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8085 program to find the element that appears once
In this program we will see how to find a number who appears only once in an array of elements.
Problem Statement
Write 8085 Assembly language program to find a number who appears only once in an array of elements. The size of the array is stored at location F100; the numbers are stored from memory location F101 onwards. The result will be stored at F150.
Discussion
The logic behind this problem is simple. Though it may find some wrong results in some cases. It can find the number if there is only one number which is appeared once, and rest of the numbers are appeared even number of times.
We are doing the XOR operation on the numbers. If one number is already present, and we XOR it with the same number it will be 0. So if the same numbers appeared even number of times we can get 0. After cumulative XOR operations, the final number will be that number whose frequency is one.
Input
| Address | Data |
|---|---|
. . . | . . . |
| F100 | 05 |
| F101 | 0C |
| F102 | 0A |
| F103 | 01 |
| F104 | 01 |
| F105 | 0C |
. . . | . . . |
Flow Diagram
Program
| Address | HEX Codes | Labels | Mnemonics | Comments |
|---|---|---|---|---|
| F000 | 21, 00, F1 | | LXI H,F100 | Point to get count of numbers |
| F003 | AF | | XRA A | Clear accumulator |
| F004 | 4E | | MOV C,M | Load count from memory to C |
| F005 | 23 | | INX H | Point to next location |
| F006 | 46 | LOOP | MOV B,M | Load memory to B |
| F007 | A8 | | XRA B | XOR B with Accumulator |
| F008 | 23 | | INX H | Point to next location |
| F009 | 0D | | DCR C | Decrease C by 1 |
| F00A | C2, 06, F0 | | JNZ LOOP | Jump to LOOP until Z = 1 |
| F00D | 32, 50, F1 | | STA F150 | Store number at F150 |
| F010 | 76 | | HLT | Terminate the program |
Output
| Address | Data |
|---|---|
. . . | . . . |
| F150 | 0A |
. . . | . . . |
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