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8085 Program to convert two-digit hex to two ASCII values
Now let us see a program of Intel 8085 Microprocessor. This program will convert 8-bit numbers to two digit ASCII values.
Problem Statement
Write 8085 Assembly language program where an 8-bit binary number is stored in memory location 8050H. Separate each nibbles and convert it to corresponding ASCII code and store it to the memory location 8060H and 8061H.
Discussion
In this problem we are using a subroutine to convert one hexa-decimal digit (nibble) to its equivalent ASCII values. As the 8-bit number contains two nibbles, so we can execute this subroutine to find ASCII values of them. We can get the lower nibble very easily by masking the upper nibble, and for the upper nibble, we have to mask the lower nibble at first, then rotate the register content dour times to the right to make, now we can change it to ASCII values.
Here we will put 3AH as input, the program will return 33 and 41. These are the ASCII values of 3 and A respectively.
Input
| Address | Data |
|---|---|
| . . . | . . . |
| 8050 | 3A |
| . . . | . . . |
Flow Diagram
Program
| Address | HEX Codes | Labels | Mnemonics | Comments |
|---|---|---|---|---|
| 8000 | 31, 00, 81 | LXI SP, 8100 | Initialize SP | |
| 8003 | 21, 50, 80 | START | LXI H, 8050H | Initialize pointer with the first location of IN-BUFFER |
| 8006 | 11, 60, 80 | LXI D, 8060H | Initialize pointer with the first location of OUT-BUFFER | |
| 8009 | 7E | MOV A, M | Move the contents of 8050H to A | |
| 800A | 47 | MOV B, A | Copy A to B | |
| 800B | 0F | RRC | Rotate accumulator right 4 times | |
| 800C | 0F | RRC | ||
| 800D | 0F | RRC | ||
| 800E | 0F | RRC | ||
| 800F | CD, 1A, 80 | CALL ASCII | This subroutine converts a binary no. toASCII | |
| 8012 | 12 | STAX D | Store the contents of the accumulator specified the contents by DE register pair | |
| 8013 | 13 | INX D | Go to next location | |
| 8014 | 78 | MOV A, B | Copy B to A | |
| 8015 | CD, 1A, 80 | CALL ASCII | This subroutine converts a binary no. toASCII | |
| 8018 | 12 | STAX D | Store the contents of the accumulator specified the contents by DE register pair | |
| 8019 | 76 | HLT | Terminate the program | |
| 801A | E6, 0F | ASCII | ANI 0FH | Converts a BCD number to its corresponding ASCII value + 48 0 To 9 -----------------à48 To 57 + 55 A To F -----------------à 65 To 70 + 48 +7 So +48 is common but if the hex digit is between A to F then +7 is additional. |
| 801C | FE, 0A | CPI 0AH | ||
| 801E | DA, 23, 80 | JC CODE | ||
| 8021 | C6, 07 | ADI 07H | ||
| 8023 | C6, 30 | CODE | ADI 30H | |
| 8025 | C9 | RET | Returning control to the calling program |
Output
| Address | Data |
|---|---|
| . . . | . . . |
| 8060 | 33 |
| 8061 | 41 |
. . . | . . . |
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