASM>debug an8add.exe

-g

AX=0B0B BX=0002 CX=0022 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000

DS=0B97 ES=0B87 SS=0B97 CS=0B98 IP=0011 NV UP EI PL NZ NA PO NC

0B98:0011 CC INT 3

-d 0B97:0000

0B97:0000 09 02 0B 0B 00 00 00 00-00 00 00 00 00 00 00 00 ................

0B97:0010 B8 97 0B 8E D8 A0 00 00-8A 1E 01 00 02 C3 A3 02 ................

0B97:0020 00 CC 86 72 FF 77 15 8A-86 70 FF 2A E4 50 B8 FD ...r.w...p.*.P..

0B97:0030 05 50 FF 36 24 21 E8 77-63 83 C4 06 FF 36 24 21 .P.6$!.wc....6$!

0B97:0040 B8 0A 00 50 E8 47 5E 83-C4 04 5E 8B E5 5D C3 90 ...P.G^...^..]..

0B97:0050 55 8B EC 81 EC 84 00 C4-5E 04 26 80 7F 0A 00 74 U.......^.&....t

0B97:0060 3E 8B 46 08 8B 56 0A 89-46 FC 89 56 FE C4 5E FC >.F..V..F..V..^.

0B97:0070 26 8A 47 0C 2A E4 40 50-8B C3 05 0C 00 52 50 E8 &.G.*.@P.....RP.

-q

C:\TASM>

2018-02-16T10:53:00.002+05:30

CENG 222 – Computer Organization Lab Work 9

Standard I/O In 8086

Standard input output operations are performed INT 21h interrupt in assembly.

Int 21h acts like a function. The operation to perform in interrupt no 21h is determined by the contents of the AH register.

Read the entry regarding Int 21h interrupt in “Supported Interrupt Functions” section of Emu8086 help (“help”->”Documentation and Tutorials”->“Supported Interrupt Functions”)

Pay attention to Int 21h functions where the value of AH is 1,2,9,0Ah

Exercises

1) Write an assembly program to read a character and print the same character back to screen

2) The user will enter a 2 digit decimal number write an assembly program to store the number in CX register.

3) Write an assembly program to print the English alphabet (capital letters only)

4) design an assembly function (that uses a variable for return value) that reads 5 characters from the keyboard. Call the function in main to get the characters and write the program to print the characters.

2018-02-16T10:52:00.004+05:30

CENG 222 – Computer Organization Lab Work 8

Variables

Syntax for a variable declaration:

name DB value

name DW value

DB - stays for Define Byte.

DW - stays for Define Word.

name - can be any letter or digit combination, though it should start with a letter. It's possible to declare unnamed variables by not specifying the name (this variable will have an address but no name).

value - can be any numeric value in any supported numbering system (hexadecimal, binary, or decimal), or "?" symbol for variables that are not initialized.

Reference Address:

http://www.yecd.com/os/8086%20assembler%20tutorial%20for%20beginners%20(part%203).htm

.model small

.stack 100h

.data

VAR1 DB 5

var2 DW 1A2Bh

.code

main proc

mov ax,@data

mov ds,ax

MOV AL, var1

MOV BX, var2

mov ax, 4c00h

int 21h

main endp

end main

Some previous Examples to remember...

MOV AX, [1BFFH] ; Copy 2-bytes (word) data stored at address 1BFF to AX

MOV [BX], 20h ; Copy the word (2-bytes) data to the memory

; block started at BX. (*BX = 0020h)

MOV WORD PTR [BX], 20h ; Exactly the same as above

MOV BYTE PTR [BX], 20h ; Copy 1 byte (*BX = 20h)

Offset command

When used with a variable offset command gives the address of the variable.

MOV BX, offset var1 ; Copy address of var1 to BX

MOV DL, [offset var2] ; Copy 1-byte data at var2 to DL

LEA (Load effective address) instruction

If we use “offset”, we need the following instruction to get the address of a data:

MOV DX, offset var1

Here, msg is defined as name of the data itself. Another way to do this without using offset:

LEA DX, var1

Both instructions above, loads the address of var1 into DX.

