Assembly Language Programming in 8085
PROGRAMS
Steps to write a program:-
- Type the starting memory address using keyboard, in this 8085kit address belongs to C000 to FFFF. We can use any memory address between them. Ex: A C000 here A stands for assemble the program.
- After that counting type the to the end.
- Whenever label is given type their corresponding memory address where we want to jump our program.
- When complete program is typed then enter key is pressed two times to execute the program.
Steps to execute a program:-
- Type G< starting address>, then press enter key, after that space key.
- Type R, then press enter key. Which given the content of registers.
- When memory address is given in program following step are performed.
B. After that step 1 is performed.
C. Type D<memory address>, then press enter key. Here memory address is the address where answer is stored which is already mention program.
Program No. 1
·
Write a program for addition of two binary numbers stored in any two
register.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
06
|
MVI B,06H
|
//initialize register B//
|
||
C001
|
06
|
||||
C002
|
16
|
MVI D,02H
|
//initialize register D//
|
||
C003
|
02
|
||||
C004
|
78
|
MOV A,B
|
//Copy the content of register B in Accumulator//
|
||
C005
|
82
|
ADD D
|
//Add the content of register D with the content
of Accumulator//
|
||
C006
|
CF
|
RST 1
|
//terminate program execution//
|
·
Input:- B=06 & D=02
·
Output:-
A=08, B=06, C=00, D=02, E=04, F=00, I=0F, H=08, L=00, M=0850, S=FFEB,
P=C007
Program
No. 2
·
Write a program for subtraction of two binary numbers stored in any two
register.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
06
|
MVI B,06H
|
//initialize register B//
|
||
C001
|
06
|
||||
C002
|
16
|
MVI D,02H
|
//initialize register D//
|
||
C003
|
02
|
||||
C004
|
78
|
MOV A,B
|
//Copy the content of register B in Accumulator//
|
||
C005
|
92
|
SUB D
|
//Subtract the content of register D with the content
of Accumulator//
|
||
C006
|
CF
|
RST 1
|
//terminate program execution//
|
·
Input:- B=06 & D=02
·
Output:-
A=04, B=06, C=00, D=02, E=00, F=10, I=0F, H=08, L=00, M=D800, S=FFEB,
P=C007
Program
No. 3
Write a program for a group of data is residing at C050 H memory
location to transfer in reverse order at memory location starting from D050 H.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
21
|
LXI H,C050 H
|
//initialize HL pair register//
|
||
C001
|
50
|
||||
C002
|
C0
|
||||
C003
|
01
|
LXI B,D054 H
|
//initialize HL pair register//
|
||
C004
|
54
|
||||
C005
|
D0
|
||||
C006
|
16
|
MVI D,05 H
|
//Load 05 in D register//
|
||
C007
|
05
|
||||
LOOP
|
C008
|
7E
|
MOV A,M
|
//Copy the content of HL pair in Accumulator//
|
|
C009
|
02
|
STAX B
|
//save in register pair BC//
|
||
C00A
|
23
|
INX H
|
//increment in HL pair//
|
||
C00B
|
0B
|
DCX B
|
//Decrement in BC pair//
|
||
C00C
|
15
|
DCR D
|
//Decrement in D//
|
||
C00D
|
C2
|
JNZ LOOP (C008)
|
//Jump to the loop//
|
||
C00E
|
08
|
||||
C00F
|
C0
|
||||
C010
|
CF
|
RST 1
|
//terminate program execution//
|
·
Input:- E_C050 [C050-1,51-2,52-3,53-4,54-5]
·
Output:-D_D054 [D054-5,53-4,52-3,51-2,50-1]
A=01 BD0 C=53 D05 E=00 F=SS I=0F H=C0
L=51 M=05 S=FFFB P=C010
Program
No. 4
Write a program to sort given 10 numbers from memory location C050 H in
the ascending order.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
06
|
MVI B,09 H
|
//Initialize counter//
|
||
C001
|
09
|
||||
START
|
C002
|
21
|
LXI H,C050 H
|
//Initialize memory pointer//
|
|
C003
|
50
|
||||
C004
|
C0
|
||||
C005
|
0E
|
MVI C,09 H
|
//initialize counter//
|
||
C006
|
09
|
||||
BACK
|
C007
|
7E
|
MOV A,M
|
//get the number//
|
|
C008
|
23
|
INX H
|
//increment memory pointer//
|
||
C009
|
BE
|
CMP M
|
//compare number with next number//
|
||
C00A
|
DA
|
