8085 Program to find the smallest number

In this program we will see how to find the smallest number from a block of bytes using 8085.

Problem Statement

Write 8085 Assembly language program to find the smallest number from a block of bytes.

Discussion

In this program the data are stored at location 8001H onwards. The 8000H is containing the size of the block. After executing this program, it will return the smallest number and store it at location 9000H.

Logic is simple, we are taking the first number at register B to start the job. In each iteration we are getting the number from memory and storing it into register A. Then if B > A, then we simply update the value of B with A, otherwise go for the next iteration. Thus we can find the smallest number in a block of bytes.

Input

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

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 00, 80		LXI H,8000H	Point to get array size
F003	4E		MOV C, M	Get the sizeof array
F004	23		INX H	Point to actual array
F005	46		MOV B, M	Load the first number into B
F006	0D		DCR C	Decrease C
F007	23	LOOP	INX H	Point to next location
F008	7E		MOV A, M	Get the next number from memory to Acc
F009	B8		CMP B	Compare Acc and B
F00A	D2, 0E, F0		JNC SKIP	if B <= A,then skip
F00D	47		MOV B,A	If CY is 1, update B
F00E	0D	SKIP	DCR C	Decrease C
F00F	C2, 07, F0		JNZ LOOP	When count is not 0, go to LOOP
F012	21, 00, 90		LXI H,9000H	Point to destination address
F015	70		MOV M,B	Store the minimum number
F016	76		HLT	Terminate the program

Output

Address	Data
...	...
9000	11
...	...

Rishi Rathor

Updated on 30-Jul-2019 22:30:24

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