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8085 program to multiply two 16-bit numbers
• Difficulty Level : Expert
• Last Updated : 19 Feb, 2021

Problem – Write an assembly language program in 8085 microprocessor to multiply two 16 bit numbers.

Assumption –

• Starting address of program: 2000

• Input memory location: 2050, 2051, 2052, 2053

• Output memory location: 2054, 2055, 2056, 2057

Example –

```INPUT:
(2050H) = 04H
(2051H) = 07H
(2052H) = 02H
(2053H) = 01H
OUTPUT:
(2054H) = 08H
(2055H) = 12H
(2056H) = 07H
(2057H) = O0H```

RESULT:
Hence we have multiplied two 16 bit numbers.

Algorithm –

1. Load the first data in HL pair.

2. Move content of HL pair to stack pointer.

3. Load the second data in HL pair and move it to DE.

4. Make H register as 00H and L register as 00H.

5. ADD HL pair and stack pointer.

6. Check for carry if carry increment it by 1 else move to next step.

7. Then move E to A and perform OR operation with accumulator and register D.

8. The value of operation is zero, then store the value else goto step 3.

Program –

Explanation – Registers B, C, D, E, H, L and accumulator are used for general purpose.

1. LHLD 2050: load HL pair with address 2050.

2. SPHL: save the content of HL in stack pointer.

3. LHLD 2052: load H-L pair with address 2052.

4. XCHG: exchange the content of HL pair with DE.

5. LXI H, 0000H: make H as 00H and L as 00H.

6. LXI B, 0000H: make B as 00h and C as 00H

7. DAD SP: ADD HL pair and stack pointer.

8. JNC 2013: jump to address 2013 if there will be no carry.

9. INX B: increments BC register with 1.

10. DCX D: decrements DE register pair by 1.

11. MOV A, E: move the content of register E to accumulator.

12. ORA D: or the content of accumulator and D register.

13. JNZ 200E: jump to address 200E if there will be no zero.

14. SHLD 2054: store the result to memory address 2054 and 2055 from HL pair register.

15. MOV L, C: move the content of register C to L.

16. MOV H, B: move the content of register B to H.

17. SHLD 2056: store the result to memory address 2056 and 2057 from HL pair register.

18. HLT: terminates the program.

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