8085 program to add two 8 bit numbers
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 –
- Load the first number from memory location 2050 to accumualtor.
- Move the content of accumulator to register H.
- Load the second number from memory location 2051 to accumaltor.
- Then add the content of register H and accumulator using “ADD” instruction and storing result at 3050
- 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 –
- LDA 2050 moves the contents of 2050 memory location to the accumulator.
- MOV H, A copies contents of Accumulator to register H to A
- LDA 2051 moves the contents of 2051 memory location to the accumulator.
- 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
- MOV L, A copies contents of A (34) to L
- MVI A 00 moves immediate data (i.e., 00) to A
- 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
- MOV H, A copies contents of A (01) to H
- SHLD 3050 moves the contents of L register (34) in 3050 memory location and contents of H register (01) in 3051 memory location
- HLT stops executing the program and halts any further execution
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