Experiment 1

Analyze the assembly code below:

title Register test program

.model small

.stack 100h

.data

msg db "AB",0dh,0ah,'$'

.code

main proc

mov ax,@data

mov ds,ax

; Add your code here

; ------------------

mov ah,9

mov dx,offset msg

int 21h

mov ax, 4c00h

int 21h

main endp

end main

The program above prints “AB” to the console. The message “AB” is stored at msg in data segment. Therefore, “offset msg” is the starting address of the string. Modify the code according to the following experiments:

1- Copy the second byte stored at “msg” (which is ‘B’ currently) to the first byte. The output should be “BB”. In order to do that, load 2-bytes data stored at “msg”into a 16-bits register. Copy one part of the register to the other part. Then, write your register back to msg.

2- Copy the first byte stored at “msg” (which is ‘A’ currently) to the second byte. The output should be “AA”. To achieve this, load the first byte stored at “msg”into one 8-bits register (high or low part) and copy it onto the other part and write the data stored in register back to msg.

3- Load the data stored at “msg” into one register and swap the high and low parts of the register using a temporary 8-bits register. Then write back your “swapped” register onto msg. The output should be “BA”.

Experiment 2

title Logical

.model small

.stack 100h

.data

varA db ?

varB db ?

varC db ?

varX db ?

.code

main proc

mov ax,@data

mov ds,ax

; Add your code below

main endp

end main

In the assembly code above, four variables are defined without initialized. Use debug mode to test your program and see registers’ and memory’s status after you make necessary modifications to compute some logical operations:

1 – Assign values to all variables except varX.

varA = 10101010b

varB = 11001100b

varC = 11110000b

2 – Compute the following logical operations:

a) varX = varA OR (varB AND varC)

b) varX = varA OR varB OR (NOT varC)

Experiment 3

Define variables facparam and facreturn. Write an assembly function that calculates the factorial of the number stored in facparam variable and write the result to facreturn variable. The registers should remain unchanged at the end of the function call.

Use the function to calculate 3! and 5!.

2018-02-16T10:52:00.002+05:30

CENG 222 – Computer Organization

Lab Work 7

CALL and RET instructions

CALL instruction is used for calling a subroutine/procedure/function.

Usage:

CALL <address or procedure name>

RET instruction is used for returning back to caller procedure from the called procedure. RET has no arguments. Analyze the code given in experiment.

Using stack

Stack is used for temporarily storing data to memory. It is also used by CALL and RET to store caller instruction address. Hence, it is critically important to use stack properly:

Leave stack as how you find!

It means that you need to get all the data out which you store in the stack. Do not forget any data in the stack. You can use it temporarily!

PUSH instruction stores a data in the stack.

PUSH <word data or word register>

POP instruction retrieves a data from the stack.

POP <word memory block or word register>

Examples:

PUSH AX

POP BX

EXPERIMENTS

Write the following C code in assembly using no jump instructions (use loop command instead) You should not use other registers to store the value of CX temporarily. You should use the stack properly to store the value of CX properly.

Hint: Don't forget that loop command only operates on CX and to use a inner loop you must remember the value of CX for the outer loop at each iteration.

int b =0;

for(int i=0; i<5;i++)
for(int j=0; j<10;j++)

b = I + 2*j;

Write an assembly function that calculates factorial. Use the function to calculate 3! and 5!.

Write an assembly program that calculates first 10 Fibonacci numbers. Numbers should be observed in dx consequently (since we did not learn how to print at the moment we can simulate it by observing the register in emulator).

Hint: Use the stack to reverse the order. Once you have pushed all the items, pop command will give you the last item.

2018-02-16T10:52:00.000+05:30

CENG 222 – Computer Organization - Lab Work 6

Exercises

Re-write the following C codes in assembly.

While loop

int a = 125;
int b = 0;
while(a>0){
  b += a;
  a -=2;
}

While loop with a condition

int a = 125;
int b = 0;
while(a>0){
  b += a;
  if(a>10)
    a -= 2;
  else
    a--;
}

For loop
```
b=0;
for(i=0;i<10;i++)
  b = b+ i*2;
```

Double for loop

b=0;
for(i=0;i<5;i++)
  for(j=0;j<5;j++)
    b = b+ j*2;

Do-while loop with a condition

int a = 125;
int b = 0;
do{
  b += a;
  if(a>10)
     a -= 2;
  else
    a--;
}while(a>0)

Collatz conjecture

int a = 21;
while(a != 1){
  if(a%2 == 1)
    a = 3*a+1;
  else
    a = a/2;
}

2018-02-16T10:51:00.004+05:30

CENG 222 – Computer Organization

Lab Work 5

Jump and Labels

JMP instruction allows to “jump” into an address of a instruction in our program. You can make loops by using JMP. Usage of JMP:

JMP <address or label name>

In assembly programs, we generally use labels to mark a starting point of a code. They end with a colon sign (“:”). Example:

L1:

…

MOV …

SUB …

…

JMP L1 ; Jump into the firstinstruction after the label “L1”.

