8085 program to find maximum and minimum of 10 numbers

In this program we will see how to find the maximum and minimum number in a block data.

Problem Statement

Write 8085 Assembly language program to find the maximum and minimum number in a block often 8-bit numbers.

Discussion

In this program we are taking the first number of the block into register D and E. The D will store the minimum number, and E will store maximum number. In each iteration we will check whether the number is smaller than D or not, if it is smaller, then update D with the new number, and then compare it again with E to check whether the number is larger than E or not. If larger, then update E with the new number.

Input

Address	Data
. . .	. . .
8000	55
8001	22
8002	88
8003	77
8004	11
8005	99
8006	44
8007	AA
8008	33
8009	66
. . .	. . .

Flow Diagram

Program

Address	HEX Codes	Label	Mnemonics	Comments
F000	21, 00, 80		LXI H,8000H	Load the initial address
F003	0E, 0A		MVI C,0AH	Load the count of numbers
F005	56		MOV D,M	Load the first number from memory
F006	5A		MOV E,D	Also load the first number to E
F007	23		INX H	Point to next location
F008	0D		DCR C	Decrease the count
F009	7E	LOOP	MOV A,M	Load the number from memory to A
F00A	BA		CMP D	Compare D with A
F00B	D2, 0F, F0		JNC SKIP	If CY = 0, A is not smaller
F00E	57		MOV D,A	Update D with A
F00F	BB	SKIP	CMP E	Compare E with A
F010	DA, 14, F0		JC DO	if CY = 1, A is not larger
F013	5F		MOV E,A	Update E with A
F014	23	DO	INX H	Point to next location
F015	0D		DCR C	Decrease C by1
F016	C2, 09, F0		JNZ LOOP	Go to loop
F019	21, 50, 80		LXI H,8050H	Point to destination address
F01C	72		MOV M,D	Store the smallest number
F01D	23		INX H	Point to next location
F01E	73		MOV M,E	Store the largest number
F01F	76		HLT	Terminate the program

Output

Address	Data
. . .	. . .
8050	11
8051	AA
. . .	. . .

Rishi Rathor

Updated on: 2019-07-30T22:30:24+05:30

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