8085 program for Linear search | Set 2

Problem – Write an assembly language program in 8085 microprocessor to find a given number in the list of 10 numbers, if found store 1 in output else store 0 in output.

Example –

Assumption – Data to be found at 2040H, list of numbers from 2050H to 2059H and output at 2060H.

Algorithm –

1. Load data byte to be searched in B register and counter in D register.
2. Load starting element in Accumulator.
3. Compare Accumulator and B register.
4. If zero flag is set then JUMP to point 8 (as CMP instruction sets Zero flag when both are equal).
5. Decrement D register
6. If D>0 take next element in Accumulator and go to point 3.
7. If D=0, this means element not found then store 00H. End the program.
8. Store 01H as element found. End the program.

Program –

Address	Label	Instruction	Comment
2000H	Data	LXI H, 2040H	Load address of data to be searched
2003H		MOV B, M	Store data to be searched in B register
2004H		LXI H, 2050H	Load starting address of list
2007H		MVI D, 0AH	Counter for 10 elements
2009H	NEXT	MOV A, M	Retrieve list element in Accumulator
200AH		CMP B	Compare element with data byte
200BH		JZ STOP	Jump if data byte found
200EH		INX H	Next element of list
200FH		DCR D	Decrement counter
2010H		JNZ NEXT	Jump to NEXT if D>0
2013H		LXI H, 2060H	Load address of output
2016H		MVI M, 00H	Store 00H
2018H		HLT	Halt
2019H	STOP	LXI H, 2060H	Load address of output
201CH		MVI M, 01H	Store 01H
201EH		HLT	Halt

Explanation –

1. One by one all elements are compared with data byte in B register
2. If element found, loop ends and 01H is stored
3. Loop executes 10 number of times
4. If at the end of 10 iterations, data byte is not found then 00H is stored

Refer for Set-1: 8085 program to search a number in an array of n numbers

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