8085 program to add two 16 bit numbers

Problem – Write an assembly language program to add two 16 bit numbers by using:

  • (a) 8 bit operation
  • (b) 16 bit operation

Example –


(a) Addition of 16 bit numbers using 8 bit operation – It is a lengthy method and requires more memory as compared to 16 bit operation.

Algorithm –

  1. Load the lower part of first number in B register
  2. Load the lower part of second number in A (accumulator)
  3. Add both the numbers and store
  4. Load the higher part of first number in B register
  5. Load the higher part of second number in A (accumulator)
  6. Add both the numbers with carry from the lower bytes (if any) and store at the next location

Program –

MEMORY ADDRESS MNEMONICS COMMENTS
2000 LDA 2050 A ← 2050
2003 MOV B, A B ← A
2004 LDA 2052 A ← 2052
2007 ADD B A ← A+B
2008 STA 3050 A → 3050
200B LDA 2051 A ← 2051
200E MOV B, A B ← A
200F LDA 2053 A ← 2053
2012 ADC B A ← A+B+CY
2013 STA 3051 A → 3051
2016 HLT Stops execution

Explanation –

  1. LDA 2050 stores the value at 2050 in A (accumulator)
  2. MOV B, A stores the value of A into B register
  3. LDA 2052 stores the value at 2052 in A
  4. ADD B add the contents of B and A and store in A
  5. STA 3050 stores the result in memory location 3050
  6. LDA 2051 stores the value at 2051 in A
  7. MOV B, A stores the value of A into B register
  8. LDA 2053 stores the value at 2053 in A
  9. ADC B add the contents of B, A and carry from the lower bit addition and store in A
  10. STA 3051 stores the result in memory location 3051
  11. HLT stops execution

(b) Addition of 16 bit numbers using 16 bit operation – It is a very short method and less memory is also required as compared to 8 bit operation.

Algorithm –

  1. Load both the lower and the higher bits of first number at once
  2. Copy the first number to another register pair
  3. Load both the lower and the higher bits of second number at once
  4. Add both the register pairs and store the result in a memory location

Program –

MEMORY ADDRESS MNEMONICS COMMENTS
2000 LHLD 2050 A ← 2050
2003 XCHG D ← H & E ← L
2004 LHLD 2052 A ← 2052
2007 DAD D H ← H+D & L ← L+E
2008 SHLD 3050 A → 3050
200B HLT Stops execution

Explanation –

  1. LHLD 2050 loads the value at 2050 in L register and that in 2051 in H register (first number)
  2. XCHG copies the content of H to D register and L to S register
  3. LHLD 2052 loads the value at 2052 in L register and that in 2053 in H register (second number)
  4. DAD D adds the value of H with D and L with E and stores the result in H and L
  5. SHLD 3050 stores the result at memory location 3050
  6. HLT stops execution


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