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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 |
Output
| Address | Data |
|---|---|
| F051 | 06 |
| Address | Data |
|---|---|
| F150 | 08 |
Updated on 30-Jul-2019 22:30:26
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