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
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.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
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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:
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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
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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
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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
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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!.
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