NOTES IN ASSEMBLY LANGUAGE
ENG. MOHAMED ZAKI, M.SC
SPRING 2004
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ASSEMBLY LANGUAGES
TOC
NUMBERING SYSTEMS __________________________________________ 5
Representation of numbers ______________________________________________
Decimal_____________________________________________________________
Binary ______________________________________________________________
Hexadecimal ________________________________________________________
Operations __________________________________________________________
Examples ____________________________________________________________
Assembly and machine languages _______________________________________
Assignments _________________________________________________________
5
5
5
5
5
6
8
9
DATA ORGANIZATION __________________________________________
Structure __________________________________________________________
Signed and Unsigned Numbers ________________________________________
Bit Fields and Packed Data ___________________________________________
The ASCII Character Set _____________________________________________
Sign extension and contraction ________________________________________
10
10
10
12
12
12
BOOLEAN VALUES AND BOOLEAN ALGEBRA _____________________
Boolean Operation on bits ____________________________________________
Boolean theories ____________________________________________________
Example ___________________________________________________________
Assignments ________________________________________________________
13
13
13
14
15
SYSTEM ORGANIZATION _______________________________________ 16
System Bus ___________________________________________________________
Data bus ___________________________________________________________
Address bus ________________________________________________________
Control Bus ________________________________________________________
System Clock _______________________________________________________
Synchronization_____________________________________________________
16
16
17
17
18
18
THE 80X86 CPUS VIEW _________________________________________
Registers ___________________________________________________________
Segment and offset __________________________________________________
Logical address to physical address ____________________________________
19
19
19
20
80X86 COMMANDS _____________________________________________ 21
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MOV : copy form source to destination ___________________________________
Example: ___________________________________________________________
ADD : add source to destination _________________________________________
Example: ___________________________________________________________
SUB _______________________________________________________________
Example: ___________________________________________________________
INC : Increment _____________________________________________________
Example: ___________________________________________________________
DEC : decrement _____________________________________________________
Example: ___________________________________________________________
AND bitwise AND ___________________________________________________
Example: ___________________________________________________________
OR ________________________________________________________________
Example: ___________________________________________________________
XOR ______________________________________________________________
Example: ___________________________________________________________
NOT: invert all bits ___________________________________________________
Example: ___________________________________________________________
SHR : shift to the right by one bit ________________________________________
Example: ___________________________________________________________
SHL : shift to the LEFT by one bit _______________________________________
Example: ___________________________________________________________
Your first assembly program __________________________________________
Assignments ________________________________________________________
CMP with JUMP (JE, JNE, JL, JNL, JLE, JG, JNG, JGE , JMP) _______________
Examples: __________________________________________________________
Assignments ________________________________________________________
21
21
21
21
22
22
22
22
22
22
23
23
23
23
23
23
23
24
24
24
24
24
24
25
26
26
30
WORKING WITH STRING AND ARRAY (ADDRESSING) _______________ 32
Addressing modes _____________________________________________________
1- Direct Addressing Modes ___________________________________________
2- Indirect Addressing Modes _________________________________________
3- Indexed Addressing Modes _________________________________________
4- Based Indexed Addressing Modes____________________________________
example ____________________________________________________________
Read a string using int21h and function 0ah_______________________________
LODSB (load string in bytes) ___________________________________________
MOVSB (move string in bytes) _________________________________________
REP (repeat until CX = 0) _____________________________________________
REPE with CMPSB (repeat if the compare equals and cx is not zero) ___________
EXAMPLES ________________________________________________________
Assignments ________________________________________________________
32
32
32
32
33
33
34
36
36
36
37
39
41
MULTIPLICATION AND DIVISION _________________________________ 42
DIV _______________________________________________________________ 42
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MUL ______________________________________________________________ 43
Examples ___________________________________________________________ 43
WORKING WITH THE STACK_____________________________________
Stack concept and mechanism _________________________________________
PUSH _____________________________________________________________
POP _______________________________________________________________
46
46
47
47
Procedures ___________________________________________________________ 50
Numeric to string _____________________________________________________ 51
String to numeric _____________________________________________________ 51
String to numeric _____________________________________________________ 52
Assignments ________________________________________________________ 53
WORKING WITH FILES __________________________________________
Open file ___________________________________________________________
Example ___________________________________________________________
Create file __________________________________________________________
Close file ___________________________________________________________
Read from a file _____________________________________________________
Assignments ________________________________________________________
54
54
54
54
55
57
60
WORKING WITH THE SCREEN ___________________________________ 61
Examples ___________________________________________________________ 61
WORKING WITH VGA ___________________________________________
Display Modes ______________________________________________________
Memory and screen mapping for mode 13h ______________________________
Examples ___________________________________________________________
Assignments ________________________________________________________
64
64
65
66
67
APPENDIX ____________________________________________________ 71
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Numbering Systems
Representation of numbers
Before we begin to understand how to program in assembly it is best to try to understand
how numbers are represented in computers. Numbers are stored in binary Numbers!.
Let’s begin with the most commonly used numbering systems and see why computers
use the binary numbers among all others numbering system.
