1. Lecture 05-07 By SHERY KHAN sherykhan186@gmail.com
Assembly Language
Lecture 05-07 By
SHERY KHAN

2. Agenda Direct Addressing Size Mismatch ERROR Indirect Addressing LOOPS
Register offset Addressing
HOW MANY WAYS TO ACCESS MEM
Segment Association
ASSEMBLY LANGUAGE

3. Addressing Example
; a program to add three numbers using memory variables
[org 0x0100]
mov ax, [num1] ; load first number in ax
mov bx, [num2] ; load second number in bx
add ax, bx ; accumulate sum in ax
mov bx, [num3] ; load third number in bx
add ax, bx ; accumulate sum in ax
mov [num4], ax ; store sum in num4
mov ax, 0x4c00 ; terminate program
int 0x21
num1: dw 5
num2: dw 10
num3: dw 15
num4: dw 0
Data if Given to Program ka sath
constant cannot be placed in Destination/ & Changed
Mov 5 ,ax is Incorrect
DB DW 8 bit ,16 bit
DB fit 1 byte DW 2 byte reserved
Memory Defined 5,10,15
Every thing is Symbol but compiler/ assembler convert into num Adress Memory Location when Translated
In ASL if we want to any particular loctaion to Reference it we attach a Symbol to it Readable Come HERE but Asm Covert it to Equivalent Adress
0100 sa start karo us na us ka baad byte Count kar liya when counting byte when reached to label it is Efective Location yeh label kitna offset pa waqi ha it is Adress of Label
[] MEMORY OPERAND is num pa memory Access Karni Ha
Illegal MEMORY TO MEMORY IS NOT ALLOWED
Starting point sa hum next position la skta hain
A1 offset is For MEMORY LOCATION B8 for Constant which Is =19
[] Memory Adresses

4. Size Mismatch ERROR:
; a program to add three numbers using byte variables
[org 0x0100]
mov al, [num1] ; load first number in al
mov bl, [num1+1] ; load second number in bl
add al, bl ; accumulate sum in al
mov bl, [num1+2] ; load third number in bl
add al, bl ; accumulate sum in al
mov [num1+3], al ; store sum at num1+3
mov ax, 0x4c00 ; terminate program
int 0x21
num1: db 5, 10, 15, 0
Source and Destination are same Byte AX------dw
Al----db
Mov Ax,bl illegal Size Nt Illegal
16 Bit MOVE mov ax,[num1]
8 bit Move mov al,[num1]
Illegal mov ax,bl
Illegal mov [num1] ,[num2]
Tricky Illegal mov [num1],5 Ambigous ?? Mov byte [num1],5 or mov word [num1],5
Label is Memory Adress no Size is Attached size is Conveyed by us

5. Indirect Addressing:
; a program to add three numbers using indirect addressing
[org 0x100]
mov bx, num1 ; point bx to first number
mov ax, [bx] ; load first number in ax
add bx, 2 ; advance bx to second number
add ax, [bx] ; add second number to ax
add bx, 2 ; advance bx to third number
add ax, [bx] ; add third number to ax
add bx, 2 ; advance bx to result
mov [bx], ax ; store sum at num1+6
mov ax, 0x4c00 ; terminate program
int 0x21
num1: dw 5, 10, 15, 0
Some Register Hold Data and Also Address from this Address we Access Data
Immediate
Direct Address Da do us Address jisa who utha la ga now the adress is fix so I cannot change I told that you can pick the content on compile time num1=0017 pa ha it is decided on compile time
Now the way I can take the address and change it on run time
Iapx88 4 register are used to access the Memory BX,BP,SI ,DI
If [] are not used then the value of label itself which is address of label not the content
The bx,si bp, di take the square brakets
[bx] it mean that the content are accessed which are present in the BX
HERE IN ABOVE
bx will contain adresses 0117
[bx] mean that it contain content so content is Num1 which is 5 in memory
Bx,2 we use word so 2 byte is added so it can point to the next word in memory
……………………..
Bx=num1
Ax=[bx]
Ax=num1=5
Bx=num1+2=second location
Ax=ax+[bx]
Ax=5+10=15
Bx=num1+4= third location
Ax=15+15=30 …………………………………………the END
Bx=num+6=4thlocation
Ax=30+0=30