JC SKIP (C015)
|
//if less, don’t interchange//
|
||
C00B
|
15
|
||||
C00C
|
C0
|
||||
C00D
|
CA
|
JZ SKIP (C015)
|
//if equal, don’t interchange//
|
||
C00E
|
15
|
||||
C00F
|
C0
|
||||
C010
|
56
|
MOV D,M
|
//move the content of memory into D//
|
||
C011
|
77
|
MOV M,A
|
//move the content of Accumulator into memory//
|
||
C012
|
2B
|
DCX H
|
//decrement in HL pair//
|
||
C013
|
72
|
MOV M,D
|
//move the content of D into memory//
|
||
C014
|
23
|
INX H
|
//increment in HL pair//
|
||
SKIP
|
C015
|
0D
|
DCR C
|
//Decrement in C//
|
|
C016
|
C2
|
JNZ BACK (C007)
|
//If not zero, repeat//
|
||
C017
|
07
|
||||
C018
|
C0
|
||||
C019
|
05
|
DCR B
|
//Decrement in B//
|
||
C01A
|
C2
|
JNZ START (C002)
|
//If not zero, repeat//
|
||
C01B
|
02
|
||||
C01C
|
C0
|
||||
C01D
|
CF
|
RST1
|
//terminate program execution//
|
·
Input:- E_C050 [C050-9,51-8,52-7,53-6,54-5,55-4,56-3,57-2,58-1,59-0]
·
Output:- D_C050 [C050-0,51-1,52-2,53-3,54-4,55-5,56-6,57-7,58-8,59-9]
A=08 B=00 C=00 D=00 E=00 F=55 I=0F
H=C059 M=C059 S=FFEB P=C01E
Program
No. 5
Write a program to multiply the 8-bit
unsigned number in memory location C050 H by the 8-bit unsigned number in
memory location C051 H. Store the 8 least significant bits of the result in
memory location C300H and the 8 most significant bits in memory location C301
H.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
LXI H, C050 H
|
// initialize the memory pointer//
|
|||
C001
|
|||||
C002
|
|||||
C003
|
MOV E,M
|
// get multiplicand//
|
|||
C004
|
MVI D,00 H
|
// initialize register D by 00//
|
|||
C005
|
|||||
C006
|
INX H
|
//increment memory pointer//
|
|||
C007
|
MOV A,M
|
// get multiplier//
|
|||
C008
|
LXI H,0000
|
// product=0//
|
|||
C009
|
|||||
C00A
|
|||||
C00B
|
MVI B,08 H
|
//initialize counter with count 8//
|
|||
C00C
|
|||||
MULT:
|
C00D
|
DAD H
|
// add the content of register pair HL with
accumulator//
|
||
C00E
|
RAL
|
// rotate accumulator left with carry//
|
|||
C00F
|
JNC SKIP
|
//jump if no carry//
|
|||
C010
|
|||||
C011
|
|||||
C012
|
DAD D
|
// add the content of register pair DE with
accumulator//
|
|||
SKIP
|
C013
|
DCR B
|
//decrement in register B//
|
||
C014
|
JNZ MULT
|
//jump on no zero//
|
|||
C015
|
|||||
C016
|
|||||
C017
|
SHLD C300
|
// store the result//
|
|||
C018
|
|||||
C019
|
|||||
C01A
|
RST 1
|
//terminate program execution//
|
·
Input:-
·
Output:-
Program
No. 6
Write a program to convert a 2-digit
BCD number stored at memory address C200 H into its binary equivalent number
& store the result in a memory location C300 H.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
3A
|
LDA C200 H
|
//get BCD number//
|
||
C001
|
00
|
||||
C002
|
C2
|
||||
C003
|
47
|
MOV B,A
|
//save it//
|
||
C004
|
E6
|
ANI OF H
|
//mask most significant four bits//
|
||
C005
|
0F
|
||||
C006
|
4F
|
MOV C,A
|
//save unpacked BCD in C register//
|
||
C007
|
78
|
MOV A,B
|
//get BCD again//
|
||
C008
|
E6
|
ANI FO H
|
//mask least significant four bit//
|
||
C009
|
F0
|
||||
C00A
|
0F
|
RRC
|
//convert most significant four bits into
unpacked BCD2//
|
||
C00B
|
0F
|
RRC
|
-“”-
|
||
C00C
|
0F
|
RRC
|
-“”-
|
||
C00D
|
0F
|
RRC
|
-“”-
|
||
C00E
|
47
|
MOV B,A
|
//save unpacked BCD2 in B register//
|
||
C00F
|
AF
|
XRA A
|
//clear accumulator (sum=0)//
|
||
C010
|
16
|
MVI D,0A H
|
//set D as a multiplier of 10//
|
||
C011
|
0A
|
||||
SUM
|
C012
|
82
|
ADD D
|
//add 10 unit (B)=0//
|
|
C013
|
05
|
DCR E
|
//decrement BCD2 by one//
|
||
C014
|
C2
|
JNZ SUM
|
//is multiplication complete? If not, go back and
add again//
|
||
C015
|
12
|
||||
C016
|
C0
|
||||
C017
|
84
|
ADD C
|
//add BCD1//
|
||
C018
|
32
|
STA C300 H
|
// store the result//
|
||
C019
|
00
|
||||
C01A
|
C3
|
||||
C01B
|
CF
|
RST 1
|
//terminate program execution//
|
·
Input:- E_C050-13
·
Output:- D_C050 [C050-0D, 02, 03.....]