Testing conditions and Conditional Jump

In order to test a condition, you can use SUB or AND instructions. For example you can test A = B, A > B, A >= B, A < B and A <=B conditions using SUB. You can also test the status of any bit of a register by using AND. These instructions will modify the flag bits according to the results they produce. Therefore, you can make a conditional jump using flag bits’ status. However, SUB and AND instructions also store the result into the destination register. If you just want to test the condition without modifying operands, you need to use CMP and TEST instructions. These instructions are used to modify just the flag bits.

CMP is the same as SUB but it does not modify the destination. TEST is the same as AND but it does not modify the destination. However they both modify the flag bits.

Analyze the code below:

…

MOV CX, 0

L1:

CMP CX, 10

JE L2 ; Go to L2 if CX = 10

INC CX

JMP L1 ; Go to L1 (CMP instruction)

L2:

…

Some of conditional jump instructions are:

JE, JNE, JZ, JNZ, JG, JNG, JL, JNL, JLE, JGE, JNLE, JNGE

JE: Jump if equal to

JNE: Jump if not equal to

JZ: Jump if zero

JNZ: Jump if not zero

JG: Jump if greater than

JNG: Jump if not greater than

JL: Jump if less than

JNL: Jump if not less than

JLE: Jump if less than or equal to

JGE: Jump if greater than or equal to

JNLE: Jump if not less than nor equal to (same as JG)

JNGE: Jump if not greater than nor equal to (same as JL)

Please note that conditional jump instructions are useless alone, but if you use an instruction that modifies flag bits, before the jump instruction, you can make IF…THEN…ELSE structures. For example:

CMP var1, var2 ; Compare var1 and var2

JG L1 ; Jump to L1 if var1>var2

… ; “IF” block

… ; These instructions will be executed if var1<=var2

JMP L2 ; End of “IF” block, we need to jump L2

L1:

… ; “ELSE” block

… ; These instructions will be executed if var1>var2

L2: ; End of “IF…ELSE” block

LOOP instruction

LOOP instruction allows creating a simple loop using CX register as a counter.

Usage:

LOOP <address of label>

You need to initialize CX first for the number of repentance of the loop. Analyze the code below:

MOV CX, 10 ; Loop 10 times

L1:

…

… ; Instructions in the loop

…

LOOP L1

LOOP instruction simply decrements CX and tests if it is zero. If not zero, then jump. Otherwise, continue.

EXPERIMENTS

1. Find the sum [1+2+3+4+5+...+10] by using both compare-jump combination and loop instruction.

2. Find the sum [1+3+5+..+17] by using both compare-jump combination and loop instruction.

3. Find the sum of numbers between 0 and 20 that are not divisible by 4

4. Write a program to find the factorial of the number given in AL register. (Assume that the number is sufficiently small so that the factorial wont be larger than 32 bits)

2018-02-16T10:51:00.002+05:30

CENG 222 – Computer Organization

Lab Work 4

Logical Operations

Some of bitwise logical operations are:

AND, OR, NOT, XOR

Bitwise logical instructions apply the operation between each corresponding bits.

Examples:

AND data, ax ; data = data AND ax

OR dx, ax ; dx = dx OR ax

NOT cl ; cl = NOT cl

XOR ax, ax ; ax = ax XOR ax

Shift Operations

Shift operations basically shift the bits of the given target (given as first parameter) by a number (given by second parameter)

Some Shift Operations are:

SHL,SHR (logical/unsigned),SAL, SAR(arithmetic/signed)

See 8086 help for detailed explanation of instructions.

Rotate Operations

Shift operations work like shift operations, they the bits of the given target (given as first parameter) by a number (given by second parameter), the shifted bits are attached to the other side of the target.

Some Rotate Operations Are:

ROL, ROR (without carry), RCL, RCR (with carry)

See 8086 help for detailed explanation of instructions.