Decimal
Base is 10
0123456789
Base is 2
01
Binary
Hexadecimal
Base is 16
0123456789ABCDEF
Operations
In all numbering systems we can do arithmetic operations like addition subtraction and
multiplication:
1-Decimal addition
1+2=3
1+3=4
1 + 9 =10
3 + 9 = 12
2- Binary addition
1+0=1
1 + 1 = 10
3- Hex addition
1+9=A
1+E=F
1 + F = 10
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Binary/ Decimal/ Hex conversion table from 0 to 15
Binary(4 digits)
Hex(1 digit)
Decimal(2 digits)
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
EXAMPLES
Given the following decimal numbers, Use binary and hex to do the following
arithmetic operation:
(1) 34+30
(2) 19+12
(3) 34-30
(1)Binary
34 (0010 0010)
30 (0001 1110)
34
17
8
4
2
1
0
0 L.O
1
0
0
0
1 H.O
30
15
7
3
1
0
0
1
1
1
1
0010 0010
0001 1110
-------------0100 0000
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(2)Binary
19
9
4
2
1
0
19 (0001 0011)
12 (0000 1100)
0001 0011
0000 1100
-------------0001 1111
(1) Hex
0010 0010 
0001 1110 
-------------0100 0000
2 2
1 E
----4 0
(2) Hex
0001 0011 
0000 1100 
-------------0001 1111
1 3
0C
----1F
1
1
0
0
1
12
6
3
1
0
0
0
1
1
(3) Binary
34-30 =34+(-30)
0010 0010 - 0001 1110 = 0010 0010 + (two’s complement of 0001 1110)
two’s complement of 0001 1110 = one’s complement +1
one’s complement == inversion of 1 to 0
(30)
(1’s comp)
(2’s comp )
0001 1110
1110 0001
1110 0001+1= 1110 0010
(-30)
1110 0010
finally
(34-30) is
0010 0010
1110 0010
-------------0000 0100
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Assembly and machine languages
Analogy to get more insight for machine and assembly languages
Red
Yellow
Blue
Switches
fig 1
The assembly language consists of the following 14 operations:
Function (Assembly instruction)
mix red
mix yellow
mix blue
mix purple
mix green
mix orange
mix black
mix white
Paint
move left
move right
move forwards
move backwards
halt
Controls value (Machine Language)
A
B
C
D
E
F
G
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
1
0
1
0
0
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
0
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
Table 1
Mix means dispense paint into the funnel, where it will mix itself.
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Paint means snap open the valve on the bottom of the funnel, so it will paint the
pixel on the piece of paper that is currently under the funnel.
Move means move the paper one pixel either left-right or forward-backward.
ASSIGNMENTS
[1] -Calculate the following operations:
1) 10011011b - 00011010b
2) 2374h + 56A3h
2) 13FFh + 0FA8h
4) 00110111b - 01001001b
4) 36 – 31 in binary
[2] – For figure #1, write a short program draw a horizontal line with orange color and
length 5 pixels. The program should be in machine and assembly language shown in
table1.
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Data Organization
Bit 0
Structure
Binary Digit (BIT)
Byte
Byte
1 BYTE
0
0
0
0
0
0
0
0
0
0
0
0
0
2 NIBBLES
8 BITS
1 WORD
0
0
0
0
0
0
0
0
0
0
0
2 BYTES
4 NIBBLES
16 BITS
Signed and Unsigned Numbers
In general, with n bits we can represent the signed values in
the range -2n-1 to +2n-1-1 or unsigned values in the range 0 to 2n
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For 16-bit numbers:
8000h is negative because the H.O. bit is one.
100h is positive because the H.O. bit is zero.
7FFFh is positive.
FFFFh is negative.
0FFFh is positive.
To convert a positive number to its negative, two’s complement form, you use
the following algorithm:
1) Invert all the bits in the number, i.e., apply the logical NOT function.
2) Add one to the inverted result.
For example, to compute the eight bit equivalent of -5:
0000 0101 Five (in binary).
1111 1010 Invert all the bits.
1111 1011
To convert the numbers above to their negative counterpart (i.e., to negate them), do the
following:
7FFFh:
0111 1111 1111 1111
1000 0000 0000 0000
1000 0000 0000 0001
+32,767
Invert all the bits (8000h)
Add one (8001h or -32,767t)
8000h:
1000 0000 0000 0000
0111 1111 1111 1111
1000 0000 0000 0000
-32,768
Invert all the bits (7FFFh)
Add one (8000h or -32768t)
4000h:
0100 0000 0000 0000
1011 1111 1111 1111
1100 0000 0000 0000
16,384
Invert all the bits (BFFFh)
Add one (0C000h or -16,384t)
Note: we don’t write the +ive equivalent in signed numbered. Instead we use the two’s
complement to use the same H/W and algorithm for both signed and unsigned
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Bit Fields and Packed Data
0100 00010 1011000 = 0100 0001 0101 1000b or 4158h
4
2
88
The ASCII Character Set
A code number to represent each character in the keyboard
ASCII Code is 69
ASCII Code is 101
The ASCII codes of the numeric digit characters:
Char (shape)
“0”
“1”
“2”
“3”
“4”
“5”
“6”
“7”
“8”
“9”
Dec
48
49
50
51
52
53
54
55
56
57
Hex
30h
31h
32h
33h
34h
35h
36h
37h
38h
39h
Sign extension and contraction
1- Examples
of sign extension:
8 Bits
80h
28h
9Ah
7Fh
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16 Bits
FF80h
0028h
FF9Ah
007Fh
32 Bits
FFFFFF80h
00000028h
FFFFFF9Ah
0000007Fh
12
2- extend an unsigned number
8 Bits
80h
28h
9Ah
7Fh
16 Bits
0080h
0028h
009Ah
007Fh
32 Bits
00000080h
00000028h
0000009Ah
0000007Fh
3- sign contraction
FF80h can be sign contracted to 80h
0040h can be sign contracted to 40h
FE40h cannot be sign contracted to 8 bits.
0100h cannot be sign contracted to 8 bits.
The rule : bytes you wish to remove must all contain either 00h or FFh.
Boolean values and Boolean algebra
True 1
False 0
Boolean Operation on bits
1- AND
0 and 0 = 0
0 and 1 = 0
1 and 0 = 0
1 and 1 = 1
2- OR
0 or 0 = 0
0 or 1 = 1
1 or 0 = 1
1 or 1 = 1
3- XOR
0 xor 0 = 0
0 xor 1 = 1
1 xor 0 = 1
1 xor 1 = 0
Boolean theories
With “ • “ represents AND
With “ + “ represents OR
With “A’ “ represents NOT A
With “ + “ represents OR
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Th1:
Th2:
Th3:
Th4:
Th5:
Th6:
Th7:
Th8:
Th15:
Th16:
A+A=A
A•A=A
A+0=A
A•1=A
A•0=0
A+1=1
(A + B)’ = A’ • B’
(A • B)’ = A’ + B’
A + A’ = 1
A • A’ = 0
W = AB
X=A+B
Y = A’B + AB’
Z = A’
A and B
A or B
A xor B
Not A
EXAMPLE
ABC+A’BC+AB’C+A’B’C+ABC’ = BC(A+A’) + B’C(A+A’) + ABC’
= BC•1 +B’C•1 + ABC’
= C(B+B’) + ABC’
= C + ABC’
in C++ we can express the above Boolean expression as
if(C || (A && B && !C) )
{
...
