1. ADDITION OF TWO 16-BIT NUMBERS
Aim : To add two 16-bit hex numbers residing in memory and store the result in memory
Apparatus : 8086 Microprocessor Kit
Example : Add two numbers in locations 8100 and 8102 and store the result at 8200
Input : [1100] = 1234
[1102] = 5678
Output: [1200] = 68AC
Flow Chart :
Aim : To add two 16-bit numbers using masm.
Apparatus : PC with masm software
data segment
num1 dw 9112h
num2 dw 9587h
sum dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,num1
add ax,num2
adc bx,00
mov sum,ax
int 3
code ends
end start
Input :
num1 = 9112h
num2 = 9587h
output :
AX = 2699h
2. 32-BIT ADDITION
Aim : Two 32-bit numbers stored as double words in a consecutive memory locations are to be added and the result stored as a double word in memory
Apparatus : 8086 Microprocessor kit
Input :
[1100] = 12345678h
[1104] = 12345678h
Outout :
[1200] = 2468ACFOh
Flow Chart :
Aim : To add two 32-bit numbers using masm.
Apparatus : PC with masm software
data segment
n1 dd 11119999h
n2 dd 22229999h
sum dd 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,n1
add ax,n2
mov bx,ax
mov ax,n1+2
adc ax,n2+2
adc dx,00
int 3
code ends
end start
Input :
n1 = 11119999h
n2 = 22229999h
output :
AX = 3334h
BX = 3332h
3. 16 – BIT SUBTRACTION
Aim : To Subtract two words in memory and place the difference in a memory location.
Apparatus : 8086 Microprocessor kit
Input :
Minuend : [1100] = 9999h
Subtrahend : [1102] = 369Ch
Result : [1200] = 62FDh
Flow Chart :
Aim : Program to implement 16-bit subtraction using masm
Apparatus : PC with masm Software
data segment
num1 dw A876h
num2 dw 5432h
res dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,num1
sub ax,num2
sbb bx,00
mov res,ax
int 3
code ends
end start
Input :
num1 = A876h
num2 = 5432h
output :
AX = 5444h
4. 32-BIT SUBTRACTION
Aim : To subtract two 32-bit numbers and store the result in memory. The input data is also to be fetched from memory.
Apparatus : 8086 microprocessor kit
Input: Minuend : [1100] = 5678h
[1102] = 123Ah
Subtrahend : [1104] = 123Ah
[1106] = 5678h
Result : [1200] = 443Dh
[1202] = BBC2h
Flow Chart :
Aim : Program to implement two 32-bit subtraction using masm.
Apparatus : PC with masm software
data segment
n1 dd 33330000h
n2 dd 22221111h
diff dd 2 dup(?)
data ends
code segment
assume cs:code, ds:data
start:
mov ax,data
mov ds,ax
mov ax,n1
sub ax,n2
mov bx,ax
mov ax,n1+2
sbb ax,n2+2
int 3
code ends
end start
Input :
n1 = 33330000h
n2 = 22221111h
output :
AX = 1111h
DX = eeeeh
5. FIXED POINT MULTIPLICATION
Aim : To multiply two 16-bit numbers in memory and store the result in memory.
Apparatus : 8086 Microprocessor kit
Example :
Input : [1100] = FFDCh
[1102] = BA98h
Result : [1200] = B9C3h
[1202] = 2AA0h
Flow Chart:
Aim : Multiplication of two 16-bit numbers using masm
Apparatus : PC with masm software
data segment
multiplicand dw 0011h
multiplier dw 0011h
res dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,multiplicand
imul multiplier
int 3
code ends
end start
Input :
multiplicand : 0011h
multiplier : 0011h
output :
AX = 0121h
DX = 0000h
6. FIXED POINT DIVISION (32-BIT DIVISION)
Aim : To perform division of a 32-bit number by a 16-bit number and store the quotient and remainder in memory
Apparatus : 8086 microprocessor kit.