EXPERIMENT 1

Given bit values

A = 10101010b

B = 11001100b

C = 11110000b

Compute the following logical operations:

A AND B
A OR (B AND C)
A OR B OR (NOT C)
(A AND B) OR (A AND (NOT C))

EXPERIMENT 2

Perform the following operations without using a MUL/ IMUL or DIV/IDIV command.

Hint: When you apply a single shift operation for a number you can multiply it or divide it by 2. Shift left for multiplying, shift right for dividing.

0FAAh *2
0FAAh / 2
FFAAh / 2
FFAAh * 2
00AAh * 5 ( Hint: 5*x = 4*x + x)
00FBh * 2 + 000Bh * 4
00AAh * 18

EXPERIMENT 3

Consider using logical and shift operations for the following assignments.

Hint:

AND AL, 11111101b; command will change the second bit of AL to 0, without changing the other bits

similarly

OR AL, 00000010b will change the second bit of AL to 1, without changing the other bits

Assign AX any value. Get the 4^th bit of AX to the first bit of BX. All the other bits of BX should be zero
Get the 3^rd bit of AL to 5^th bit of AH. Other bits of AH should be unchanged.
Get the 2^nd and 3^rd bits of AH to 1^st and 2^nd bits of AL

2018-02-16T10:51:00.000+05:30

CENG 222 – Computer Organization

Lab Work 3

Special Register: FLAGS

The flags register is unlike the other registers on the microprocessor. The other registers hold eight or 16 bit values. The flags register is simply a collection of one bit values which help determine the current state of the processor. Although the flags register is 16 bits wide, some of them are used. Of these flags, four flags you use all the time: zero, carry, sign, and overflow. These flags are the 8086 condition codes. The flags register appears below:

Zero flag bit is set when last instruction produces 0 as a result.

Carry flag bit is set when last instruction produces a number with a carry.

Sign flag bit is set when last instruction produces a negative number.

Overflow flag bit is set when last instruction produces a number which cannot fit in register.

Flag bits are generally used for testing the conditions. For example, if you want to test whether A = B, you need to subtract A from B, then if you have a result 0, zero flag will be set. Therefore, you need to check zero flag to obtain the result of the condition.

Arithmetic operations

Some of the arithmetic operations that are supported by Intel microprocessors are:

ADD, ADC, INC, DEC, SUB, MUL, IMUL, DIV, IDIV

ADD <destination>, <addend> ; destination = destination + addend

ADD instruction adds <addend> to <destination>.

ADD ax, bx ; ax = ax + bx

ADD var, 2 ; var = var + 2

ADD al, cl ; al = al + cl

ADC <destination>, <addend> ; destination = destination + addend + C

ADC instruction adds <addend> + Carry to <destination>. It is used to add large sized numbers.

Let AX:BX be 32 bits number. Let CX:DX be another 32 bits number. In order to add these numbers together using 16-bits microprocessor, we need to use ADC.

ADD BX, DX ; Add low order digits

ADC AX, CX ; Add high order digits together with carry

The instructions above simply calculates: (AX:BX) = (AX:BX) + (CX:DX)

INC <destination> ; destination = destination + 1;

DEC <destination> ; destination = destination - 1;

INC instruction increments and DEC instruction decrements the destination.

INC dx

DEX al

INC var

SUB <destination>, <subtrahend> ; destination = destination – subtrahend

SUB instruction subtracts <subtrahend> from <destination>.

SUB var, 2 ; var = var - 2;

SUB bl, cl ; bl = bl – cl;

SUB bx, ax ; bx = bx – ax;

MUL <multiplicand>

MUL instruction multiplies AX or AL register by <multiplicand>. Destination is AX and DX. If <multiplicand> is an 8-bits number, it will be multiplied by AL. The destination is AX. If <multiplicand> is a 16-bits number, it will be multiplied by AX. The destination is (DX:AX). Here are some examples to make it clear:

MUL DL ; AX = AL * DL

MUL BX ; (DX:AX) = AX * BX

MUL var ; AX = AL * var (Assuming var is a byte)

MUL var ; (DX:AX) = AX * var (Assuming var is a word)

IMUL is the same as MUL, but it is used for signed numbers which are stored in 2’s complement format.

DIV <divisor>

DIV instruction divides AX or (DX:AX) by <divisor>.

If <divisor> is an 8-bits number,

AL = AX / <divisor>

AH = AX mod <divisor>

If <divisor> is a 16-bits number,

AX = (DX:AX) / <divisor>

DX = (DX:AX) mod <divisor>

Here is piece of code for division.

