Program to perform bubble sort based on choice in 8085 Microprocessor

Here we will see one 8085 microprocessor program that will help to arrange the numbers in ascending or descending order based on our choice. Sort using bubble sort technique.

Problem Statement −

Write an 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