A=0F B=00 C=05 D=0A
E=00 F=04 I=0F H=C0 L=59 M=C059 S=FFEB P=C01C
Program -07
Write a main Program and a conversion subroutine
to convert the binary number stored at D000H into its equivalent BCD number.
Store the result from memory location D100H.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
LXI SP,C7FF H
|
//Initialize stack pointer//
|
|||
C001
|
|||||
C002
|
|||||
C003
|
LDA D000 H
|
//get the binary number in accumulator//
|
|||
C004
|
|||||
C005
|
|||||
C006
|
CALL Subroutine (C00A)
|
//call subroutine//
|
|||
C007
|
|||||
C008
|
|||||
C009
|
RST 1
|
//terminate program execution//
|
|||
Subroutine to convert binary number into its
equivalent BCD number.
|
|||||
C00A
|
PUSH B
|
//save BC register pair content//
|
|||
C00B
|
PUSH D
|
//save DE register pair content//
|
|||
C00C
|
MVI B, 64
|
//load divisor decimal 100 in B register//
|
|||
C00D
|
|||||
C00E
|
MVI C,0A
|
//load divisor decimal 10 in C register//
|
|||
C00F
|
|||||
C010
|
MVI D,00
|
//initialize digit 1//
|
|||
C011
|
|||||
C012
|
MVI E,00
|
//initialize digit 2//
|
|||
C013
|
|||||
STEP1
|
C014
|
CMP B
|
//check if number <decimal 100//
|
||
C015
|
JC STEP2 (C01D)
|
//if yes go to step 2//
|
|||
C016
|
|||||
C017
|
|||||
C018
|
SUB B
|
//subtract decimal 100//
|
|||
C019
|
INR E
|
//increment register E by 1//
|
|||
C01A
|
JMP STEP1 (C014)
|
// go to step 1//
|
|||
C01B
|
|||||
C01C
|
|||||
STEP2
|
C01D
|
CMP C
|
//check if number < decimal 10//
|
||
C01E
|
JC STEP3 (C026)
|
//if yes go to step 3//
|
|||
C01F
|
|||||
C020
|
|||||
C021
|
SUB C
|
//subtract decimal 10//
|
|||
C022
|
INR D
|
//increment register by 1//
|
|||
C023
|
JMP STEP2 (C01D)
|
//go to step 2//
|
|||
C024
|
|||||
C025
|
|||||
STEP3
|
C026
|
STA D100 H
|
//store digit 0//
|
||
C027
|
|||||
C028
|
|||||
C029
|
MOV A,M
|
//get digit 1//
|
|||
C02A
|
STA D101 H
|
//store digit 1//
|
|||
C02B
|
|||||
C02C
|
|||||
C02D
|
MOV A,E
|
//get digit 2//
|
|||
C02E
|
STA D102 H
|
//store digit 2//
|
|||
C02F
|
|||||
C030
|
|||||
C031
|
POP D
|
//restore DE register pair//
|
|||
C032
|
POP B
|
//restore BC register pair//
|
|||
C033
|
RST 1
|
//terminate program execution//
|
|||
·
Input:-
·
Output:-
Program
No. -08
Write a program to convert the ASCII
number in memory to equivalent decimal.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
LXI H,C050H
|
//point to data//
|
|||
C001
|
|||||
C002
|
|||||
C003
|
MOV A,M
|
//get operand//
|
|||
C004
|
SUI 30H
|
//convert to decimal//
|
|||
C005
|
|||||
C006
|
CPI 0AH
|
//check whether it is valid decimal number//
|
|||
C007
|
|||||
C008
|
JC LOOP (C00D)
|
//yes, store result//
|
|||
C009
|
|||||
C00A
|
|||||
C00B
|
MVI A,FFH
|
//no, make result=FF H//
|
|||
C00C
|
|||||
LOOP:
|
C00D
|
INX H
|
//increment in register pair HL//
|
||
C00E
|
MOV M,A
|
//move content of accumulator into memory//
|
|||
C00F
|
RST 1
|
//terminate program execution//
|
·
Input:-
·
Output:-
Program
No. 9
Two ten bytes are residing at location
starting from C050 & D050 respectively. Write a program to add them up
& store the result starting from C090.
