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Transfer Block of N Bytes from Source to Destination

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Transfer block of N bytes from source to destination 

Algorithm for Transfer block of N bytes
Step I          :   Initialize the data in the source memory and destination memory.
Step II         :   Initialize SI and DI with source and destination address.
Step III       :   Initialize CX register with the count.
Step IV       :   Initialize the direction flag to zero.
Step V         :   Transfer the data block byte by byte to destination.
Step VI       :   Decrement CX.
Step VII      :   Check for count in CX, if not zero goto step V else goto step VIII.
Step VIII    :   Display the bytes in destination location.
Step IX       :   Stop.
 

Transfer Block of N Bytes Code   

 

.model small  
 .data  
        src_blk db 01, 02, 03, 04, 05, 06, 07, 08, 09, 0AH  
        dest_blk db 10 dup(?)  
        count dw 0AH  
 .code   
        mov  ax, @data          ; initialize data  
        mov  ds, ax  
        mov  es, ax  
        mov  si, offset src_blk     ; si to point to source block  
        mov  di, offset dest_blk    ; di to point to destination block  
        mov  cx, count           ; initialize counter  
        cld                    ; df=0  
 again :  rep   movsb                   ; transfer contents  
        mov  di, offset dest_blk    ; di to point to  
                               ; destination block  
        mov  bh, 0Ah            ; initialize counter  
 up:     mov  bl, [di]             ; store result in bl  
        mov  cx, 0204h    ; Count of digits to be  
                         displayed in    
                         ; ch and digits to be  
                         mrolled in cl   
 l1:     rol   bl, cl        ; roll bl so that msb comes  
                         to lsb  
        mov  dl, bl        ; load dl with data to be  
                         displayed  
        and  dl, 0fh       ; get only lsb  
        cmp  dl, 09h      ; check if digit is 0-9    
                         ; or letter A-F  
        jbe   l12  
        add  dl, 07h            ; if letter add 37H      
                         ; else only add 30H  
 l12:    add  dl, 30h  
        mov  ah, 02       ; Function 02 under  
                         ; INT 21H                        
        int   21h  
        dec  ch          ; Decrement Count  
        jnz   l1  
        dec  bh          ; decrement counter  
        inc   di  
        mov  ah, 02h            ; display space        
                         ; between bytes  
        mov  dl, ' '  
        int   21h  
        cmp  bh, 00h            ; repeat till all bytes  
                         ; are displayed  
        jne   up  
        mov  ah, 4ch      ; normal termination  
                         ; to dos  
        int   21h  
 end

 

 

Other Projects to Try:

  1. Add 8 Bit BCD Numbers
  2. BCD to 7 Segment Code Conversion
  3. 2s Complement of a Numbers Assembly Code
  4. How to find 1’s complement of a number
  5. Multiply Two 8 Bit Numbers using Add and Shift Method

2s Complement of a Numbers Assembly Code

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2s Complement of a Numbers Assembly Program

2s Complement of a Numbers Algorithm

Step I          :    Initialize the data memory.
Step II        :    Load the number in AX.
Step III       :    Initialize counter = 16 in CX.
Step IV       :    Rotate number by 1 bit position to left with carry.
Step V         :    Complement carry.
Step VI       :    Decrement count.
Step VII     :    Check if count = 0. If not goto step IV
Step VIII    :    Rotate once again by 1 bit to restore original carry back.
Step IX       :    2’s complement = 1’s complement in AX + 1
Step X        :    Display result.
Step XI       :    Stop.
    

2s Complement of a Numbers Assembly Code 
 .model small  
 .data  
 a dw 1234H  
 .code  
     mov  ax, @data  ; Initialize data section  
     mov  ds, ax  
     mov  ax, a      ; Load number1 in ax  
     mov  cx, 16         ; Load CX with count  
 up: rcl    ax, 1      ; Rotate ax by 1bit to left with carry  
     cmc            ; find 1's complement of bit  
     loop  up        ; check if all bits complemented, if not goto up  
     rcl    ax, 1      ; return carry back to original position  
     add   ax, 1      ; 2's complement=1's complement+1  
     mov  ch, 04h    ; Count of digits to be displayed  
     mov  cl, 04h    ; Count to roll by 4 bits                  
     mov  bx, ax         ; Result in reg bx  
 l2:  rol    bx, cl     ; roll bl so that msb comes to lsb                
     mov  dl, bl      ; load dl with data to be displayed  
     and   dl, 0fH    ; get only lsb  
     cmp  dl, 09     ; check if digit is 0-9 or letter A-F            
     jbe   l4  
     add   dl, 07     ; if letter add 37H else only add 30H  
 l4:  add   dl, 30H  
     mov  ah, 02         ; Function 2 under INT 21H  
                    ; (Display character)  
     int    21H  
     dec   ch        ; Decrement Count  
     jnz   l2  
     mov  ah, 4cH    ; Terminate Program  
     int    21H  
     end

 

Other Projects to Try:

  1. How to find 1’s complement of a number
  2. Add 8 Bit BCD Numbers
  3. Add Two 16 Bit BCD Numbers Code
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  5. Program to Add Two 16 Bit Numbers Assembly Code

BCD to 7 Segment Code Conversion

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BCD to 7 segment code conversion

BCD to 7 segment code conversion Algorithm

Step I       :    Initialize the data segment.
Step II     :    Load the offset of look up table in BX.
Step III   :    Load AL with key i.e. digit whose 7             
                      segment code is to be found.
Step IV    :    Compute 7 segment code using XLAT
                      instruction.
Step V     :    Display the result.
Step VI    :    Stop.

