8085 program to find the set bit of accumulator

Here we will see how to find the place of the set bit of the Accumulator data.

Problem Statement

Write 8085 Assembly language program to find the position where the bit is 1. In the accumulator all bits are 0, but only one bit is 1. We have to get the position of the bit which is 1. The position will be displayed in decimal from 1 to 8.

Discussion

We are taking the number like (0010 0000). The place value is 6. So we are rotating the number to the right through carry. If the carry bit is 1, then we break the loop and get the result at memory location F051.

Input

Address	Data
F050	20

Address	Data
F050	80

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	3A, 50 F0		LDA F050	Take the number into accumulator into memory
F003	0E, 00		MVI C,00H	Clear C flag
F005	1F	LOOP	RAR	Rotate Acc right with carry
F006	0C		INR C	increase C by 1
F007	D2, 05, F0		JNC LOOP	If C is not 1, jump to LOOP
F00A	79		MOV A,C	Take value from C to A
F00B	32, 51, 50		STA F051	Store result into memory
F00E	76		HLT	Terminate the program