}
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ASSIGNMENTS
- Sign extend the following hex numbers to 16-bits.
a) 5F
b) 9F
c) F3
d) D1
e) 04
f) F2
- Sign contract the following hex numbers to 16-bits and indicate the invalid contraction.
a) FF8F
b) 8080
c) 0DF3
d) 209D
e) 007D
f) FFFF
1-Indicate which one of the following 16-bit numbers is positive and which one is
negative.
a) 8000h
b) 100h
c) 7FFFh
d) 0FFFFh
e) 0FFFh
- Perform the bitwise AND operation on the following pairs of hex numbers. Present the
output in hex
a) 0FF00, 0FF0
b) 0F00F, 1234
d) 2F41, F241
c)FFFF,EDCB
f) 1111, 5789
- Simplify the following Boolean expression
ABC+A’BC+AB’C+B’C+BC’+A’
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System Organization
The typical Von-Newman architecture is shown below indicating three main units CPU,
Memory and I/O Devices. The connection between them is called System Bus
System bus
System Bus
System Bus consists of
 Data Bus
 Control Bus
 Address Bus
Data bus
To convey data between CPU, memory and I/O devices
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Address bus
To identify the source and/or destination
Control Bus
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System Clock
Synchronization
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The 80x86 CPUs View
Registers
2345-
General purpose registers
Segment registers
Pointer register
Flag register
Code Segment Register
16-bits
registers
CS
Data Segment Register
DS
Stack Segment Register
SS
Extra Segment Register
ES
Segment and offset
Segment on 80x86 can form 64K byte to 4 gigabyte. Offset is used to point inside the
segment. The memory can be seen as 2-D array as shown in the figure below
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Logical address to physical address
The following figure shows how to calculate the physical address from the logical
address.
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80X86 commands
MOV : copy form source to destination
mov
mov
mov
mov
mov
mov
mov
mov
reg(8,16),reg(8,16)
reg,mem
reg, value
mem,reg
segreg,mem
segreg,reg
reg,segreg
mem,segreg
In all cases, check the size (8,16,32)
EXAMPLE:
mov ax,12
mov ax,12h
mov al,0
mov var1,90
mov var2,00010100b
ADD : add source to destination
add
add
add
add
add
reg,reg
reg,mem
mem,reg
mem,value
reg, value
In all cases, check the size
EXAMPLE:
add ax,12
add ax,12h
add al,ah
add var1,90
add var2,00010100b
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SUB
sub
sub
sub
sub
sub
reg,reg
reg,mem
mem,reg
mem,value
reg, value
In all cases, check the size
EXAMPLE:
sub ax,12
sub ax,12h
sub al,ah
sub var1,90
sub var2,00010100b
INC : Increment
inc reg
inc mem
EXAMPLE:
inc ax
inc var1
DEC : decrement
dec reg
dec mem
EXAMPLE:
dec ax
dec var1
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AND bitwise AND
and
and
and
and
reg,reg
mem,reg
reg,mem
reg,value
EXAMPLE:
and al,bl
and al,00010010b
and var,32h
OR
or
or
or
or
reg,reg
mem,reg
reg,mem
reg,value
EXAMPLE:
or al,bl
or al,00010010b
or var,32h
XOR
xor
xor
xor
xor
reg,reg
mem,reg
reg,mem
reg,value
EXAMPLE:
xor al,bl
xor al,00010010b
xor var,32h
NOT: invert all bits
not reg
not mem
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EXAMPLE:
not al
not var1
SHR : shift to the right by one bit
shr
shr
shr
shr
reg,number
mem,number
reg,cl
mem,cl
EXAMPLE:
shr
shr
shr
shr
al,1
ax,1
var1,1
var1,cl
SHL : shift to the LEFT by one bit
shl
shl
shl
shl
reg,number
mem,number
reg,cl
mem,cl
EXAMPLE:
shl
shl
shl
shl
al,1
ax,1
var1,1
var1,cl
Your first assembly program
Write a program to calculate the following:
W=x + y – z
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.model small
.data
x db 10
y db 3
z db 5
w db ?
d
.code
main:
mov ax,@data
mov ds,ax
Mov
Add
sub
Mov
al,x
al,y
al,z
w,al
mov ah,4ch
int 21h
end main
ASSIGNMENTS
[1]- Write a statement for each of the following tasks:
- Clear all bits of AX except nibble 3
- Change bit 0 and bit 7 of AH
- Set all bits of AH except bit 3
[2] Write the following expressions in assembly language using shifts, additions and
subtraction only.
