Here we will see one 8085 Microprocessor program. That program will convert HEX to ASCII values.
Problem Statement −
Write an 8085 Assembly language program to convert Hexadecimal characters to ASCII values.
We know that the ASCII of number 00H is 30H (48D), and ASCII of 09H is 39H (57D). So all other numbers are in the range 30H to 39H. The ASCII value of 0AH is 41H (65D) and ASCII of 0FH is 46H (70D), so all other alphabets (B, C, D, E, F) are in the range 41H to 46H.
Here we are providing hexadecimal digit at memory location 8000H, The ASCII equivalent is storing at location 8001H.
The logic behind HEX to ASCII conversion is very simple. We are just checking whether the number is in range 0 – 9 or not. When the number is in that range, then the hexadecimal digit is numeric, and we are just simply adding 30H with it to get the ASCII value. When the number is not in range 0 – 9, then the number is range A – F, so for that case, we are converting the number to 41H onwards.
In the program at first we are clearing the carry flag. Then subtracting 0AH from the given number. If the value is numeric, then after subtraction the result will be negative, so the carry flag will be set. Now by checking the carry status we can just add 30H with the value to get ASCII value.
In other hand when the result of subtraction is positive or 0, then we are adding 41H with the result of the subtraction.
|F000||21, 00, 80||LXI H, 8000H||Load address of the number|
|F003||7E||MOV A,M||Load Acc with the data from memory|
|F004||47||MOV B,A||Copy the number into B|
|F005||37||STC||Set Carry Flag|
|F006||3F||CMC||Complement Carry Flag|
|F007||D6, 0A||SUI 0AH||Subtract 0AH from A|
|F009||DA, 11, F0||JC NUM||When carry is present, A is numeric|
|F00C||C6, 41||ADI 41H||Add 41H for Alphabet|
|F00E||C3, 14, F0||JMP STORE||Jump to store the value|
|F011||78||NUM||MOV A, B||Get back B to A|
|F012||C6||ADI 30H||Add 30H with A to get ASCII|
|F014||23||STORE||INX H||Point to next location to store address|
|F015||77||MOV M,A||Store A to memory location pointed by HL pair|
|F016||76||HLT||Terminate the program|