MOV AX, 1000

MOV DL, 30

DIV DL

The code above divides 1000 by 30. The quotient (33) will be in AL. The remainder (10) will be in AH. If the number is in <divisor> is too small or zero, the quotient will be too large to fit in AL. Therefore you need to select the registers according to their sizes. For example if you want to divide 1000 by 3, you will get 333 as a quotient which cannot be stored in 8-bits register AL! In order to do this, you need to use 16-bits division. Here is how:

MOV AX, 1000 ; The number is 1000.

MOV DX, 0 ; DX = 0, so (DX:AX) is 0000:1000 = 1000.

MOV BX, 3 ; Divisor is 3

DIV BX ; AX = (DX:AX) / 3 = 1000 / 3 = 333. AX is capable to store 333.

; DX = (DX:AX) mod 3 = 1000 mod 3 = 1.

IDIV is the same as DIV, but it is used for signed numbers which are stored in 2’s complement format.

EXPERIMENTS Advanced Arithmetic Operations

I) Write assembly program to do the following substractions.For each value, what is the hexadecimal result? What is the decimal result? Investigate Carry and Sign and Zero Bits.

01h – 01h
00h + 00h
265Ch - 1AFFh.
0000h - 0001h

II) Write assembly programs to find the sum of the following hexadecimal numbers. For each summation answer following questions. Does the result fit in a 16 bit register? Why? Investigate the Carry and sign bits.

0Fh + 01h
FFFFh + 0001h
0FAAh + 1B04h + F0A0h.

III) Write assembly programs to do the following computations. Chose registers according to the data and state how the resulting number is represented.

03h * 05h
A1A1h * 00F0h
(A5C4h * 0040h) + 00FFh
(A5C4h / 0004h) – 0A0Ah
02h * AAh * F2h

2018-02-16T10:50:00.004+05:30

CENG 222 – Computer Organization

Lab Work 2

Arithmetic operations

Some of the arithmetic operations that are supported by Intel microprocessors are:

ADD, ADC, INC, DEC, SUB, MUL, IMUL, DIV, IDIV

ADD <destination>, <addend> ; destination = destination + addend

ADD instruction adds <addend> to <destination>.

ADD ax, bx ; ax = ax + bx

ADD var, 2 ; var = var + 2

ADD al, cl ; al = al + cl

INC <destination> ; destination = destination + 1;

DEC <destination> ; destination = destination - 1;

INC instruction increments and DEC instruction decrements the destination.

INC dx

DEX al

INC var

SUB <destination>, <subtrahend> ; destination = destination – subtrahend

SUB instruction subtracts <subtrahend> from <destination>.

SUB var, 2 ; var = var - 2;

SUB bl, cl ; bl = bl – cl;

SUB bx, ax ; bx = bx – ax;

EXP1 – Direct Addressing and Arithmetic Operations

1. Convert your year of birth to a hexadecimal number and store write it to memory address 0200h. Is 1 byte enough to store the number? Why/why not? Do the writing to memory accordingly.

2. Get the last 4 digits of your student number and write it to next memory address. Consider which memory address you should be using according to the answer of the first question.

3. Find the sum of these two numbers. Store the result in register DX.

4. Subtract first number from second store the result in register CX.

EXP2 – Indirect Addressing and Arithmetic Operations

1. Store 50 (decimal) in memory address 0300h, this time use indirect addressing (access with [BX] ). Is 1 byte enough to store the number? Why/why not? Do the writing to memory accordingly.

2. Store 51 in next address. Consider which memory address you should be using according to the answer of the first question.

3. Store 52 in next address.

4. Find the sum of 3 numbers. Store the result in the next adress.

EXP3 – Advanced Data Transfer

Analyze the assembly code below:

title Register test program

.model small

.stack 100h

.data

msg db "AB",0dh,0ah,'$'

.code

main proc

mov ax,@data

mov ds,ax

; Add your code here

; ------------------

mov ah,9

mov dx,offset msg

int 21h

mov ax, 4c00h

int 21h

main endp

end main

1- Copy the second byte stored at “msg” (which is ‘B’ currently) to the first byte. The output should be “BB”. In order to do that, load 2-bytes data stored at “msg” into a 16-bitsregister. Copy one part of the register to the other part. Then, write your register back to msg.

