8085 code to convert binary number to ASCII code

• Difficulty Level : Expert
• Last Updated : 22 May, 2018

Problem – Assembly level program in 8085 which converts a binary number into ASCII number.

Example –

Assumptions – Binary number which have to convert in ASCII value is stored at memory location 2050 and output will be displayed at memory location 3050 and 3051.

Algorithm –

1. Load the content of 2050.
2. Then separate the LSB of the no. using ANI 0F instruction and MSB of the number by again loading the content of 2050 and rotate it by one bit 4 times to get reverse of the number and then again use ANI 0F to separate the digit.
3. If the digit is more than or equal to 0A (in hexadecimal) then add 37 otherwise add 30 to convert into ASCII value (For checking the number is greater than or equal to A then use instruction: CPI 0A and then check the carry flag, if it is 0 then it means digit is greater than or equal to A and if 1 digit is less than A).
4. Now Store the ASCII values of both the digits in 3050 and 3051 respectively.

Program –

Main routine:

2000 LDA 2050 A<-[2050]
2003 CALL 2500 go to address 2500
2006 STA 3050 A->[3050]
2009 LDA 2050 A<-[2050]
200C RLC Rotate the number by one bit to left without carry
200D RLC Rotate the number by one bit to left without carry
200E RLC Rotate the number by one bit to left without carry
200F RLC Rotate the number by one bit to left without carry
2010 CALL 2500 go to address 2500
2013 STA 3051 A->[3051]
2016 HLT Terminates the program

Sub routine:

2500 ANI 0F A<-[A] AND 0F
2502 CPI 0A [A]-0A

Explanation –

Main routine:

1. LDA 2050: This instruction will load the number from address 2050 to the accumulator.
2. CALL 2500: This instruction will stop executing the main routine instructions after it and will move to the subroutine address 2500 for performing the subtask and after performing subroutine instructions it will come back to mainroutine and execute the instructions after CALL 2500.
3. STA 3050: This instruction will store the result (performed in subroutine) of Accumulator to address 3050.
4. LDA 2050: This instruction will again load the number from address 2050 to the accumulator as the earlier loaded number is changed in accumulator.
5. RLC: Rotate the contents of Accumulator by one bit left side without carry.
6. RLC: Rotate the contents of Accumulator by one bit left side without carry.
7. RLC: Rotate the contents of Accumulator by one bit left side without carry.
8. RLC: Rotate the contents of Accumulator by one bit left side without carry.
(Applying RLC 4 times it will reverse the contents of the Accumulator)
9. 9. CALL 2500: This instruction will stop executing the main routine instructions after it and will move to the subroutine address 2500 for performing the subtask and after performing subroutine instructions it will come back to mainroutine and execute the instructions after CALL 2500.
10. 10. STA 3051: This instruction will store the result (performed in subroutine) of Accumulator to address 3051.
11. 11. HLT: This instruction will terminate the program.

Sub routine:

1. ANI 0F: This instruction will separate the LSB of the number present in Accumulator and stores the result back in Accumulator.
2. CPI 0A: This instruction will compare the content of Accumulator with 0A i.e. [A]-0A.
3. JNC 205A: If the carryflag becomes 0 then it will jump to 205A otherwise move to the next instruction.
4. ADI 30: It will add 30 to the content of Accumulator and again store the result back in Accumulator.
5. RET: Now it will move back to the main routine after the next instruction of CALL and start executing instructions of main routine.
6. It will add 37 to the content of Accumulator and again store the result back in Accumulator.
7. RET: Now it will move back to the main routine after the next instruction of CALL and start executing instructions of main routine.
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