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a 8085 Program to perform bubble sort based on choice
Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to perform bubble sort in based on choice.
Problem Statement:
Write 8085 Assembly language program to perform bubble sorting operation on a set of data, and arrange them into ascending or descending order based on choice.
Discussion:
In this program we are arranging some numbers into ascending or descending order based on some choice. We are storing the choice at location A000H. If the choice value is 00H, then data will be sorted in ascending order, otherwise it will be sorted in descending order. The 8000H is holding the block size, and 8001H onwards is holding the data.
Input:
First input
| Address | Data |
|---|---|
| . . . | . . . |
| 8000 | 06 |
| 8001 | 22 |
| 8002 | 55 |
| 8003 | 33 |
| 8004 | 66 |
| 8005 | 44 |
| 8006 | 11 |
| . . . | . . . |
| A000 | 00 |
| . . . | . . . |
Second input
| Address | Data |
|---|---|
| . . . | . . . |
| 8000 | 06 |
| 8001 | 22 |
| 8002 | 55 |
| 8003 | 33 |
| 8004 | 66 |
| 8005 | 44 |
| 8006 | 11 |
| . . . | . . . |
| A000 | 45 |
| . . . | . . . |
Flow Diagram:
Program:
| Address | HEX Codes | Labels | Mnemonics | Comments |
|---|---|---|---|---|
| F000 | 31, 00, 90 | LXI SP,9000H | Initialize Stack | |
| F003 | 21, 00, 80 | LXI H,8000H | Point to get the block size | |
| F006 | 4E | MOV C,M | Get the count | |
| F007 | 0D | DCR C | Decrease C by 1 | |
| F008 | 1E, 01 | L1 | MVI E,01H | E will store one more than number of swaps in a pass |
| F00A | 41 | MOV B,C | Store number of comparisons | |
| F00B | 23 | INX H | Point to next location | |
| F00C | 7E | L2 | MOV A,M | Load Memory to A |
| F00D | 23 | INX H | Point to next location | |
| F00E | BE | CMP M | Compare memory element with A | |
| F00F | F5 | PUSH PSW | Store AF into stack | |
| F010 | 3A, 00, A0 | LDA A000 | Get the choice | |
| F013 | FE, 00 | CPI 00H | Compare choice with 00H | |
| F015 | CA, 1F, F0 | JZ ASC | If Z = 1, sort in Ascending order | |
| F018 | F1 | POP PSW | Pop AF from stack | |
| F019 | D2, 29, F0 | JNC SKIP | Jump to SKIP if CY = 0 | |
| F01C | C3, 23, F0 | JMP EXG | otherwise Jump to exchange | |
| F01F | F1 | ASC | POP PSW | Pop AF from stack |
| F020 | DA, 29, F0 | JC SKIP | If CY = 1, go to skip | |
| F023 | 56 | EXG | MOV D,M | Load memory to D |
| F024 | 77 | MOV M,A | Load A to memory | |
| F025 | 2B | DCX H | Point to previous location | |
| F026 | 72 | MOV M,D | Load D to Memory | |
| F027 | 23 | INX H | Point to next location | |
| F028 | 1C | INR E | Increase number of exchanges | |
| F029 | 05 | SKIP | DCR B | Decrease B by 1 |
| F02A | C2, 0C, F0 | JNZ L2 | if Z = 0, go to L2 loop | |
| F02D | 1D | DCR E | Decrease E register by 1 | |
| F02E | CA, 38, F0 | JZ DONE | If Z = 1, terminate the program | |
| F031 | 21, 00, 80 | LXI H,8000H | Point to initial address of block | |
| F034 | 0D | DCR C | Decrease the count | |
| F035 | C2, 08, F0 | JNZ L1 | If Z = 0, go to L1 | |
| F038 | 76 | DONE | HLT | Terminate the program |
Output:
First output
| Address | Data |
|---|---|
| . . . | . . . |
| 8000 | 06 |
| 8001 | 11 |
| 8002 | 22 |
| 8003 | 33 |
| 8004 | 44 |
| 8005 | 55 |
| 8006 | 66 |
| . . . | . . . |
Second Output
| Address | Data |
|---|---|
| . . . | . . . |
| 8000 | 06 |
| 8001 | 66 |
| 8002 | 55 |
| 8003 | 44 |
| 8004 | 33 |
| 8005 | 22 |
| 8006 | 11 |
| . . . | . . . |
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