Problem – Write an assembly language program in 8085 microprocessor to subtract two 16 bit numbers.
- Starting address of program: 2000
- Input memory location: 2050, 2051, 2052, 2053
- Output memory location: 2054, 2055
INPUT: (2050H) = 19H (2051H) = 6AH (2052H) = 15H (2053H) = 5CH OUTPUT: (2054H) = 04H (2055H) = OEH
Hence we have subtracted two 16 bit numbers.
- Get the LSB in L register and MSB in H register of 16 Bit number.
- Exchange the content of HL register with DE register.
- Again Get the LSB in L register and MSB in H register of 16 Bit number.
- Subtract the content of L register from the content of E register.
- Subtract the content of H register from the content of D register and borrow from previous step.
- Store the result in memory location.
|2000||LHLD 2050||Load H-L pair with address 2050|
|2003||XCHG||EXCHANGE H-L PAIR WITH D-E PAIR|
|2004||LHLD 2052||Load H-L pair with address 2052|
|2007||MVI C, 00||C<-00H|
|2009||MOV A, E||A<-E|
|200E||MOV A, D||A<-D|
|200F||SBB H||SUBTRACT WITH BORROW|
|2013||HLT||TERMINATES THE PROGRAM|
- LHLD 2050: load HL pair with address 2050.
- XCHG: exchange the content of HL pair with DE.
- LHLD 2052: load HL pair with address 2050.
- MOV A, E: move the content of register E to A.
- SUB L: subtract the content of A with the content of register L.
- STA 2054: store the result from accumulator to memory address 2054.
- MOV A, D: move the content of register D to A.
- SBB H: subtract the content of A with the content of register H with borrow.
- STA 2055: store the result from accumulator to memory address 2055.
- HLT: stops executing the program and halts any further execution.
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- 8085 program to add two 16 bit numbers
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- 8085 program to divide two 8 bit numbers
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