8085 Programming : store the nos in a memory loc. In the reverse order in another memory location

               MVI C, 0AH  ; Initialize counter  

               LXI H, 2200H  ; Initialize source memory pointer  

               LXI D, 2309H  ; Initialize destination memory pointer  

BACK:    MOV A, M  ; Get byte from source memory block  

               STAX D  ; Store byte in the destination memory block  

               INX H  ; Increment source memory pointer  

               DCX D  ; Decrement destination memory pointer   

               DCR C  ; Decrement counter   

               JNZ BACK  ; If counter  0 repeat  

               HLT ; Terminate program execution

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8085 Programming : In a set of nos add only the odd nos

                 LDA 2200H  

           MOV C, A  ; Initialize counter  

           LXI H, 2201H  ; Initialize pointer  

           MVI E, 00  ; Sumlow = 0  

           MOV D, E ; Sumhigh = 0  

BACK:      MOV A, M  ; Get the number  

           ANI 01H   ; Mask Bit1 to Bit7  

           JZ SKIP                  ; Do not add if number is even  

           MOV A, E  ; Get the lower byte of sum  

                 ADD M     ; Sum = sum + data  

           MOV E, A   JNC SKIP  ; Store result in E register  

           INR D   ; Add carry to MSB of SUM  

SKIP:        INX H   ; Increment pointer  

           DCR C     ; Decrement counter  

          JNZ BACK  ; Check if counter 0 repeat  

          MOV A, E  

          STA 2300H  ; Store lower byte  

          MOV A, D  

          STA 2301H ; Store higher byte  

         HLT ; Terminate program execution  

  

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8085 Programming : In a set of nos add only the even nos

            LDA 2200H  

      MOV C, A  ; Initialize counter  

            MVI B, 00H ; sum  =  0  

      LXI H, 2201H ; Initialize pointer  

BACK: MOV A, M  ; Get the number  

      ANI, 01H   ; Mask Bit1 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 2210H ; store sum  

      HLT    ; Terminate program execution  

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8085 Programming : To find the no of 1’s in the byte

            MVI B,00H   

            MVI C,08H  

            MOV A,D  

BACK: RAR  

            JNC SKIP  

            INR B  

SKIP:   DCR C  

            JNZ BACK  

            HLT  

  

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8085 Programming : Block data transfer

              MVI C, 0AH  ; Initialize counter  

              LXI H, 2200H  ; Initialize source memory pointer  

              LXI D, 2300H  ; Initialize destination memory pointer  

BACK:   MOV A, M  ; Get byte from source memory block  

              STAX D  ; Store byte in the destination memory block  

              INX H  ; Increment source memory pointer  

              INX D  ; Increment destination memory pointer   

              DCR C  ; Decrement counter   

              JNZ BACK  ; If counter  0 repeat  

              HLT ; Terminate program execution  

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8085 Programming : Adding 2 Hexa-Decimal nos with carry

             LDA 2200H  

             MOV C,A ; Initialize counter  

             LXI H, 2201H ; Initialize pointer  

             SUB A  ; Sum low = 0  

             MOV B,A ; Sumhigh = 0  

BACK:  ADD M ; Sum = sum + data  

             JNC SKIP  

             INR B   ; Add carry to MSB of SUM  

SKIP:    INX H   ; Increment pointer  

            DCR C ; Decrement counter  

            JNZ BACK  ; Check if counter 0 repeat  

            STA 2300H 

            MOV A,B  ; Store lower byte  

            STA 2301H  ; Store higher byte  

            HLT   ; Terminate program execution

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8085 Programming : Adding 2 Hexa-decimal nos without carry

  LDA 2200

             MOV C, A ;  Initialize counter

             SUB A  ;  sum = 0

             LXI H, 2201H ;  Initialize pointer

BACK:  ADD M   ; SUM = SUM + data

             INX H  ;  increment pointer

             DCR C ;  Decrement counter

             JNZ BACK ;  if counter  0 repeat

             STA 2300H ;  store sum

             HLT

         

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8085 Programming: Adding 2 BCD nos without carry

MOV A, L ; Get lower 2 digits of no. 1  

ADD E ; Add two lower digits  

DAA  ; Adjust result to valid BCD  

STA 2300H ; Store partial result  

MOV A, H ; Get most significant 2 digits of no. 2  

ADC D ; Add two most significant digits   

DAA  ; Adjust result to valid BCD   

STA 2301H ; Store partial result  

HLT ; Terminate program execution  

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8085 Programming : ADDITION OF 16BIT NUMBERS

ADDITION OF TWO 16BIT NUMBERS SUM 16 BITS OR MORE

Manually strore 1st 16 bit no in the memory location C050 & C051 in reverse order

Manually store 2nd 16 bit no in the memory location C052 & C053 in reverse order

Result is stored in C053, C054 & C055 in reverse order



          LHLD C050

         XCHG

         LHLD C052

         MVI C,00

         DAD D

         JNC AHEAD

         INR C



AHEAD:   SHLD C054

        MOV A,C

        STA C056

        HLT



EXAMPLE-> A645+9B23=014168

STORE-> C050=45,C051=A6,C052=23,C053=9B

Answer-> C054=68,C055=41,C056=01

// 45H,A6H,23H,9BH

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8085 Programming : 8 BIT DECIMAL SUBSTRACTION

