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8085 program to count the number of ones in contents of register B
In this program, we will see how to count the number of 1’s in an 8-bit number stored in register B.
Problem Statement
Write 8085 Assembly language program to count the number of 1s in 8-bit number stored in register B.
Discussion
In this program, we are using the rotate operation to count the number of 1’s. As there are 8 different bits in 8-bit number, then we are rotating the number eight times. we can use RRC or RLC. Here we have used the RRC instruction. This instruction sends the LSb to MSb also to carry flag. So after each iteration, we can check the carry status to get the count of 1s.
If the number is DA (1101 1010) then the answer will be 5, as there are five 1s in the number.
Input
| Register | Data |
|---|---|
| B | DA |
Flow Diagram
Program
| Address | HEX Codes | Labels | Mnemonics | Comments |
|---|---|---|---|---|
| F000 | 06, DA | | MVI B, DA | Take number 1101 1010 |
| F002 | 37 | | STC | Set Carry |
| F003 | 3F | | CMC | Complement Carry |
| F004 | 78 | | MOV A,B | Load the number into A from B |
| F005 | 0E, 08 | | MVI C,08H | Initialize the counter to 08H |
| F007 | 06, 00 | | MVI B,00H | Clear B register |
| F009 | 0F | LOOP | RRC | Rotate right |
| F00A | D2, 0E, F0 | | JNC SKIP | If CY = 0, skip the next step |
| F00D | 04 | | INR B | If CY = 1, increase B |
| F00E | 0D | SKIP | DCR C | Decrease C by 1 |
| F01F | C2, 09, F0 | | JNZ LOOP | If Z = 0, jump to loop |
| F012 | 78 | | MOV A,B | Load the count to A |
| F013 | 32, 50, 80 | | STA 8050H | Store the result at 8050H |
| F016 | 76 | | HLT | Terminate the program |
Output
| Address | Data |
|---|---|
| 8050 | 05 |
Updated on: 30-Jul-2019
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