Input :
Dividend : [1100] = FFFFh
Dividend : [1102] = EFFFh
Divisor : [1104] = FFFEh
Result : Quotient : [1200] = F001h
Remainder : [1202] = E001h
Flow Chart :
Aim : Program to implement 32-bit division
Apparatus : PC with masm software
data segment
n1 dd 0000000bh
n2 dw 0002h
re dw 2 dup(?)
qu dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,n1
mov dx,n1+2
mov cx,n2
div cx
mov re,ax
mov qu,dx
int 3
code ends
end start
Input :
n1 = 0000000bh
n2 = 0002h
Output :
AX = 0005h
DX = 0001h
Aim : Program to implement 16-bit division (division of word by byte)
Apparatus : PC with masm software
data segment
dividend dw 2223h
divisor dw 02h
res dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,dividend
div divisor
int 3
code ends
end start
Input :
Dividend : 2223h
Divisor : 02h
Output :
AX = 1111h
DX = 0001h
7. FLOATING POINT MULTIPLICATION AND DIVISION
Aim : To carry out the floating point multiplication / division on two three – digit numbers.
Apparatus: 8086 Microprocessor kit along with 8087 co processr
Convert the three digit multiplicand into short real form and store the multiplicand locations 0100, 0101, 0102, 0103.
Convert the multiplier also into short real form and store it in ,locations 0104, 0105 and 0107.
FINIT
FLD 0100
FMUL 0104
FST 0108
HLT
Input :
Operand 1: 52.5 0100:00 0101:00 0102:52 0103:42
Operand 2: 35.25 0104:00 0105:00 0106:32 0107:42
Output :
0108 :
0109:
010A:
010B:
Note : Floating point division is implemented in the same manner by replacing FMUL with FDIV.
8. CODE CONVERSION – 16-BIT HEX TO DECIMAL
Aim : To convert the number FFFF to its decimal equivalent and store the result in memory.
Apparatus : 8086 Microprocessor kit
Input : (AX) = FFFF
Result : [1200] = 06
[1201] = 05
[1202] = 05
[1203] = 03
[1204] = 05
Flow Chart :
Aim : To convert a given hexadecimal number to its equivalent decimal number
Apparatus : PC with masm software
data segment
num dw 0100h
tt dw 10000
ot dw 1000
hund dw 100
one dw 10
res db 5 dup(?)
data ends
code segment assume ds:data, cs:code
start
mov ax,data
mov ds,ax
mov dx,[num]
mov cx,5
lea si,tt
lea di,res
calc: mov ax,dx
mov dx,0
div word ptr[si]
or al,30h
mov [di],al
add si,2
add di,1
loop calc
mov al,’$’
mov [di],al
lea dx,res
mov ah,9
mov ah,1
int 3
code ends
end start Output : AX = 0256
9. CALCULATING THE LENGTH OF A STRING
Aim : To find the number of characters in a string
Apparatus : 8086 microprocessor kit
Example : The sting is assumed to start at 1200 and end of string is the value FF. Let the content of location 1300 be FF. Then the result, which is the length of the string , is checked in 1100 to be 0100. Please note that FF is not part of the string.
Result : [1100] = 0100
Flow Chart :
Aim : Program to find the string length using masm
Apparatus : PC with masm software
data segment
msg db 'anand karnati$'
data ends
code segment
assume ds:data, cs:code
start:
mov ax, data
mov ds, ax
lea si, msg
mov cx,0
mov ah,'$'
label1: inc cx
mov al,[si]
inc si
cmp al,ah
jne label1
int 3
code ends
end start
Output : CX = 14
10. SUM OF NUMBERS IN A WORD ARRAY
Aim : To obtain the sum of a 16-bit array in memory, using index register SI and store the result in memory
Apparatus : 8086 microprocessor kit
Example : Data in array from START = 1110
[1110] = 0FFF
[1112] = 0FFF
[1114] = 0FFF
[1116] = 0FFF
[1118] = 0FFF
Result : [1200] = 4FFB
Flow Chart :
Aim : Program to find the sum of n numbers using masm.
