a 8085 Program to perform bubble sort based on choice

MicrocontrollerMicroprocessor8085

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to perform bubble sort in based on choice.

Problem Statement:

Write 8085 Assembly language program to perform bubble sorting operation on a set of data, and arrange them into ascending or descending order based on choice.

Discussion:

In this program we are arranging some numbers into ascending or descending order based on some choice. We are storing the choice at location A000H. If the choice value is 00H, then data will be sorted in ascending order, otherwise it will be sorted in descending order. The 8000H is holding the block size, and 8001H onwards is holding the data.

Input:

First input

Address
Data
.
.
.
.
.
.
8000
06
8001
22
8002
55
8003
33
8004
66
8005
44
8006
11
.
.
.
.
.
.
A000
00
.
.
.
.
.
.

Second input

Address
Data
.
.
.
.
.
.
8000
06
8001
22
8002
55
8003
33
8004
66
8005
44
8006
11
.
.
.
.
.
.
A000
45
.
.
.
.
.
.

Flow Diagram:

Program:

Address
HEX Codes
Labels
Mnemonics
Comments
F000
31, 00, 90

LXI SP,9000H
Initialize Stack
F003
21, 00, 80

LXI H,8000H
Point to get the block size
F006
4E

MOV C,M
Get the count
F007
0D

DCR C
Decrease C by 1
F008
1E, 01
L1
MVI E,01H
E will store one more than number of swaps in a pass
F00A
41

MOV B,C
Store number of comparisons
F00B
23

INX H
Point to next location
F00C
7E
L2
MOV A,M
Load Memory to A
F00D
23

INX H
Point to next location
F00E
BE

CMP M
Compare memory element with A
F00F
F5

PUSH PSW
Store AF into stack
F010
3A, 00, A0

LDA A000
Get the choice
F013
FE, 00

CPI 00H
Compare choice with 00H
F015
CA, 1F, F0

JZ ASC
If Z = 1, sort in Ascending order
F018
F1

POP PSW
Pop AF from stack
F019
D2, 29, F0

JNC SKIP
Jump to SKIP if CY = 0
F01C
C3, 23, F0

JMP EXG
otherwise Jump to exchange
F01F
F1
ASC
POP PSW
Pop AF from stack
F020
DA, 29, F0

JC SKIP
If CY = 1, go to skip
F023
56
EXG
MOV D,M
Load memory to D
F024
77

MOV M,A
Load A to memory
F025
2B

DCX H
Point to previous location
F026
72

MOV M,D
Load D to Memory
F027
23

INX H
Point to next location
F028
1C

INR E
Increase number of exchanges
F029
05
SKIP
DCR B
Decrease B by 1
F02A
C2, 0C, F0

JNZ L2
if Z = 0, go to L2 loop
F02D
1D

DCR E
Decrease E register by 1
F02E
CA, 38, F0

JZ DONE
If Z = 1, terminate the program
F031
21, 00, 80

LXI H,8000H
Point to initial address of block
F034
0D

DCR C
Decrease the count
F035
C2, 08, F0

JNZ L1
If Z = 0, go to L1
F038
76
DONE
HLT
Terminate the program

Output:

First output

Address
Data
.
.
.
.
.
.
8000
06
8001
11
8002
22
8003
33
8004
44
8005
55
8006
66
.
.
.
.
.
.

Second Output

Address
Data
.
.
.
.
.
.
8000
06
8001
66
8002
55
8003
44
8004
33
8005
22
8006
11
.
.
.
.
.
.



raja
Published on 28-Feb-2019 15:06:11
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