6. LOOP CONTROL [org 0x0100] mov bx, num1 ; point bx to first number
mov cx, 10 ; load count of numbers in cx
mov ax, 0 ; initialize sum to zero
l1: add ax, [bx] ; add number to ax
add bx, 2 ; advance bx to next number
sub cx, 1 ; numbers to be added reduced
jnz l1 ; if numbers remain add next
mov [total], ax ; write back sum in memory
mov ax, 0x4c00 ; terminate program
int 0x21
num1: dw 10, 20, 30, 40, 50, 10, 20, 30, 40, 50
Jump if NT Zero : Jump to the Location Li1 if the Zero flag is Not Set
ZF: the Z flag is Set if the Last mathematical or logical has produced zero in its Destination
Mov is nt mathematical or logical so zf is not set
But ADD SUB is LOGICAL OR MATHEMATICAL
AND OR XOR ETC if Zero in its destination then ZF is Set

7. Register + offset Addressing
a program to add ten numbers using register + offset addressing
[org 0x0100]
mov bx, 0 ; initialize array index to zero
mov cx, 10 ; load count of numbers in cx
mov ax, 0 ; initialize sum to zero
l1: add ax, [num1+bx] ; add number to ax
add bx, 2 ; advance bx to next index
sub cx, 1 ; numbers to be added reduced
jnz l1 ; if numbers remain add next
mov [total], ax ; write back sum in memory
mov ax, 0x4c00 ; terminate program
int 0x21
num1: dw 10, 20, 30, 40, 50, 10, 20, 30, 40, 50
total: dw 0
JNZ : Jump If the Zero flag is Not Set
Cx=0 then zero flag is Set
Bx is Increased by 2 and Cx is decreased by 1 according to the Command
Conditional Jump is Decision maker in computer
oFFSEt+ register= INDIRECT
We use Bx,Bp which is Base register class
When we use SI,DI INDEX FORMAT for Memory Access
MIX MATCH
(indirect +base) = Base +offset
BX work like a Array Adress Function ka Sath adress ma 2 add karta gya
INDEX CHANGED KARTA JAO aur Array ka jo base ha who constant raha & we can pick this from Index
Lower BYTE on Lower Adress and Higher Byte In Higher Adress

8. How many ways we access Memory
Direct Mode: offset Addressing:
Register Indirect:
Register Indirect +offset
Base +index
Base+index+offset
Mov ax,[num1] // Address is Fixed and on the Address the Content is Access
Mov ax,[bx] //that Address is fit in Register & through this Register Memory is Access
Mov ax,[bx+num1]: base +offset
Mov ax,[si+Num1]: index+ offset
Mov ax,[bx+si] the two adresses one in index and other in base. Both are added from the resultant adress we can pick the data
Mov ax,[bx+si+num1]

9. Total Seven Adressing Modes
SEVEN WAYS TO ACCESS THE MEMORY
OFFSET addressing
Base Addressing
INDEX Addressing
Base +offset
Index+offset
Base+index
Base+index+offset
EFFECTIVE ADRESS: were our data is present

10. Segment Association Ip Associate by default CS(TIED)
SI,DI,BX By def DS
BP with SS(TIED)
Sp with SS(TIED)
SEGMENT OVER RIDE
Mov ax,[cs:bx]
Every thing is in segment when we go in Data Window we Do DS:
& Code window is CS:
Segment are performing there function at there desire places every thing is Related to there Segment
Every segment can calculate his her address
Now we know about Effective Adresses generally if we use OFFSETS so as we tell you that No address is Complete until we can attach his Segment part. till no meaning
Until we attach the segment part that in which segment the offset is . As Every Segment has its own Default Segment
IP(instruction pointer) by Default Code Segment (CS) related or Associated & this Default segment cannot be separated and it cannot access any other segment
IN data Case is different in direct addressing the OFFSET Form BY DEFAULT Access Data Segment (DS) we cannot Associate it as we can do .as we cannot do it standard goes to DS . And it see that DS relative that in this particular offset what is there
SI,DI ,BX these Three REGISTER INDIRECT FORM by default DS Associated
BP Cannot Associate By Default DS. It By Default SS Associated
SP is also Tied with SS
By Default: not Tied it can be used if we want Bx to Associate with ES then Segement overide is Done As
Mov ax,[cs:bx] for this Instruction we say that bx is now Associated to Cs instead DS 2E num is called override Byte
All the Segment value is Same so the offset from any segment the answer is Same
COM File is Single Segment in which code ,data ,stack all are in it the standard in which all 4 segments are present on each other