 

If 2nd no is greater than 1st no then the answer will in 2's complement

Manually strore 1st 8 bit no in the memory location C050

Manually store 2nd 8 bit no in the memory location C051

Result is stored in C052



  LXI H,C051

  MVI A,99

  SUB M

  INR A

  DCX H

  ADD M

  DAA

  STA C052

  HLT

 EXAMPLE-> 99-48=51

 STORE-> C050=99,C051=48



OUTPUT:

C052=51

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8085 Programming: Exchange the contents of memory locations

Exchange the contents of memory locations 2000H and 4000H.

Program 1:



LDA 2000H : Get the contents of memory location 2000H into accumulator



MOV B, A : Save the contents into B register



LDA 4000H : Get the contents of memory location 4000H into accumulator



STA 2000H  : Store the contents of accumulator at address 2000H



MOV A, B : Get the saved contents back into A register



STA 4000H : Store the contents of accumulator at address 4000H









Program 2:



LXI H 2000H : Initialize HL register pair as a pointer to memory location 2000H.



LXI D 4000H : Initialize DE register pair as a pointer to memory location 4000H.



MOV B, M  : Get the contents of memory location 2000H into B register.



LDAX D : Get the contents of memory location 4000H into A register.



MOV M, A : Store the contents of A register into memory location 2000H.



MOV A, B : Copy the contents of B register into accumulator.



STAX D : Store the contents of A register into memory location 4000H.



HLT : Terminate program execution.



Note: In Program 1, direct addressing instructions are used, whereas in Program 2, indirect addressing instructions are used.

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8085 Programming: Store 8 bit data in Memory

Program 1:

MVI A, 52H : Store 32H in the accumulator



STA 4000H : Copy accumulator contents at address 4000H



HLT : Terminate program execution

Program 2:

LXI H : Load HL with 4000H



MVI M : Store 32H in memory location pointed by HL register pair (4000H)



HLT : Terminate program execution



Note: The result of both programs will be the same. In program 1 direct addressing instruction is used, whereas in program 2 indirect addressing instruction is used.

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8085 Programming: 8 BIT MULTIPLICATION: PRODUCT 16-BIT

       LHLD 2501

       XCHG

       LDA 2503

       LXI H,0000

       MVI C,08



LOOP:     DAD H

       RAL

       JNC AHEAD

       DAD D



AHEAD:  DCR C

       JNZ LOOP

       SHLD 2504

       HLT





LSB OF MULTIPLICAND - 84H, MSB OF MULTIPLICAND - 00H ,MULTIPLIER - 56H





ANSWER

AT ADDRESS 2504 - 58H, LSBs OF PRODUCT

AT ADDRESS 2505 - 2CH, MSB sOF PRODUCT

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8085 Programming: 8 BIT DIVISION

      LHLD 2501

      LDA 2503

      MOV B,A

      MVI C,08



LOOP:   DAD H

     MOV A,H

     SUB B

     JC AHEAD

     MOV H,A

     INR L



AHEAD: DCR C

      JNZ LOOP

      SHLD 2504

      HLT







INPUT:



LSB OF DIVIDEND - 9BH, MSB OF DIVIDEND - 48H , DIVISOR - 1AH







ANSWER

AT ADDRESS 2504 - F2H, QUOTIENT

AT ADDRESS 2505 - 07H, REMAINDER

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8085 Programming: 2's COMPLEMENT OF AN 8-BIT NUMBER

The number to be complemented is stored in C050. Answer is stored in C051



  LDA C050

  CMA

  INR A

  STA C051

  HLT



EXAMPLE-> C050=96

Answer-> C051=6A

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8085 Programming: 2's COMPLEMENT OF A 16-BIT NUMBER

The 16bit number is stored in C050,C051. The answer is stored in C052,C053



  LXI H,C050

  MVI B,00

  MOV A,M

  CMA

  ADI 01

  STA C052

  JNC GO

  INR B



GO:   INX H

  MOV A,M

  CMA

  STA C053

  HLT

EXAMPLE-> C050=8C,C051=5B

Answer-> C052=74,C053=A4

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8085 Programming: 1's COMPLEMENT OF AN 8-BIT NUMBER

The number to be complemented is stored in C050. Answer is stored in C051



  LDA C050

  CMA

  STA C051

  HLT





EXAMPLE-> C050=96

Answer-> C051=69

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8085 Programming: 1's COMPLEMENT OF A 16-BIT NUMBER

The 16bit number is stored in C050,C051

The answer is stored in C052,C053



  LXI H,C050

  MOV A,M

  CMA

  STA C052

  INX H

  MOV A,M

  CMA

  STA C053

  HLT

 EXAMPLE-> C050=85,C051=54

 Answer-> C052=7A,C053=AB

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8085 Programming :

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