a) CX * 7
b) BX * 129 c) AX * 512 d) AX / 8
[3] Write a statement for each of the following tasks
-clear all bits of AX except nibble 3
-change bit 0 and bit 7of AH
e) BX * 5
[3] Write a program to compute the result of
ABC’ + ABC + A’B + B’A + C
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Flow control instructions
CMP with JUMP (JE, JNE, JL, JNL, JLE, JG, JNG, JGE , JMP)
cmp reg,number
je lable1
cmp mem,number
jle lable
jmp lable
EXAMPLES:
loop1:
EndLoop:
mov
cmp
jle
dec
jmp
-----
al,10
al,0
EndLoop
al
loop1
int x=10;
while(x>0)
{
x--;
}
Some examples using CMP , JUMP and LOOP
for(int i=0;i<100;i++)
{
var1++;
}
loop1:
mov cl,100
inc var1
loop loop1
In this case we know the limit of
the loop
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for(int i=start;i<stop;i++)
{
var1++;
}
mov cl,stop
sub cl,start
jl SkipFor
inc var1
loop loop1
loop1:
SkipFor:
General case
int i=0;
while(i<100)
{
var1++;
i++;
}
loop1:
mov
cmp
jge
inc
inc
jmp
al,0
al,100
LoopDone
var1
al
loop1
Loop1Done:
Some other loops (using JZ, JS, JC, JNZ, JNS, JNC)
For(int i = 1 ;i<= 8; i++)
{
k = k + i - j;
}
FLP:
mov
mov
add
sub
mov
inc
cmp
jle
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i, 1
ax, k
ax, i
ax, j
k, ax
i
i, 8
FLP
For(int i = 8 ;i>=1; i++)
{
k = k + i - j;
}
FLP:
mov
mov
add
sub
mov
dec
jnz
i, 8
ax, k
ax, i
ax, j
k, ax
i
FLP
27
IF THEN ELSE control
IF (a==b)
{
c = d ;
}
mov
cmp
jne
mov
mov
ax, a
ax, b
EndOfIf
bx, d
c, bx
mov
cmp
jne
mov
mov
jmp
ax, a
ax, b
ElseBlk
bx, d
c, bx
EndOfIf
EndOfIf:
IF (a==b)
{
c = d ;
}
else
{
b = b + 1;
}
ElseBlk:
inc b
EndOfIf:
Control examples
IF ((X > Y) && (Z < T))
{
C = D;
}
IF (X > Y)
{
IF(Z < T)
{
C = D;
}
}
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mov
cmp
jng
mov
cmp
jnl
mov
mov
ax, x
ax, y
EndOfIf
ax, z
ax, t
EndOfIf
bx,d
c,bx
EndOfIf:
28
IF ((X > Y) && (Z < T))|| (A!=B)
{
C = D;
}
mov
cmp
jne
mov
cmp
jng
mov
cmp
jnl
ax ,a
ax,b
DoIf
ax, x
ax, y
EndOfIf
ax, z
ax, t
EndOfIf
DoIf:
mov bx,d
mov c,bx
EndOfIf:
1- Check if a given number is odd or even
int num ;
cin>>num
if((num%2)==0)
{
cout<<”it is an even ”;
}
else
{
cout<<”it is an even ”;
}
mov al,num ;
and al,00000001b
jnz OddNum
code to Print even
OddNum:
Code to Print odd
2- Print a character on screen using interrupt
mov ah,02h
mov dl,’A’
int 21h
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3- Print a number within 0-9
mov
mov
add
int
dl,5
ah,02h
dl,30h
21h
ASSIGNMENTS
[1]- Convert the following C statement to assembly language
a. if (x==0) && ((y-2) > 1)
{
y --;
}
b. if ((X > Y) && (Z < T))|| (A!=B)
{
C = D;
}
c. if ((X *5)>100) && (Y / 4))<100))
{
C = D;
}
[2]- Trace DL and Y
.model small
.data
stop db 10
start db 5
y db 0
.code
prog:
mov ax,@data
mov ds,ax
mov cl,stop
sub cl,start
jl SkipFor
loop1:
inc y
loop loop1
SkipFor:
mov ah,02
mov dl,y
or dl, 30h
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int 21h
mov ah,4ch
int 21h
end prog
[3] - Print character from a-z
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Working With String and Array (Addressing)
Addressing modes
1- Direct Addressing Modes
MOV al,ds:[14F0h]
Or simply
MOV ax,[14F0h]
2- Indirect Addressing Modes
Using register (BX, SI, DI and BP only) the first three assume to use DS by default and
last one uses SS as its default segment
Mov
Mov
Mov
Mov
al,
al,
al,
al,
[bx]
[bp]
[si]
[di]
However, you can use another segment instead of the default segment
Mov
Mov
Mov
Mov
al,
al,
al,
al,
cs:[bx]
ds:[bp]
ss:[si]
es:[di]
3- Indexed Addressing Modes
The indexed addressing modes use the following syntax:
mov
mov
mov
mov
al,
al,
al,
al,
disp[bx]
disp[bp]
disp[si]
disp[di]
If bx contains 1000h, then the instruction mov cl,20h[bx] will load cl from
memory location ds:1020h. Likewise, if bp contains 2020h, mov dh,1000h[bp]
will load dh from location ss:3020.
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32
4- Based Indexed Addressing Modes
mov
mov
mov
mov
al,
al,
al,
al,
[bx][si]
[bx][di]
[bp][si]
[bp][di]
EXAMPLE
.model small
.data
x DB 1,2,3,4,5,6
y DB 3
var1 DB 5
.code
START:
MOV AX,@data
MOV DS,AX
LEA BX,x
; load Effective address
;indirect mode
MOV DL,DS:[BX]
MOV AH,02h
ADD DL,'0'
INT 21h
1
;indexed mode
MOV DL,[BX+3]
MOV AH,02h
ADD DL,'0'
INT 21h
4
;indexed mode
MOV DL,DS:05h[BX]
MOV AH,02h
ADD DL,'0'
INT 21h
6
;based indexed mode
MOV SI,0h
MOV DL,[BX][SI]
MOV AH,02h
ADD DL,'0'
INT 21h
1
MOV AH,4ch
INT 21h
end start
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33
-
Read a string using int21h and function 0ah
Define label (structure)
REP, REPE,REPNE
LODSB
MOVSB
CMPSB
READ A STRING USING INT21H AND FUNCTION 0AH
plist1 label byte
mlen1 db 10
alen1 db ?
str1 db 10 dup(' '), 'S'
...
...
mov ah,0ah
lea dx,plist1
int 21h
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34
- Write a program to read a string and then search about given character
.data
plist1 label byte
mlen1 db 10
alen1 db ?
str1 db 10 dup('$')
msg1 db 'found','$'
msg2 db 'not found','$'
.code
main:
mov ax,@data
mov ds,ax
mov es,ax
mov ah,0ah
lea dx,plist1
int 21h
mov al, ‘s’
lea bx,str1
wlp: cmp [bx],al
je found
inc bx
loop wlp
mov
lea
int
jmp
ah,09h
dx,msg2
21h
endprog
found:
mov ah,09h
lea dx,msg1
int 21h
endprog:
mov ah,4ch
int 21h
end main
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35
LODSB (LOAD STRING IN BYTES)
plist1 label byte
mlen1 db 10
alen1 db ?
str1 db 10 dup(' '), 'S'
...
...
lea si,str1
loadsb  mov al,[si]
inc si
MOVSB (MOVE STRING IN BYTES)
Copy one string to another
...