2- Copy the first byte stored at “msg” (which is ‘A’ currently) to the second byte. The output should be “AA”. To achieve this, load the first byte stored at “msg” into one 8-bits register (high or low part) and copy it onto the other part and write the data stored in register back to msg.

Advanced usage of MOV in MASM

Previously we used “offset” to point a memory location, but MASM supports a more user friendly way. As explained in previous lab, MASM does not support a moving operation to a memory location directly pointed by an immediate data but actually, currently used Intel microprocessors support it! Fortunately, some advanced usage of MOV instructions in MASM make it possible. Analyze the source and assembled code below:

Source Code	Assembled code
title Register test program .model small .stack 100h .data msg db "AB",0dh,0ah,'$' .code main proc mov ax,@data mov ds,ax mov al,msg mov msg+1,al mov ah,9 mov dx,offset msg int 21h mov ax, 4c00h int 21h main endp end main	3BA6:0000 B8A73B MOV AX,3BA7 3BA6:0003 8ED8 MOV DS,AX 3BA6:0005 A00800 MOV AL,BYTE PTR [0008] 3BA6:0008 A20900 MOV BYTE PTR [0009],AL 3BA6:000B B409 MOV AH,09 3BA6:000D BA0800 MOV DX,0008 3BA6:0010 CD21 INT 21 3BA6:0012 B8004C MOV AX,4C00 3BA6:0015 CD21 INT 21

Since “msg” is defined in data segment, if we use it without “offset”, it denotes the data pointed by msg (not the address of it). As we can see in the assembled code the address is inside [] brackets. ([0008]). Note that if we use “msg+1”, it means that the data stored in the memory location msg+1 that is [0009]. Here it is clear that MASM automatically calculates offsets and uses [] to point the data.

You can choose to use “offset” and denote pointer operations by yourself or use MASM to interpret the instructions. It is important to learn the usage of offset in order to understand how the operation is done exactly. Please do not go any further without understanding the code above.

LEA (Load effective address) instruction

If we use “offset”, we need the following instruction to get the address of a data:

MOV DX, offset msg

Here, msg is defined as name of the data itself. Another way to do this without using offset:

LEA DX, msg

Both instructions above, loads the address of msg into DX.

2018-02-16T10:50:00.001+05:30

CENG 222 – Computer Organization

Lab Work 1

16-bits Intel microprocessors have some capabilities of moving data from register to register; register to memory, and vice versa.

Commonly used 16-bits registers are:

AX, BX, CX, DX, SI, DI, SP, DS

First four registers (AX, BX, CX, and DX) consist of two 8-bits parts called high part (denoted by ‘H’) and low part (denoted by ‘L’).

AX = AH, AL BX = BH, BL CX = CH, CL DX = DH, DL

Note that the high part register is the most significant byte and the low part is the least significant byte of the whole 16-bits register.

Assembly instructions of moving data using memory and registers are given below:

§ Parameters of instructions are denoted by <…> symbols for explaining the usage of the instruction.

§ [] braces in Macro assembly has the same function with * operator in C language. For example, [AX] denotes the data or memory location where AX points to.

1- Immediate data to register

MOV <destination register>, <data>

MOV AX, 4C00H ; AX = 4C00h

MOV DL, 120 ; DL = 120

Size of data and destination register must be equal!

2- Register to register

MOV <destination register>, <source register>

MOV AX, CX ; AX=CX

MOV CL, BH ; CL=BH

Note that size of destination and source registers must be the same!

3- Memory to register (Direct addressing)

MOV <destination register>, [<address>]

MOV AX, [1BFFH] ; Copy 2-bytes (word) data stored at address 1BFF to AX

MOV DL, [offset msg] ; Copy 1-byte data at msg to DL

Important: Size of the data read from memory is determined by the size of the destination register!

4- Memory to register (Indirect addressing)

MOV <destination register>, [<register>]

MOV AX, [BX] ; AX = (*BX)

MOV DL, [BX] ; DL = (*BX)

Important: Size of the data read from the memory location pointed by register, is determined by the size of the destination register!

5- Register to memory

MOV [<register>], <source register>

MOV [BX], DX ; *BX = DX (Writes 2 bytes)

MOV [BX], DL ; *BX = DL (Writes 1 byte)

MOV [1BFFH], AX ; *(1BFFH) = AX; (Writes 2 bytes)

Note: Size of the data written is the size of register.

