OBSERVATIONS of Assembly Lab 2

OBSERVATIONS

I am not writing the whole code but I am discussing the logics which are used to execute the program properly.

TASK 1:

We have to add 5 bytes.To understand this concept ,we have to review that a general register in 8086 can store 16 bits means 2 bytes .So to store 5 bytes we need maximum 3 registers .

MOV AL,1H

MOV AH,2H

MOV  BL,3H

MOV BH,4H

MOV DL,5H

We can perform addition on these registers accordingly.

ADD BL,DL

ADD BH,BL

ADD AL,BH

ADD AH,AL

The sum will be stored in AH and you can see the result by debugging

TASK 2:

To add 5 bytes using memory locations,We first have to define bytes in the data segment

.DATA section

DATA1 DB 1H

We will perform same task for other 4 bytes as well.

.CODE section

MOV AL,DATA1

MOV AH, DATA2

MOV  BL, DATA3

MOV BH, DATA4

MOV DL, DATA5

We can perform addition on these registers accordingly.

ADD BL,DL

ADD BH,BL

ADD AL,BH

ADD AH,AL

TASK 3:

.DATA

DATA1 DW ‘T’

DATA2 DW ‘t’

.CODE

MOV AX,DATA1

MOV DX,DATA2

MOV DATA1,DX

MOV DATA2,AX

The above processes will exchange the ASCII values of ‘T’ and ‘t’

TASK 4:

.DATA

DATA DB ‘JKML$’

.CODE

MOV AH,[BX]             ;CONVERTING TO LOWER CASE

ADD AH,20H

The above process move value J into AH and adding 20H will convert it into lower case

Then for the next character K

INC BX                       ;POINT TO NEXT CHARACTER IN BX

This will go to next character K and lower case of K can be taken by the above process

We will perform same process for remaining characters.

TASK 5:

.DATA

A DB 2D      ;WE ARE DEFINING DECIMALS

B DB 3D

In code section we can simply apply ADD and MUL instructions according to the formula.

TASK 6:

Assemble the file:

Masm TASK6.asm

Linking the object file:

Link Task6.obj

Loading the file in debugger:

Debugger Task6.exe

To check the status of flags we use “t”

By executing each instruction one by one we can check the status of each flag

After execution of instruction 1: Sf=0,Zf=0,Af=0,Pf=0,Of=0,Cf=0

After execution of instruction 2: Sf=0,Zf=0,Af=0,Pf=0,Of=0,Cf=0

After execution of instruction 3: Sf=0,Zf=0,Af=1,Pf=0,Of=0,Cf=0

After execution of instruction 4: Sf=0,Zf=0,Af=1,Pf=1,Of=0,Cf=1

After execution of instruction 5: Sf=0,Zf=0,Af=1,Pf=1,Of=0,Cf=1

Task 7:

First in data segment ,we have to define to memory locations we are using in the program.As in the required output ,we have to display a row  with proper spaces and on new line the next row.Similarly,the result of them in another row.

IN .DATA section of the program,we define bytes ,For new line and space ,as in ASCII table LF(line feed) and CR(carriage return) are used .We will define them in bytes like

LF EQU 0AH

CR EQU 0DH

MATRIX   DB LF,CR,’FOR MATRIX ONE$’           ;PROMPT MESSAGE

NL DB LF,CR, ’$’                    ;TO MOVE TO NEXT LINE

TAB DB ‘  $’                ;FOR SPACE

We will define other location similarly.These memory locations will be used when to input data from the user.

To store the result of their addition ,we have to define bytes like this

NF DB ?

NS DB ?

RESULT DB ?                ;STORE RESULT OF CERTAIN ELEMENTS RESPECTIVELY

Interrupts are used in this program .Interrupts are used when we are taking user input .

An important thing to notice is that when using interrupts ,the data stored in DX is used.Means the data which is stored last in the DX will be used in interrupts.

In .CODE section

MOV DX,OFFSET MATRIX

MOV AH,09H                                        ;THIS WILL DISPLAY THE STRING LAST STORED IN DX

INT 21H                                                 ;09H IS FOR INPUT  FOR STRINGAND 21H IS INTERUPT

To input numbers from user

MOV AH,1H

INT 21H

The data input from user is then stored in memory location for further addition

MOV DH,AL

MOV NF,DH

When to add numbers,important thing to notice that numbers we input are stored in hexadeximal so first we have to convert it into decimal

MOV CL,NS

SUB DH,30H

SUB CL,30H

ADD DH,CL

Then they are added ,But to show them we have to convert them into hexadecimal again

ADD DH,30H

MOV SUM,DH                            ;STORING RESULT IN SUM

Similarly for other elements input we will perform the above actions repeatedly.

To output on the screen

MOV DL,NF                               ;DISPLAY CONTENTS OF NF

MOV AH,2

INT 21H

To show other outputs and spaces and newline.We will perform  the appropriate tasks accordingly.

TASK 8:

For new line and position of cursor ,the process is described above.

MSG DB LF,CR, ‘ENTER CHARACTER IN LOWER CASE$’

SPACE EQU 20H

Important thing to notice about EQU ,is that EQU doesn’t occupy memory location .It only provides a constant to a label.

.DATA SECTION

CHAR1 DB ?

In .CODE SECTION

MOV AH,1H     ;TO USER INPUT

INT 21H

MOV DH,AL

SUB DH,20H        ;TO CONVERT LETTER TO UPPERCASE

Similarly the inputs for other characters are taken respectively and we can display the output by the process