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8085 program to add two 8 bit numbers

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Problem – Write an assembly language program to add two 8 bit numbers stored at address 2050 and address 2051 in 8085 microprocessor. The starting address of the program is taken as 2000. 

Example – 


Algorithm – 

  1. Load the first number from memory location 2050 to accumulator.
  2. Move the content of accumulator to register H.
  3. Load the second number from memory location 2051 to accumulator.
  4. Then add the content of register H and accumulator using “ADD” instruction and storing result at 3050
  5. The carry generated is recovered using “ADC” command and is stored at memory location 3051

Program – 


Memory Address Mnemonics Comment
2000 LDA 2050 A<-[2050]
2003 MOV H, A H<-A
2004 LDA 2051 A<-[2051]
2007 ADD H A<-A+H
2008 MOV L, A L←A
2009 MVI A 00 A←00
200B ADC A A←A+A+carry
200C MOV H, A H←A
200D SHLD 3050 H→3051, L→3050
2010 HLT  

Explanation – 


  1. LDA 2050 moves the contents of 2050 memory location to the accumulator.
  2. MOV H, A copies contents of Accumulator to register H to A
  3. LDA 2051 moves the contents of 2051 memory location to the accumulator.
  4. ADD H adds contents of A (Accumulator) and H register (F9). The result is stored in A itself. For all arithmetic instructions A is by default an operand and A stores the result as well
  5. MOV L, A copies contents of A (34) to L
  6. MVI A 00 moves immediate data (i.e., 00) to A
  7. ADC A adds contents of A(00), contents of register specified (i.e A) and carry (1). As ADC is also an arithmetic operation, A is by default an operand and A stores the result as well
  8. MOV H, A copies contents of A (01) to H
  9. SHLD 3050 moves the contents of L register (34) in 3050 memory location and contents of H register (01) in 3051 memory location
  10. HLT stops executing the program and halts any further execution


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Last Updated : 28 Jun, 2022
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