Programming in 8085

Let’s see some simple example to demonstrate the use of some important instructions of 8085.

The memory addresses given in the program are for a particular microprocessor kit. These addresses can be changed to suit the microprocessor kit available in your system.

Store 8-bit data in memory

Program

Store 8-bit data in memory using direct addressing

  MVI A, 49H  : “Store 49H in the accumulator”  STA 2501H       : “Copy accumulator contents at address 2501H”  HLT             : “Stop”  

Store 8-bit data in memory using indirect addressing

  LXI H   : “Load H-L pair with 2501H”  MVI M      : “Store 49H in memory location pointed by H-L register pair (2501H)”  HLT     : “Stop”  

Add two 8-bit numbers

Example

(2501 H) = 99H   (2502 H) = 39H   Result (2503 H) = 99H + 39H = D2H    Since,     1 0 0 1 1 0 0 1 (99H)    + 0 0 1 1 1 0 0 1 (39H)      1 1 0 1 0 0 1 0 (D2H)

Program

  LXI H, 2501H  : “Get address of first number in H-L pair. Now H-L points to 2501H”  MOV A, M      : “Get first operand in accumulator”  INX H         : “Increment content of H-L pair. Now, H-L points 2502H”  ADD M        : “Add first and second operand”  INX H       : “H-L points 4002H”  MOV M, A      : “Store result at 2503H”  HLT           : “Stop”  

Subtract two 8-bit numbers

Example

(2501 H) = 49H   (2502 H) = 32H   Result (2503 H) = 49H - 32H = 17H

Program

  LXI H, 2501H   :  “Get address of first number in H-L pair. Now H-L points to 2501H”  MOV A, M       :  “Get first operand in accumulator”  INX H          :  “Increment content of H-L pair. Now, H-L points 2502H”  SUB M         :  “Subtract first to second operand”  INX H        :  “H-L points 4002H”  MOV M, A       :  “Store result at 2503H”  HLT            :  “Stop”  

Add two 16-bits numbers

Add the 16-bit number in memory locations 2501H and 2502H to the 16-bit number in memory locations 2503H and 2504H. The most significant eight bits of the two numbers to be added are in memory locations 2502H and 4004H. Store the result in memory locations 2505H and 2506H with the most significant byte in memory location 2506H.

Example

(2501H) = 15H  (2502H) = 1CH  (2503H) = B7H  (2504H) = 5AH    Result = 1C15 + 5AB7H = 76CCH    (2505H) = CCH  (2506H) = 76H

Program

Add two 16-bits number with ADD and ADC instruction

  LHLD 2501H  : “Get 1st 16-bit number in H-L pair”  XCHG         : “Save 1st 16-bit number in DE”  LHLD 2503H : “Get 2nd 16-bit number in H-L pair”  MOV A, E   : “Get lower byte of the 1st number”  ADD L      : “Add lower byte of the 2nd number”  MOV L, A   : “Store result in L-register”  MOV A, D    : “Get higher byte of the 1st number”  ADC H       : “Add higher byte of the 2nd number with CARRY”  MOV H, A    : “Store result in H-register”  SHLD 4004H  : “Store 16-bit result in memory locations 2505H and 2506H”  HLT         : “Stop”   

Add two 16-bits numbers with DAD instruction

  LHLD 2501H : “Get 1st 16-bit number”  XCHG : “Save 1st 16-bit number in DE”  LHLD 2503H : “Get 2nd 16-bit number in H-L”  DAD D     : “Add DE and HL”  SHLD 2505H : “Store 16-bit result in memory locations 2505H and 2506H”.  HLT : “Stop”  

Subtract two 16-bit numbers

Example

(2500H) = 19H  (2501H) = 6AH  (2504H) = 15H (2503H) = 5CH  Result    = 6A19H ? 5C15H = OE04H  (2504H) = 04H  (2505H) = OEH