Apparatus : PC with masm software
data segment
array dw 0001h,0002h,0003h,0004h,0005h
m dw 5h
sum dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov cx,m
mov ax,0
mov si,ax
start_loop:
add ax,array[si]
add si,2
loop start_loop
mov sum,ax
int 3
code ends
end start
Output: 000Fh
11. LARGEST OF GIVEN NUMBERS IN A WORD ARRAY
Aim : To find the largest of given numbers using masm
Apparatus : PC with masm software
data segment
array dw 1h,3h,9h,5h,7h
m dw 5h
data ends
code segment
assume ds:data,cs:code
start:
mov ax,data
mov ds,ax
dec m
mov cx,m
mov si,0
mov bx,array[si]
add si,2
start_loop:
mov dx,array[si]
cmp dx,bx
jbe bigg
mov bx,array[si]
bigg:
add si,2
loop start_loop
int 3
code ends
end start
Output: BX = 0009h
12. BCD ADDITION
Aim : Program to implement BCD Addition
Apparatus : PC with masm Software
data segment
v1 dw 0009h
v2 dw 0008h
total dw ?
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov ax,v1
add ax,v2
daa
int 3
code ends
end start
Output:
AX = 0017
13. AVERAGE OF N NUMBERS
Aim : Program to find average of n numbers
Apparatus : PC with masm software
data segment
m dw 0005h
array dw 0002h,0002h,0003h,0004h,0005h
v_avg db ?
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov cx,m
mov ax,0
mov dx,0
mov si,ax
start_loop:
add ax,array[si]
add si,2
loop start_loop
div m
mov v_avg,al
int 3
code ends
end start
Output:
AL = 03h
14. PROGRAM TO PRINT THE STRING
Aim : Program to print the string
Apparatus : PC with masm software
data segment
str db 'cbitcse$'
data ends
code segment
assume ds:data,cs:code
start:
mov ax,data
mov ds,ax
mov ah,09h
mov dx,offset(str)
int 21h
int 3
code ends
end start
Output:
cbitcse
15. PROGRAM TO COUNT THE NUMBERS IN AN ARRAYAim : program to count the numbers in an array
Apparatus : PC with masm software
data segment
array dw 0001h,0002h,0003h,0004h,0005h,0010h,0040h,0011h
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov dx,0011h
lea si,array
mov cx,0
loop1:
inc cx
mov ax,[si]
add si,2
cmp ax,dx
jnz loop1
int 3
code ends
end start
Output:
CX = 0008h
16. SORTING OF AN ARRAY
Aim : To arrange an array of unsigned words in descending order
Apparatus : 8086 Microprocessor kit
Example :
Number of elements : 8
Unsorted array : [1100] = 00FF
[1102] = 0100
[1104] = 0101
[1106] = 00FE
[1108] = 00CC
[110A] = CDEF
[110C] = ABCD
[110E] = 1234
Sorted array : [1100] = CDEF
[1102] = 0100
[1104] = 1234
[1106] = 0101
[1108] = 0100
[110A] = 00FF
[110C] = 00FE
[110E] = 00CC
Flow Chart :
data segment
array dw 2h,7h,1h,3h,6h
n dw 5h
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov cx,n
dec cx
loop1:
mov di,cx
mov bx,0
loop2:
mov ax,array[bx]
cmp ax,array[bx+2]
jge conti
xchg ax,array[bx+2]
mov array[bx],ax
conti:
add bx,2
loop loop2
mov cx,di
loop loop1
int 3
code ends
end start
Note : For Ascending Order jge instruction should be replaced by jle instruction.
18. WAVEFORM GENERATING PROGRAM
Aim : Program for Digital to Analog conversion
Apparatus : 8086 Microprocessor kit with D/A converter Module.
MOVW DX #CMD55
MOVB AL #MODE
OUTB DX
MOVB AL #00H
LOOP:
MOVW DX, #PORTA
OUTB DX
MOVW DX, #PORTB
OUTB DX
INCW AX
JMP LOOP
OPCODE:
8000 BA E7 FF
BO 80
EE
BO 00
BA 00
BA E1 FF
EE
BAA E3 FF
EE
40
EB F5
Output:
The output waveform is as shown in figure below. The amplitude and time period of the waveform are to be noted.
