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8086 program to divide a 16 bit number by an 8 bit number

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Problem – Write an assembly language program in 8086 microprocessor to divide a 16 bit number by an 8 bit number.

Example –
division of 16 bit number by an 8 bit number

Algorithm –

  1. Assign value 500 in SI and 600 in DI
  2. Move the contents of [SI] in BL and increment SI by 1
  3. Move the contents of [SI] and [SI + 1] in AX
  4. Use DIV instruction to divide AX by BL
  5. Move the contents of AX in [DI].
  6. Halt the program.

Assumption – Initial value of each segment register is 00000.

Calculation of physical memory address –
Memory Address = Segment Register * 10(H) + offset,
where Segment Register and Offset is decided on the basis of following table.

Instruction fetchingCode SegmentInstruction Pointer
Data operationData SegmentBase Register [BX], Displacement [DISP]
Stack operationStack SegmentStack Pointer (SP), Base Pointer (BP)
String as a sourceData SegmentSource Indexed (SI)
String as a destinationExtra SegmentDestination Indexed (DI)

Program –

0400MOV SI, 500SI <- 500
0403MOV DI, 600DI <- 600
0406MOV BL, [SI]BL <- [SI]
0408INC SISI <- SI + 1
0409MOV AX, [SI]AX <- [SI]
040DMOV [DI], AX[DI] <- AX
040FHLTEnd of program

Explanation – Registers used AX, BL, SI, DI

  1. MOV SI, 500 assigns 500 to SI
  2. MOV DI, 600 assigns 600 to DI
  3. MOV BL, [SI] moves the content of [SI] to BL register i.e. value of divisor will be stored in BL
  4. INC SI increment the content of SI by 1
  5. MOV AX, [SI] moves the content of [SI] and [SI + 1] to AX register i.e. value of dividend will be stored in AX
  6. DIV BL divide the content of AX by BL, after execution of this instruction the quotient get stored in AL and remainder in AH
  7. MOV [DI], AX moves the content of AX to [DI]
  8. HLT stops executing the program and halts any further execution
Last Updated : 23 May, 2018
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