6- Immediate data to memory

MOV [<register>], <data>

MOV [BX], 20h ; *BX = 0020h

MOV [BX], 4C00h ; *BX = 4C00h

MOV [1BFFH], 20h ; *(1BFFH) = 20h;

If the source data is 1-byte, size of the data written to the memory is 16-bits unless you state the size. Leading zeros will be added automatically.

MOV [BX], 20h ; Copy the word (2-bytes) data to the memory

; block started at BX. (*BX = 0020h)

MOV WORD PTR [BX], 20h ; Exactly the same as above

MOV BYTE PTR [BX], 20h ; Copy 1 byte (*BX = 20h)

If the source data is 2-bytes, size of the data written to the memory is 16-bits and you cannot state the size.

Experiments

Analyze the assembly code below:

title Register test program

.model small

.stack 100h

.data

msg db "AB",0dh,0ah,'$'

.code

main proc

mov ax,@data

mov ds,ax

; Add your code here

; ------------------

mov ah,9

mov dx,offset msg

int 21h

mov ax, 4c00h

int 21h

main endp

end main

1- Copy the second byte stored at “msg” (which is ‘B’ currently) to the first byte. The output should be “BB”. In order to do that, load 2-bytes data stored at “msg” into a 16-bitsregister. Copy one part of the register to the other part. Then, write your register back to msg.

2- Copy the first byte stored at “msg” (which is ‘A’ currently) to the second byte. The output should be “AA”. To achieve this, load the first byte stored at “msg” into one 8-bits register (high or low part) and copy it onto the other part and write the data stored in register back to msg.

3- Load the data stored at “msg” into one register and swap the high and low parts of the register using a temporary 8-bitsregister. Then write back your “swapped” register onto msg. The output should be “BA”.

2018-02-16T10:43:00.002+05:30

Unified Modelling Language Practicals. Below is the case study of it for the construction of different UML diagrams

CASE STUDY: Hospital Management System

Hospital management system helps in registering information about patients and handles patient’s query. A unique ID is generated for each patient after registration. This helps in implementing customer relationship management and also maintains medical history of patient. This system also monitors the doctor appointments, when the ID is generated the patient receives the appointment time and number from the receptionist and accordingly visit the doctor. This system also deals with testing appointments as and when ID is generated the patient receives the appointment time and number and accordingly undergoes the test.

It also deals with bed allotments to various patients by checking their ID. It also undergoes various operations by diagnosing the patients. The system identifies whether the person is a doctor or staff and handles various activities such as draw salary and give salary, also it adds doctor/staff information into database. This system is responsible for handling various other activities like deleting, editing doctor/staff information into the database. As per doctor diagnoses the patient, gives treatment and gives suggestions to patients and prescribe laboratory tests and medicines. This system also takes care of medical equipment, doctor visit, vitals recording, patient case sheet, diet ordering, blood requisition, transfer information and discharge information, maintenance of wards, inter and intrawards transfers also it generates patient’s discharge summary which includes patients health at the time of discharge, medical history, various diagnosis and drug prescriptions, history of patients illness and course in hospital. Patient can pay bill through credit card, cash or cheque whose information is maintained by this system.

Use Case Diagram:-

Activity Diagram:- Patient Payment

Activity Diagram:- Setting Doctor Appointment

Class Diagram:-

State Chart Diagram :- Patient

State Chart Diagram :- Doctor

State Chart Diagram:- Receptionist

Sequence Diagram:- Test Appointments

Sequence Diagram:- Bed Allotment

Collaboration Diagram:- Test Appointments

Collaboration Diagram:- Bed Allotments

Component Diagram:-

Deployment Diagram:-

2018-02-16T10:42:00.002+05:30

Unified Modelling Language Practicals. Below is the case study of it for the construction of different UML diagrams

CASE STUDY:- ATM MACHINE AND SYSTEM

ATM or rather Automated Teller Machine is also called as ANY TIME MONEY by many. The ATM system is connected to banks.

The ATM is given the utmost security in terms of technology because its a stand alone system and easily prone to malicious attacks.

The Below diagrams are has come up in Mumbai University MCA exams.