...
lea si,str1
lea di,str2
movsb  mov [di],[si]
inc si
inc di
REP (REPEAT UNTIL CX = 0)
...
mov
lea
lea
rep
D:\612925830.doc
cx,8
si,str1
di,str2
movsb  lp: mov
inc
inc
dec
cmp
jg
[di],[si]
si
di
cx
cx,0
lp
36
REPE WITH CMPSB (REPEAT IF THE COMPARE EQUALS AND CX IS NOT
ZERO)
Write a program to compare 2-strings of equal length
.data
str1 db 'assembly','$'
str2 db 'assembly','$'
msg1 db 'equal','$'
msg2 db 'not equal','$'
.code
st:
mov ax,@data
mov ds,ax
mov es,ax
mov cx,8
lea si,str1
lea di,str2
mov cx,8
repe cmpsb
je equal
notequal:
lea dx,msg2
mov ah,09
int 21h
jmp endprog
equal:
lea dx,msg1
mov ah,09
int 21h
endprog:
mov ah,4ch
int 21h
end st
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37
- Print a string
.model small
.data
str1 db ‘Hello world’,’$’
.code
start:
mov ax,@data
mov ds,ax
lea dx, str1
mov ah,09
int 21h
mov ah,4ch
int 21h
end start
- Working with a string (print string using a loop)
.model small
.data
str1 db 'Hello world','$'
.code
start:
mov ax,@data
mov ds,ax
lea di, str1
mov cx,11
mov ah,02
flp: mov dl,[di]
int 21h
inc di
loop flp
mov ah,4ch
int 21h
end start
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38
4- Working with array
Array definition
x db 0,1,2,3,4,5,6,7,8,9
x db size dup(value)
EXAMPLES
.model small
.data
x db 0,1,2,3,4,5,6,7,8,9
.code
st:
mov ax,@data
mov ds,ax
lea bx,x
mov cx,10
l1:
mov
mov
add
int
ah,02h
dl,[bx]
dl,'0'
21h
cont: inc bx
loop l1
mov ah,4ch
int 21h
end st
.model small
.data
x db 10 dup(3)
.code
st:
mov ax,@data
mov ds,ax
lea bx,x
mov cx,10
l1: mov ah,02h
sub [bx],1
mov dl,[bx]
add dl,'0'
int 21h
cont: inc bx
loop l1
mov ah,4ch
int 21h
end st
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39
- Write a program to print the binary repetition of the 2’s complement of a given
number
.model small
.data
num db 07
mask db 10000000b
.code
st:
mov ax,@data
mov ds,ax
mov cx,8
mov al,num
not al
add al,1
flp:
and al,mask
jz printzero
printone:
mov dl,'1'
mov ah,02h
int 21h
shr mask,1
loop flp
jmp endprog
printzero:
mov dl,'0'
mov ah,02h
int 21h
shr mask,1
loop flp
endprog:
mov ah,4ch
int 21h
end st
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40
ASSIGNMENTS
- Write a program to print 16-bits number in hexadecimal format
- As shown below the ASCII of character ‘R ‘ differs only from character ‘r’ at bit 5,
how can you quickly convert from upper to lower and from lower to upper based on the
above fact.
R
7
0
6
1
5
0
4
0
3
0
2
1
1
1
0
1
r
7
0
6
1
5
1
4
0
3
0
2
1
1
1
0
1
- Write a program that reads a string from the user, reverses the 2nd half of the string and
prints it on the screen.
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41
Multiplication and division
DIV
1- word/byte
Quotient
AH:AL
AL
DIV
AH
MEM/REG
int x=230;
int y=5;
int z=x/y;
2- word/word
Reminder
X dw 230
Y db 5
...
mov ax,x
div y
mov z, al
DX:AX
AX
DIV
DX
MEM/REG
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42
MUL
1- Byte * byte
ax = al * reg/mem
AL
MUL
AX
MEM/REG
2- word *word
dx:ax=ax * reg/mem
DX:AX
MUL
DX:AX
MEM/REG
int x=130;
int y=5;
int z=x*y;
x db 130
y db 5
...
mov ax,x
mul y
mov z, ax
EXAMPLES
- Write a program to calculate x = (a+b)*(c/d)
.model
.data
x dw
a db
b db
c db
d db
small
?
2
3
10
2
.code
st:
mov ax,@data
mov ds,ax
mov bl,a
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43
add bl,b
mov al,c
mov ah,0
div d
mul bl
mov x,ax
mov ah,4ch
int 21h
end st
- Write a program to calculate W=(x+y)/(z+t) (use w as 16-bits)
Xor
Mov
Add
Mov
Add
Div
Mov
dx,dx
ax,x
ax,y
bx,t
bx,z
bx
w,ax
- Write a program to print only the even number from a given array (assume the
numbers if the array within 0-9)
.model small
.data
x db 0,1,2,3,4,5,6,7,8,9
x1 db 2
.code
st:
mov ax,@data
mov ds,ax
lea bx,x
mov cx,10
l1:
mov al,[bx]
mov ah, 0
div x1
cmp ah,0
jne cont
mov
mov
add
int
ah,02h
dl,[bx]
dl,'0'
21h
cont: inc bx
loop l1
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44
mov ah,4ch
int 21h
end st
- Write a program to calculate the factorial
res=n*(n-1)*(n-2)…. * (1)
title fact
.model small
.data
x db 5
res dw 1
.code
start:
mov ax , @data
mov ds,ax
l1:
mov al,x
mov ah,0h
mul res
mov res,ax
dec x
cmp x,1
jg l1
end1:
mov ah,4ch
int 21
end start
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45
Working with the Stack
Stack concept and mechanism
Stack is a special memory area used to store data temporarily when calling a
subprogram and nested loops. Stack mechanism works in style “Last In First Out”
LIFO. The following figure shows the stack mechanism.
PUSH AX
Stack is empty
PUSH AX
3
8
3
3
8
Sp=2
Sp=1
Sp=0
POP BX
3
POP BX
8
3
Sp=1
Sp=0
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46
The process of calling a subprogram is shown below
mov al,x
mov ah,0h
mul res
mov res,ax
dec x
cmp x,1
jg l1
call subprog
mov ah,0h
mul res
mov res,ax
call subprog
cmp x,1
jg l1
mov
mov
mul
mov
dec
al,y
ah,0h
res
res,ax
y
PUSH
push
REG/MEM 16-bits only
push ax
 Increment SP register and store AX in the top of stack
memory area
POP
pop
REG/MEM 16-bits only
pop bx
D:\612925830.doc
 copy the value in the top of stack to BX then remove it and
decrement SP register
47
-
write a program for the c code
.model small
.data
res db 0
.code
st:
mov cx,10
flp1:
Res=0;
for(int i=0;i<10;j++)
{
for(int j=0;j<5;j++)
{
res+=j;
}
}
push cx
mov cx,5
flp2:
add res,cx
loop flp2
pop cx
loop flp1
...