19. SCREEN PROCESSING
Aim : Program to clear the Screen
Apparatus : PC with masm software
data segment
rows dw 19h
cols dw 50h
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
mov cx,rows
loop1:
push cx
mov cx,cols
loop2:
mov ah,02h
mov dl,' '
int 21h
loop loop2
pop cx
loop loop1
int 3
code ends
end start
Output : The Screen is cleared
20. PRINT SCREEN USIN BIOS INTERRUPTS
Aim : Program to perform print screen operation
Apparatus : PC with masm software
.model small
.stack 64
.data
cr equ 0dh
lf equ 0ah
prnstat dd 00500000h
msgsuc db "print screen operation succesful",cr,lf,"$"
norights db "'print screen operation is disabled",cr,lf,"$"
msgerr db "error during print screen operation",cr,lf,"$"
.code
;para1
prinscr:
mov ax,@data ;[move to ds the segment address of the data
mov ds,ax ; area. invoke bios interrupt 05h, to perform
int 05h ; print screen operation]
; para2
push ds ;[save ds on stack top. load ds with 0050h and
lds si,prnstat ; si with 0000h. get print screen status from
mov al,[si] ; byte location 0050:0000 = 00500h and load it
pop ds ; in al. get back segment address of data area
;para3 in ds from the stack top
cmp al,0 ;[if al=0, the print screen operation is
je sucmsg ; a success. in such a case go to para 6
;para4
cmp al,01 ;[if al=0, it means print screen operation
je disabled ; is disabled. in such a case go to para7]
;para5
cmp al,0ffh ;[if al = ffh, it means therer is an error in print
je errmsg ;screen operation. it could be because printer is
jmp exit ;offline, or printer is switched off, etc., in such
; a case go to para 8. if al value is none of
;para6 these, go to para10 to terminate the program]
sucmsg :
lea dx,msgsuc ;load dx with offset of sucmsg and goto prar9
jmp disp ;for display or success message on the crt
;para7
disabled:
lea dx,norights ;[load dx with offset of norights and go to
jmp disp ;para 9 for display of appropriate failure
;para8 ;message on crt]
errmsg:
;[load dx with offset of msgerr, for display
lea dx,msgerr ; of appropriate error message on crt in para9]
;para9
disp:
mov ah,09h ;[display success / appropriate failure message
int 21h ;on the crt using dos function call 09h]
;para10
exit:
mov ah,4ch ;[terminate the program using
int 21h ;dos function call 4ch
end prinscr
output:
Display of result on the CRT (for success case)
PRINT SCREEN OPERATION SUCCESSFUL
Display of result on the CRT (for print screen disabled case)
PRINT SCREEN OPERATION IS DISABLED
Display of result on the CRT (for error case)
ERROR DURING PRINT SCREEN OPERATION
21. READING AND WRITING CONTENTS OF A FILE
Aim : Program for reading and writing to a file
Apparatus : PC with masm software
data segment
filen db "infile",0
filex db "outfile",0
buf db 2048 dup(?)
data ends
code segment
assume ds:data, cs:code
start
mov ax,data
mov ds,ax
mov ah,3dh ; access permissions to first file
mov dx, offset filen
int 21h
mov bx,ax
mov ax,4200h ; for setting pointer-00 starting position
mov dx,0
mov cx,0
int 21h
mov ah,3fh ;reading a file
mov cx,2048
mov dx,offset buf
int 21h
mov ah,3eh ; closing a file
int 21h
mov ah,3dh
mov dx,offset(filex)
int 21h
mov bx,ax
mov ax,4200h
mov dx,0
mov cx,0
int 21h
mov ah,40h ; write a file
mov dx,offset(buf)
mov cx,2048
int 21h
mov ah,3eh
int 21h
int 3
code ends
end start
Input : Infile
This is an input file which is opened in read mode and the contents will be copied on the output file which is opened in write mode.
Output: outfile
This is an input file which is opened in read mode and the contents will be copied on the output file which is opened in write mode.
21. PROGRAM TO APPEND A GIVE FILE
Aim : Program to append a file
Apparatus : PC with masm software
data segment
filen db "infile",0
mesg db "this line to be appended to the existing file"
data ends
code segment
assume ds:data, cs:code
start
mov ax,data
mov ds,ax
mov ax,3d02h
mov dx,offset(filen)
int 21h
mov bx,ax
mov ax,4202h
mov dx,0
mov cx,0
int 21h
mov ah,40h
mov cx,45
mov dx,offset(mesg)
int 21h
mov ah,3eh
int 21h
int 3
code ends
end start
Output :
This is an input file which is opened in read mode and the contents will be copied on the output file which is opened in write mode. this line to be appended to the existing file