Class Diagram for ATM Machine

Use Case Diagram ATM Machine

Activity Diagram for ATM Machine 1

Activity Diagram for ATM Machine 2

Activity Diagram for Overall ATM Machine:-

Activity Diagram Verify Password ATM Machine:-

Sequence Diagram for ATM Machine :-

Sequence Diagram for Invalid Pin ATM Machine:-

Sequence Diagram ATM withdrawal:-

Collaboration Diagram For ATM machine:-

Collaboration Diagram for ATM withdrawal :-

State Diagram for One Transaction ATM machine:-

State Diagram one Session ATM Machine:-

Component Diagram ATM Machine:-

Deployment Diagram ATM Machine :-

2018-02-16T10:41:00.003+05:30

Unified Modelling Language Practicals. Below is the case study of it for the construction of different UML diagrams

CASE STUDY: Hospital Management System

The hospital has several specialized departments like Cardiology, Gynecologic, Orthopedics, Pediatrics, ENT etc. OPD is another independent department. A doctor is only associated with one specialized department at a time though he/she can be a member of the OPD(Outside Patients Department) department. Each doctor has a visiting time and day in a week.

At reception the patient details are entered and the fees are also taken and the patient is tracked on the basis of the Id generated.

In routine a patient can visit the doctors either directly selecting a doctor or by getting admitted to the hospital and then a doctor visits the patients.

A doctor can prescribe tests for the patient to perform. The patient visits the lab to get done the tests prescribed by his/her doctor. The reports are given to the patient. The payments pertaining to the tests are done at the reception. Referring the reports, the doctor prescribes the patient medicines or further tests or is asked to get admitted.

A patient is admitted into a ward of a specialized department (if available) as per the doctor’s prescription. The number of wards is limited and if there is no vacant ward the admission of the patient is rescheduled.

As per the prescription of the doctor the patient is operated on a specified date and time as decided by the doctor who is doing the operation.

After the completion of the treatment a patient may get discharged on an advice of a doctor and upon the complete payment of all due charges at the reception. On payment of full dues the reception generates a discharge ticket for the patient

Use Case Diagram:-

Class Diagram:-

Activity Diagram Registration :-

Activity Diagram for Ward Allocation:-

Activity Diagram for Tests to Perform:-

Activity Diagram for Treatment and Operations:-

Activity Diagram Discharge:-

State Chart Diagram for Patient:-

State Diagram for Doctor:-

State Diagram for Ward Object:-

Sequence Diagram for Patient Admit / Registration:-

Sequence Diagram Test & Operation:-

Sequence Diagram Discharge from Hospital :-

Collaboration Diagram Admit to Hospital:-

Collaboration Diagram for Treatment at Hospital:-

Collaboration Diagram for Discharge from Hospital :-

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Computer Organization Lab

MASM

CENG 222 – Computer Organization Lab Work 9

Standard I/O In 8086

Exercises

CENG 222 – Computer Organization Lab Work 8

Variables

Some previous Examples to remember...

Offset command

LEA (Load effective address) instruction

Experiment 1

Experiment 2

Experiment 3

CENG 222 – Computer Organization - Lab Work 6

Exercises

EXPERIMENT 1

EXPERIMENT 2

EXPERIMENT 3

Arithmetic operations

EXP1 – Direct Addressing and Arithmetic Operations

EXP2 – Indirect Addressing and Arithmetic Operations

EXP3 – Advanced Data Transfer

Advanced usage of MOV in MASM

LEA (Load effective address) instruction

CASE STUDY:- ATM MACHINE AND SYSTEM

Class Diagram for ATM Machine

Use Case Diagram ATM Machine

Activity Diagram for ATM Machine 1

Activity Diagram for ATM Machine 2

Activity Diagram for Overall ATM Machine:-

Activity Diagram Verify Password ATM Machine:-

Sequence Diagram for ATM Machine :-

Sequence Diagram for Invalid Pin ATM Machine:-

Sequence Diagram ATM withdrawal:-

Collaboration Diagram For ATM machine:-

Collaboration Diagram for ATM withdrawal :-

State Diagram for One Transaction ATM machine:-

State Diagram one Session ATM Machine:-

Component Diagram ATM Machine:-

Deployment Diagram ATM Machine :-

10TB SOLID STATE DRIVES MAY SOON BE POSSIBLE
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GARMIN VIVO ACTIVE : GPS WATCH REVIEW
FOR DETAILS:CLICK HERE

THE WORLD'S MOST POWERFUL ATOM SMASHER RESTARTS WITH A BIG BANG
FOR DETAILS:CLICK HERE

SALARY OF GOOGLE'S NEW CFO PORAT REVEALED
FOR DETAILS:CLICK HERE