Program:-
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
21
|
LXI H,C050 H
|
//initialize memory pointer one//
|
||
C001
|
50
|
||||
C002
|
C0
|
||||
C003
|
01
|
LXI B,D050 H
|
//initialize memory pointer two//
|
||
C004
|
50
|
||||
C005
|
D0
|
||||
C006
|
11
|
LXI D,C090 H
|
//initialize result pointer//
|
||
C007
|
90
|
||||
C008
|
C0
|
||||
BACK:
|
C009
|
0A
|
LDAX B
|
//get the number from array two//
|
|
C00A
|
86
|
ADD M
|
//add it eight number in array one//
|
||
C00B
|
12
|
STAX D
|
//store the addition in array three//
|
||
C00C
|
23
|
INX H
|
//increment pointer one//
|
||
C00D
|
03
|
INX B
|
//increment pointer two//
|
||
C00E
|
13
|
INX D
|
//increment result pointer//
|
||
C00F
|
7D
|
MOV A,L
|
//move the content of L in accumulator//
|
||
C010
|
FE
|
CPI 0AH
|
//check pointer one for last number//
|
||
C011
|
0A
|
||||
C012
|
C2
|
JNZ BACK (C009)
|
//if not repeat step four//
|
||
C013
|
09
|
||||
C014
|
C0
|
||||
C015
|
CF
|
RST 1
|
//terminate program execution//
|
·
Input:- E_C050 [C050-1,51-2,52-3,53-4,54-5,55-6,56-7,57-8,58-9,59-A]
·
Output:- D_D050 [D050-A,51-9,52-8,53-7,54-6,55-5,56-4,57-3,58-2,59-1]
A=0A B=D C=0A D=C E=4A F=54 I=0F H=C1
L=0A M=C10A S=FFEB P=C016
Program
No. 10
Write a program for a block of 16
signed binary numbers is residing at location starting from C050 add them up
& store the result in D050.
Program:
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
21
|
LXI H,C050 H
|
//initialize the pointer//
|
||
C001
|
50
|
||||
C002
|
C0
|
||||
C003
|
01
|
LXI B,D050 H
|
//initialize second pointer//
|
||
C004
|
50
|
||||
C005
|
D0
|
||||
C006
|
16
|
MVI D,10 H
|
//Initialize the counter//
|
||
CO07
|
10
|
||||
C008
|
7E
|
MOV A,M
|
//loading the number in accumulator//
|
||
LOOP
|
C009
|
23
|
INX H
|
//incrementing the pointer//
|
|
C00A
|
86
|
ADD M
|
//adding the binary number to the last number//
|
||
C00B
|
15
|
DCR D
|
//decrementing the counter//
|
||
C00C
|
C2
|
JNZ LOOP (C009)
|
//if not zero jump to loop//
|
||
C00D
|
09
|
||||
C00E
|
C0
|
||||
C00F
|
02
|
STAX B
|
//saving the addition to a new memory location//
|
||
C010
|
CF
|
RST 1
|
//terminate program execution//
|
·
Input:- E C050- 1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,10
·
Output:-
A=89 B=D0 C=50 D=00 E=00 F=54 I=0F H=C0
L=60 M=C060 S=FFEB P=C011
Program
No. 11
Write a program to calculate the sum of
series of number. The length of the series is in memory location C050 & the
series begins from memory location C090.
Program:
Label
|
Address
|
Hex
Code
|
Nemonics
|
Comment’s
|
Remarks
|
C000
|
LDA C050 H
|
//load accumulator with number//
|
|||
C001
|
50
|
||||
C002
|
C0
|
||||
C003
|
MOV C,A
|
//initialize counter//
|
|||
C004
|
LXI H,C050 H
|
//Initialize pointer//
|
|||
C005
|
50
|
||||
C006
|
C0
|
||||
C007
|
SUB A
|
//sum low=0//
|
|||
C008
|
MOV B,A
|
//sum high=0//
|
|||
BACK:
|
C009
|
ADD M
|
//sum=sum + data//
|
||
C00A
|
JNC SKIP (C00E)
|
//jump on no carry//
|
|||
C00B
|
|||||
C00C
|
|||||
C00D
|
INR B
|
//increment in B//
|
|||
SKIP
|
C00E
|
INX H
|
//increment pointer //
|
||
C00F
|
DCR C
|
//decrement counter//
|
|||
C010
|
JNZ BACK (C009)
|
//check if counter zero repeat//
|
|||
C011
|
|||||
C012
|
|||||
C013
|
STA C300 H
|
//store lower byte//
|
|||
C014
|
00
|
||||
C015
|
C3
|
||||
‘
|
C016
|
MOV A,B
|
//move counting of register B into accumulator//
|
||
C017
|
STA D050 H
|
//store higher byte//
|
|||
C018
|
|||||
C019
|
|||||
C01A
|
RST 1
|
//terminate program execution//
|
·
Input:-
·
Output:-
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