-Write a program to use the stack to determine if a given string is palindrome
.model small
.stack 46h
.data
str1 db 'mooom'
msg1 db 'is pal','$'
msg2 db 'is not pal','$'
st:
mov ax,@data
mov ds,ax
lea si,str1
mov cx,5
flp1:
mov al,[si]
mov ah,0
push ax
inc si
loop flp1
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48
lea si,str1
mov cx,5
flp2:
pop ax
cmp al,[si]
jne notpal
inc si
loop flp2
pal:
mov
lea
int
jmp
notpal:
mov
lea
int
endprog:
mov
int
end
D:\612925830.doc
ah,09h
dx,msg1
21h
endprog
ah,09h
dx,msg2
21h
ah,4ch
21h
st
49
Procedures
- Write a program to print a string
.model small
.data
num db 4
str1 db ‘hello’,’$’
.code
prog:
main proc near
mov ax,@data
mov ds,ax
call printme
mov ah,4ch
int 21h
endp main
printme proc
lea dx,str1
mov ah,09h
int 21h
ret
endp printme
end prog
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50
Numeric to string
AL 251 
11111011
Print AL 

How to convert to string  ‘2’ ‘5’ ‘1’
-
Repeat integer division by 10 and take the remainder
o 251/10 = 25 and r = 1
o 25/10 = 2
and r = 5
o 2/10
and r = 2
=0
3267
o 3267/10 = 326 and r = 7
o 326/10 = 32
and r = 6
o 32/10
=3
and r = 2
o 3/10
=0
and r = 3
.model small
.data
stringval db 4 dup(' '),'$'
binval dw 1034
tostring
lea
mov
mov
proc near
si, stringval+3
cx,10
ax,binval
l1:
cmp ax,cx
jb l2
xor dx,dx
div cx
or dl,30h
mov [si],dl
dec si
jmp l1
l2: or al,30h
mov [si],al
ret
tostring endp
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51
String to numeric
‘5245’  to 5245
5245= 5 * (10)3 + 2 * (10)2 + 4 * (10)1 + 5 * (10)0
‘5245’= [35h-30h] * (10)3 + [32h-30h] * (10)2 + [34h-30h] * (10)1 + [35h-30h] * (10)0
.model small
.data
strvaldb 10 dup('1234')
binval dw 0
mulfactor dw 1
.code
prog:
mov ax,@data
mov ds,ax
mov es,ax
mov bx, 10
mov cx,4
lea si,asciival+3
lp1: mov
sub
mov
mul
add
al[si]
al,30h
ah,0
mulfactor
binval,ax
mov ax,mulfactor
mul bx
mov mulfactor,ax
dec si
loop lp1
mov ah,4ch
int 21h
end prog
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52
ASSIGNMENTS
- Write a program to read 2-numbers form the user then add them and print the result
- Write a program to read a number from the user and then print two’s complement in
binary format
- Write a program to check if a given string is well formed or ill formed according to the
following table. The table shows some examples of ill formed expression in terms
of (), {}, [] constructions.
Well-formed
Ill-formed
For(i=0; i<10; i++){y++;}
For(i=0; i<10; i++){y++;}}
While(true);
while(true){;
If(i=9){{res+=10;};}
If(i=9)}{res+=10;};}
int i[100]; int y=0;;;
int i[100); int y=0;;;
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53
Working with Files
Open file
Function (AH): 3Dh
Entry parameters:
AH - file access value
0- File opened for reading
1- File opened for writing
2- File opened for reading and writing
DS:DX- Point at a zero terminated string containing the filename.
Exit parameters: If the carry is set, AX contains one of the following error codes:
1- File not found
4- Too many open files
5- Access denied
12- Invalid access
If the carry is clear, AX contains the file handle value assigned by DOS.
A file must be opened before you can access it. The open command opens a file that
already exists. Attempting to open a file that doesn’t exist produces an error.
EXAMPLE
Filename DB 'test.txt',0
FileHandle DW ?
...
lea dx, Filename
mov ah, 3dh
mov al, 0
;Open for reading.
int 21h
jc OpenError
mov FileHandle, ax
Create file
Function (ah): 3Ch
Entry parameters: ds:dx- Address of zero terminated pathname
cx- File attribute
Exit parameters: If the carry is set, ax contains one of the following error codes:
3- Path not found
4- Too many open files
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54
5- Access denied
If the carry is clear, ax is returned containing the file handle
Create opens a new file for output. As with the OPEN command, ds:dx points at a zero
terminated string containing the filename. Since this call creates a new file, DOS assumes
that you’re opening the file for writing only. Another parameter, passed in cx, is the
initial
file attribute settings. The L.O. six bits of cx contain the following values:
Bit Meaning if equal to one
0 File is a Read-Only file
1 File is a hidden file
2 File is a system file
3 File is a volume label name
4 File is a subdirectory
5 File has been archived
In general, you shouldn’t set any of these bits. Most normal files should be created
with a file attribute of zero. Therefore, the cx register should be loaded with zero before
calling the create function. Upon exit, the carry flag is set if an error occurs. The “Path
not found” error requires some additional explanation. This error is generated, not if the
file isn’t found (which would be most of the time since this command is typically used to
create a new file), but if a subdirectory in the pathname cannot be found.
If the carry flag is clear when DOS returns to your program, then the file has been
properly opened for output and the ax register contains the file handle for this file.
Close file
Function (ah): 3Eh
Entry parameters: bx- File Handle
Exit parameters: If the carry flag is set, ax contains 6, the only possible error, which is an
invalid handle error. This call is used to close a file opened with the Open or Create
commands above. It is passed the file handle in the bx register and, assuming the file
handle is valid, closes the specified file.
You should close all files your program uses as soon as you’re through with them to
avoid disk file corruption in the event the user powers the system down or resets the
machine while your files are left open.
Note that quitting to DOS (or aborting to DOS by pressing control-C or control-break)
automatically closes all open files. However, you should never rely on this feature since
doing so is an extremely poor programming practice.