Program

  LHLD 2500H  : “Get first 16-bit number in HL”  XCHG        : “Save first 16-bit number in DE”  LHLD 2502H  : “Get second 16-bit number in HL”  MOV A, E    : “Get lower byte of the first number”  SUB L       : “Subtract lower byte of the second number”  MOV L, A    : “Store the result in L register”  MOV A, D    : “Get higher byte of the first number”  SBB H       : “Subtract higher byte of second number with borrow”  MOV H, A    : “Store l6-bit result in memory locations 2504H and 2505H”  SHLD 2504H  : “Store l6-bit result in memory locations 2504H and 2505H”  HLT         : “Terminate program execution”  

Add contents of two memory locations

Example

(2500H) = 7FH  (2501H) = 89H  Result    = 7FH + 89H = lO8H  (2502H) = 08H  (2503H) = 01H

Program

  LXI H, 2500H    : “HL Points 2500H”  MOV A, M        : “Get first operand”  INX H         : “HL Points 2501H”  ADD M        : “Add second operand”  INX H         : “HL Points 2502H”  MOV M, A    : “Store the lower byte of result at 2502H”  MVIA, 00     : “Initialize higher byte result with 00H”  ADC A         : “Add carry in the high byte result”  INX H          : “HL Points 2503H”  MOV M, A     : “Store the higher byte of result at 2503H”  HLT            : “Terminate program execution”  

Finding 1’s complement of a number

To obtain one’s complement of a number its 0 bits are replaced by 1 and 1 by 0.

Example 1

(2501H) = 96 H = 1001 0110  (9)    (6)  One's complement = 0110 1001 = 69 H  Result = (2502H) = 69H

Example 2

(2501H) = E4H  Result = (2502H) = 1BH

Program

The number is placed in the memory location 2501 H.

The result is stored in the memory location 2502 H.

  LDA 2501H : “Get the number IN accumulator”  CMA       : “take its complement”  STA 2502H : “Store result in 2502H”  HLT       : “Stop”  

Finding 2’s complement of a number

2’s complement of a number is obtained by adding 1 to the 1’s complement of the number.

Example

To find the two's complement of 96.  96 = 1001 0110   1's complement = 0110  1001 = 69   +     0000  0001   2's complement = 0110  1010 = 6A

Program

The number is placed in the memory location 2501 H.

The result is to be stored in the memory location 2502 H.

  LDA 2501 H  : “Get data in accumulator”  CMA        : “Take its 1’s complement”  ADI, 01 H  : “Add one in the number”  STA 2502 H  : “Store the result in 2502 H”  HLT        : “Stop”  

Count number of 1’s in a number

Example

2501 H = 04  2502 H = 34 H  2503 H = A9H  2504 H = 78H  2505 H = 56H  Result = 2503 H = A9H

Program

Count number of 1’s of the content of the register D and store the count in the register B.

  MVI B, 00H  MVI C, 08H  MOV A, D  BACK: RAR  JNC SKIP  INR B  SKIP: DCR C  JNZ BACK  HLT  

Find larger of two numbers

Example

2501H = 98 H  2502H = 87H  Result = 98H (2503H)

Program

The first number 98H is placed in the memory location 2501 H.
The second number 87H is placed in the memory location 2502H.
The result is stored in the memory location 2503 H.

LXI H, 2501H : "Address of first number in H-L pair"  MOV A, M : "1stt number in accumulator"  INX H     : "Address of 2^nd number in H-L pair"  CMP M     : "compare 2^nd number with 1^st number"  JNC AHEAD : "No, larger is in accumulator. Go to AHEAD"  MOV A, M : "Yes, get 2^nd number in the accumulator"  STA 2503 H : "Store larger number in 2503H"  HLT         : "Stop"

Find smaller of two numbers

Example

2501H = 84 H  2502H = 99 H  Result = 84 H(2503H)

Program

The first number 84H is placed in the memory location 2501 H.

The second number 99H is placed in the memory location 2502H.

The result is stored in the memory location 2503 H.

LXI H, 2501H : "Address of first number in H-L pair"  MOV A, M : "1stt number in accumulator"  INX H     : "Address of 2^nd number in H-L pair"  CMP M     : "compare 2^nd number with 1st number"  JC AHEAD : "Yes, smaller number is in accumulator. Go to AHEAD"  MOV A, M : "No, get 2^nd number in the accumulator"  STA 2503 H : "Store smaller number in 2503H"  HLT         : "Stop"

Calculate the sum of series of even numbers

Example

2500 H = 4H  2501 H = 20H  2502 H = 15H  2503 H = 13H  2504 H = 22H  Result = 2505 H = 20+22= 42H

Program

The numbers are placed in the memory locations 2501 to 2504H.

