8085 Program to perform selection sort in descending order

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to sort a sequence of numbers in reverse order using selection sort.

Problem Statement:

Write 8085 Assembly language program to sort a given sequence using selection sort in descending order. The numbers are stored at 8001H onwards. 8000H is holding the block size.

Discussion:

In the selection sorting technique, we will choose the minimum or the maximum term from a set of numbers. In this case we are considering the sorting in descending order, so we are choosing the maximum number. By taking the maximum number, we are swapping it with the element of currently pointed location. In each pass the maximum number of the sub block is placed at the correct position.

This program has two subroutines. The first subroutine will check the maximum number in a set of numbers, and the second subroutine will be used to swap the numbers.

Input:

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

Flow Diagram:

Program:

Address	HEX Codes	Labels	Mnemonics	Comments
F000	31, 00, 90		LXI SP,9000H	Initialize Stack Pointer
F003	21, 00, 80		LXI H,8000H	Point to get the block size
F006	4E		MOV C,M	Get the count
F007	23	DO	INX H	Point to next location to get block
F008	7E		MOV A,M	Load the element to A
F009	CD, 18, F0		CALL MAX	Find the maximum
F00C	BE		CMP M	Compare M and A
F00D	CA, 13, F0		JZ GO	if Z = 1, they are same, skip swapping
F010	CD, 2A, F0		CALL SWAP	Swap minimum and current content
F013	0D	GO	DCR C	Decrease C by 1
F014	C2, 07, F0		JNZ DO	If Z = 0, go to Do
F017	76		HLT	Terminate the program
F018	E5	MAX	PUSH H	Push HL into Stack
F019	C5		PUSH B	Push BC into stack
F01A	0D		DCR C	Decrease C by 1
F01B	23	LOOP	INX H	Point to next location
F01C	BE		CMP M	Compare memory data with A
F01D	D2, 23, F0		JNC SKIP	If CY = 0, jump to SKIP
F020	7E		MOV A,M	Update the value of A
F021	54		MOV D,H	Copy H to D
F022	5D		MOV E,L	Copy L to E
F023	0D	SKIP	DCR C	Decrease C by 1
F024	C2, 1B, F0		JNZ LOOP	If Z = 0, go to Loop
F027	C1		POP B	Pop BC from stack
F028	E1		POP H	Pop HL from stack
F029	C9		RET	Return from subroutine
F02A	F5	SWAP	PUSH PSW	Store AF into stack
F02B	C5		PUSH B	Push BC into stack
F02C	1A		LDAX D	A = get data from location pointed by DE
F02D	47		MOV B,A	Copy A to B
F02E	7E		MOV A,M	get data from location pointed by HL
F02F	12		STAX D	Store A content into memory pointed by DE
F030	70		MOV M,B	Store B content to memory pointed by HL
F031	C1		POP B	Pop BC from Stack
F032	F1		POP PSW	Pop AF from stack
F033	C9		RET	Return from swap

Output:

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

Ankith Reddy

Published on 28-Feb-2019 10:46:13

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