D:\612925830.doc
55
-
Create a new file and close file
.model small
.data
Filename db 'test00.txt',0
FHndl dw ?
.code
prog:
mov ax,@data
mov ds,ax
mov cx,0
mov ah, 3ch
;Open the file
mov al, 0
;Open for reading
lea dx, Filename
int 21h
jc BadOpen
mov FHndl, ax
;Save file handle
EOF: mov bx, FHndl
mov ah, 3eh
int 21h
jc CloseError
;Close file
CloseError:
...
BadOpen:
...
mov ah,4ch
int 21h
end prog
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56
Read from a file
Function (ah): 3Fh
Entry parameters:
bx- File handle
cx- Number of bytes to read
ds:dx- address of array large enough to hold bytes read
Exit parameters: If the carry flag is set, ax contains one of the following error codes
5- Access denied
6- Invalid handle
If the carry flag is clear, ax contains the number of bytes actually read from the file.
The read function is used to read some number of bytes from a file. The actual number
of bytes is specified by the cx register upon entry into DOS. The file handle, which
specifies the file from which the bytes are to be read, is passed in the bx register. The
ds:dx register contains the address of a buffer into which the bytes read from the file are
to be stored.
On return, if there wasn’t an error, the ax register contains the number of bytes actually
read. Unless the end of file (EOF) was reached, this number will match the value passed
to DOS in the cx register. If the end of file has been reached, the value return in ax will
be somewhere between zero and the value passed to DOS in the cx register.
-
write a program to read file and print its contents
.model small
.data
Filename db 'test.txt',0
FHndl dw ?
Buffer db ?
.code
prog:
mov ax,@data
mov ds,ax
mov ah, 3dh
;Open the file
mov al, 0
;Open for reading
lea dx, Filename;Presume DS points at filename
int 21h
jc BadOpen
mov FHndl, ax ;Save file handle
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57
LP:
mov ah,3fh
lea dx, Buffer
mov cx, 1
mov bx, FHndl
int 21h
jc ReadError
;Read data from the file
;Address of data buffer
;Read one byte
;Get file handle value
cmp ax, cx
jne EOF
;EOF reached?
mov ah,02h
mov dl,Buffer
int 21h
;print charcrter
;the character to print
jmp LP
;Read next byte
EOF:
mov bx, FHndl
mov ah, 3eh
int 21h
jc CloseError
;Close file
CloseError:
...
ReadError:
...
BadOpen:
...
mov ah,4ch
int 21h
end prog
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58
-
write a program to read a file and put the content in a string
.model small
.data
Filename db 'test.txt',0
FHndl dw ?
Buffer db ?
filedata db 1024 dup('$')
.code
prog:
mov ax,@data
mov ds,ax
lea di,filedata
mov ah, 3dh
;Open the file
mov al, 0
;Open for reading
lea dx, Filename;Presume DS points at filename
int 21h
jc BadOpen
mov FHndl, ax
;Save file handle
LP:
mov ah,3fh
lea dx, Buffer
mov cx, 1
mov bx, FHndl
int 21h
jc ReadError
;Read data from the file
;Address of data buffer
;Read one byte
;Get file handle value
cmp ax, cx
jne EOF
;EOF reached?
mov dl,Buffer
mov [di],dl
inc di
jmp LP
;Read next byte
EOF:
mov bx, FHndl
mov ah, 3eh
int 21h
jc CloseError
;Close file
CloseError:
...
ReadError:
...
BadOpen:
...
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59
mov ah,09h
lea dx,filedata
int 21h
mov ah,4ch
int 21h
end prog
ASSIGNMENTS
1- Write a program to display a file on the screen page by page based on the user hits
on the keyboard
2- Write a program to copy an existing file into a new file
3- Write a program to display only the last 10 lines of a file
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60
Working with the screen
col
1- Setting the courser
AH
 02h
int 10h
0
0
 row
 col
BH
DH
DL
0
79
row
25
2- Clear screen area
AH
 06H
int
10h
 00
 attribute
 (row:col) start
 (row:col) end
AL
BH
CX
DX
Attributes
7
BL
6
R
5
G
Background
4
B
3
I
2
R
1
G
0
B
Foreground
EXAMPLES
-
Write a program to clear screen and color it with red
.model small
.data
.code
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61
prog:
mov
mov
mov
mov
mov
int
ah,06h
al,00
bh,01001000b
cx,0000h
dx,184eh
10h
mov ah,4ch
int 21h
end prog
3- Print character N-times
AH
int
 09H
10h
AL
BH
BL
CX
 character to be printed
 page #
 Attribute
N
-
Write a program to display character # 20 times start from col:30 at row 12
.model small
.data
.code
prog:
mov
mov
mov
mov
mov
int
mov
mov
mov
mov
int
D:\612925830.doc
ah,06h
;clear screen and color it
al,00
bh,01001000b
cx,0000h
dx,184eh
10h
ah,02h
bh,0
dh, 12
dl ,30
10h
;set cursor at pos 12,30
62
mov
mov
mov
mov
mov
int
ah,09h
;print # 20 times
al,'#'
bh,0
bl, 01001001b
cx,20
10h
mov ah,4ch
int 21h
end prog
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63
Working with VGA
Display Modes
VGA has several modes:
Mode
0
1
2
3
4
5
6
7
Graphics (pixels)
640x200
320x200
Same as mode 1
None
320x200
Same as mode 4
None
None
Logical Colours
2 colour display
4 colour display
Text (chars)
80x25
40x25
16 colour text only
4 colour display
80x25
40x25
16 colour text only
Monochrome text only
Viewdata emulation
16 colour
16 colour
40x25
80x25
40x25
80x25
40x25
8
9
10
11
12
13
14
15
16
640x200
320x200
Same as mode 9
Same as mode 3
Same as mode 9
Same as mode 9
Same as mode 6
640x350
640x350
monochrome display
16 colour
80x25
80x25 or
80x43
17
18
Reserved for expansion.