The sum is to be stored in the memory location 2450H.

As there are 4 numbers in the series, count = 04

The initial value of the sum is made 00. The even number of the series are taken one by one and added to the sum.

  LDA 2500H  MOV C, A        : “Initialize counter”  MVI B, 00H         : “sum = 0”  LXI H, 2501H    : “Initialize pointer”  BACK: MOV A, M  : “Get the number”  ANI 01H         : “Mask Bit l to Bit7”  JNZ SKIP        : “Don’t add if number is ODD”  MOV A, B        : “Get the sum”  ADD M           : “SUM = SUM + data”  MOV B, A        : “Store result in B register”  SKIP: INX H     : “increment pointer”  DCR C           : “Decrement counter”  JNZ BACK        : “if counter 0 repeat”  STA 2505H       : “store sum”  HLT             : “Stop”  

Calculate the sum of series of odd numbers

Example

2500 H = 4H  2501 H = 9AH  2502 H = 52H  2503 H = 89H  2504 H = 3FH  Result = 2505 H = 89H + 3FH= C8H

Program

The numbers are placed in the memory locations 2501 to 2504H.

The sum is to be stored in the memory location 2450H.

As there are 4 numbers in the series, count = 04

The initial value of the sum is made 00. The odd number of the series are taken one by one and added to the sum.

  LDA 2500H  MOV C, A       : “Initialize counter”  LXI H, 2501H    : “Initialize pointer”  MVI E, 00       : “Sum low = 0”  MOV D, E           : “Sum high = 0”  BACK: MOV A, M  : “Get the number”  ANI 01H         : “Mask Bit 1 to Bit-7”  JZ SKIP         : “Don’t add if number is even”  MOV A, E        : “Get the lower byte of sum”  ADD M           : “Sum = sum + data”  MOV E, A        : “Store result in E register”  JNC SKIP  INR D           : “Add carry to MSB of SUM”  SKIP: INX H     : “Increment pointer”  

Find the square of given number

Program

Find the square of 07 (decimal) using lookup table techniques.

The number 07 D is in the memory.

The result is to be stored in the memory location 2501H.

The table for square is stored from 2600 to 2609 H.

  LDA 2500H: “Get data in accumulator”  MOV L, A: “Get data in register L”  MVI H, 26H: “Get 26 in register H”  MOV A, M: “Square of data in accumulator”  STA 2501 H: “Store Square in 2501 H”  HLT: “Stop”  

Separate even numbers from given numbers

Program

Let’s see the program to separate even numbers from the given list of 50 numbers and store them in other location starting from 2600H.

Starting location of 50 number list is 2500H.

  LXI H, 2500H     : “Initialize memory pointer 1”  LXI D, 2600H    : “Initialize memory pointer2”  MVI C, 32H        : “Initialize counter”  BACK:MOV A, M     : “Get the number”  ANI 01H          : “Check for even number”  JNZ SKIP         : “If ODD, don’t store”  MOV A, M         : “Get the number”  STAX D           : “Store the number in result list”  INX D            : “Increment pointer 2”  SKIP: INX H      : “Increment pointer 1”  DCR C            : “Decrement counter”  JNZ BACK         : “If not zero, repeat”  HLT              : “Stop”  

Next TopicMicroprocessor Applications

Programming in 8085

Programming in 8085

Store 8-bit data in memory

Add two 8-bit numbers

Subtract two 8-bit numbers

Add two 16-bits numbers

Subtract two 16-bit numbers

Add contents of two memory locations

Finding 1’s complement of a number

Finding 2’s complement of a number

Count number of 1’s in a number

Find larger of two numbers

Find smaller of two numbers

Calculate the sum of series of even numbers

Calculate the sum of series of odd numbers

Find the square of given number

Separate even numbers from given numbers

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