640x480
16 colour
80x30 or
80x60
Resolution
Colours
Memory Required
640 x 400
256
250Kb
640 x 480
256
300Kb
800 x 600
256
469Kb
1024 x 768
256
768Kb
1280 x 1024
256
1.28Mb
640 x 400
65536
500Kb
640 x 480
65536
600Kb
800 x 600
65536
930Kb
1024 x 768
65536
1.5Mb
1280 x 1024
65536
2.5Mb
640 x 400
16.7 million
750Kb
640 x 480
16.7 million
900Kb
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800 x 600
16.7 million
1.4Mb
1024 x 768
16.7 million
2.25Mb
1280 x 1024
16.7 million
3.75Mb
Memory and screen mapping for mode 13h
Screen Layout
X
Memory Location = (Y * Width) + X
Y
Width
Video RAM Layout
0xA000h
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EXAMPLES
1-Write a program to plot one pixel on location (x,y) using VGA mode 13h
Answer
.model small
.386
.stack 100h
.data
color db ?
.code
start:
mov ax,@data
mov ds,ax
Mov ax,0013h
int 10h
; start vga mode 13h with BIOS interrupt 10h
mov ax,0a000h
mov es,ax
mov cx,100
mov dx,100
; set video RAM location(address) into es
; x coordinate
; y coordinate
mov color,120 ; color in al
call point
call keyd
mov
int
mov
int
; call procedure to plot point at cx,dx coords
; wait for the user to hit any key
ax,0003h ; reset to text mode 3
10h
ax,4c00h
21h
point:
pusha
mov ax,320
mul dx
add ax,cx
mov bx,ax
mov al,color
mov es:[bx],al
popa
ret
; key delay procedure - wait for keypress before continuing
KEYD: push ax
; save ax register before local use
mov ah,1
;1 DOS interrupt for keyboard input
int 21h
pop ax
; restore ax
ret
; pop saved instruction address and jump
end start
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ASSIGNMENTS
1- Write a program to draw a multi-colored (16 colors only) line in VGA mode 31h
2- Write a program which has trhee procedure to draw :
a. horizontal line
b. vertical line
c. rectangle
3- Write a program to move a rectangle from left to right and reverse
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4- Assignment # 2
Print a given 16-bits number in hexadecimal format
Answer 1
.model small
.386
.stack 100h
.data
; ascii values of hex digits 0-F
asciis db 48,49,50,51,52,53,54,55,56,57,65,66,67,68,69,70
endl
db ‘H’,13,10,'$'
.code
prog:
mymain proc
mov ax,@data
mov ds,ax
mov cx,26
; sample value decimal value
call printhex
mov cx,0abcdh ; simple hex value
call printhex
mov ax,4c00h
int 21h
mymain endp
;printhex proc
printhex:
push ax
push bx
push dx
mov
shr
lea
add
mov
mov
int
mov
and
shr
mov
add
mov
int
mov
and
shr
mov
add
mov
int
mov
and
mov
add
mov
int
; save registers
dx,cx
dx,12
; first 4 bits now can be used as offset
bx,asciis ; load the address in bx
bx,dx
dl,[bx]
ah,2
21h
; print the first char
dx,cx
dx,0F00h ; mask out other 12 bits before shift
dx,8
bx,offset asciis
bx,dx
dl,[bx]
21h
; print second char
dx,cx
dx,00F0h
dx,4
bx,offset asciis
bx,dx
dl,[bx]
21h
; print the third char
dx,cx
dx,000Fh
bx,offset asciis
bx,dx
dl,[bx]
21h
; print the last char
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mov dx,offset endl
mov ah,9
int 21h
pop dx
pop bx
pop ax
ret
;printhex endp
end prog
; print "\n"
; restore registers and return
Answer 2 Using a loop
.model small
.386
.stack 100h
.data
; ascii values of hex digits 0-F
asciis db 48,49,50,51,52,53,54,55,56,57,65,66,67,68,69,70
endl
db 'H',13,10,'$'
bits db 16
num dw 26
mask1 dw ?
.code
prog:
mymain proc
mov ax,@data
mov ds,ax
; assume value to be printed is in cx
mov cx,48
; sample value decimal value
call printhex
mov cx,0abcdh ; simple hex value
call printhex
mov ax,4c00h
int 21h
mymain endp
;printhex proc
printhex:
push ax
push bx
push dx
mov
mov
mov
mov
mov
; save registers
mask1, 0F000h
num,cx
dx,cx
cx,4
bits,16
l1:
push cx
sub bits,4
mov cl,bits
and dx,mask1
shr dx,cl
; first 4 bits now can be used as offset
lea bx,asciis ; load the address in bx
add bx,dx
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mov dl,[bx]
mov ah,2
int 21h
shr mask1,4
mov dx,num
pop cx
loop l1
; print the first char
mov dx,offset endl
mov ah,9
int 21h
pop dx
pop bx
pop ax
ret
;printhex endp
end prog
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; print "\n"
; restore registers and return
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Appendix
To read a character:
Function: 10h
in AH
Int:
16h
Character read will be in AL
To print a character:
Function: 02h
in AH
Int:
21h
Character to print should be in DL
To print a string terminated by $:
Function: 09h
in AH
Int
21h
Starting address of the string to print in DX
To print a string :
Function:: 40h
in AH
Int:
21h
Starting address of the string to print in DX
Number of characters to print in CX
Setting the cursor at a certain position :
Function: 02h
in AH
Int:
10h
Page no. BH
Row
DH
Column DL
To scroll a number of lines in the screen:
Function: 06h
in AH
Int:
10h
No. of lines to scroll AL
Attribute BH
Start point CX (row : column)
End point DX (row : column)
The attribute field is as follows :
0
BLINKING
1
R
2
G
3
B
4
INTENSITY
5
R
6
G
7
B
Bits 1, 2, 3 are for the background color
Bits 5, 6, 7 are for the foreground color
To read a string from the user :
Function: 0ah
in AH
Int:
21h
Starting address of the paramter list DX
The parameter list should conatin:
1) Maximum length
2) Actual length
3) A place to store the read string in
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Multiplication:
Byte by Byte:
Multiplicand AL
Multiplier
Register/Memory
Result
AX
Word by Word:
Multiplicand AX
Multiplier
Register/Memory
Result
DX : AX
Division:
a. Word/ Byte:
Dividend
AX
Divisor
Register/Memory
Quotient
AL
Remainder AH
b. Double Word/Word:
Divident
DX :AX
Divisor
Register/Memory
Quotient
AX
Remainder DX
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