BCD to 7 segment code conversion Code

 .model small                                             
 .data  
 lookup db 01h, 04h, 1Ah, 09h, 02H  
 key db 02h  
 .code  
     mov  ax, @data      ; Initialize data section  
     mov  ds, ax  
     mov  bx, offset lookup ; Load offset of lookup in bx  
     mov  al, key             ; key no. in al  
     xlat                 ; translate byte in al  
     mov  bh, al          ; al = lookup(key)  
     mov  ch, 02h        ; Count of digits to be displayed  
     mov  cl, 04h         ; Count to roll by 4 bits  
 l2:  rol    bh, cl          ; roll bl so that msb comes to lsb  
     mov  dl, bh          ; load dl with data to be displayed  
     and   dl, 0fH         ; get only lsb  
     cmp  dl, 09          ; check if digit is 0-9 or letter A-F  
     jbe   l4  
     add   dl, 07          ; if letter add 37H else only add 30H  
 l4:  add   dl, 30H  
     mov  ah, 02              ; Function 2 under INT 21H (Display character)  
     int    21H   
     dec   ch            ; Decrement Count  
     jnz   l2     
     mov  ah, 4cH        ; Terminate Program  
     int    21H  
     end

 

Other Projects to Try:

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  3. BCD to decimal Conversion in Assembly Language
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  5. Add Two 16 Bit BCD Numbers Code

Hex to BCD Conversion in Assembly Language Code

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Hex to BCD Conversion in Assembly Language Code

Write a Program For Hex to BCD Conversion in Assembly language . 

 

We  have a 4 digit  Hex  number  whose equivalent binary number is to be found i.e. FFFF H. Initially we compare FFFF H with decimal 10000 ( 2710 H in Hex ). If number is greater than 10,000 we add it to DH register. Also, we subtract decimal 10,000 from FFFF H, each time comparison is made. Then we compare the number obtained in AX by 1000 decimal. Each time we subtract 1000 decimal from AX and add 1000 decimal to BX. Then we compare number obtained in AX by 100 decimals. Each time we subtract 100 decimal from AX and add 100 decimal to BX to obtain BCD equivalent. Then we compare number obtained in AX with 10 decimal. Each time we subtract 10 decimal from AX and we add 10 decimal to BX. Finally we add the result in BX with remainder in AX. The final result is present in register DH with contains the 5th bit if present and register AX.

Algorithm For Hex to BCD Conversion

Step I              :    Initialize the data segment.
Step II             :    Initialize BX = 0000 H and DH = 00H.
Step III           :    Load the number in AX.
Step IV           :    Compare number with 10000 decimal. If below goto step VII else goto step V.
Step V             :    Subtract 10,000 decimal from AX and add 1 decimal to DH
Step VI           :    Jump to step IV.
Step VII          :    Compare number in AX with 1000, if below goto step X else goto step VIII.
Step VIII        :    Subtract 1000 decimal from AX and add 1000 decimal to BX.
Step IX           :    Jump to step VII.
Step X             :    Compare the number in AX with 100 decimal if below goto step XIII
Step XI           :    Subtract 100 decimal from AX and add 100 decimal to BX.
Step XII         :    Jump to step X
Step XIII        :    Compare number in AX with 10. If below goto step XVI
Step XIV        :    Subtract 10 decimal from AX and add 10 decimal to BX..
Step XV          :    Jump to step XIII.
Step XVI        :    Add remainder in AX with result in BX.
Step XVII      :    Display the result in DH and BX.
Step XVIII     :    Stop.
Hex to BCD Conversion in Assembly Language Code 1

 Program For Hex to BCD Conversion Code

.model small  
 .stack 100  
 .code  
      mov ax, 0ffffh            ; hex number to find it's bcd  
      mov      bx, 0000  
      mov      dh, 0  
 l9 :     cmp     ax, 10000     ; if ax>10000  
      jb      l2  
      sub      ax, 10000        ; subtract 10000  
      inc      dh               ; add 1 to dh  
      jmp      l9  
 l2 :     cmp      ax, 1000     ; if ax>1000  
      jb      l4  
      sub      ax, 1000  
      add      bx, 1000h        ; add 1000h to result  
      jmp      l2  
 l4 :     cmp      ax, 100      ; if ax>100  
      jb      l6  
      sub      ax, 100  
      add      bx, 100h         ; add 100h to result  
      jmp      l4  
 l6 :     cmp      ax, 10       ; if ax>10  
      jb      l8  
      sub      ax, 10  
      add      bx, 10h          ; add 10h to result  
      jmp      l6  
 l8 :     add      bx, ax       ; add remainder   
                                ; to result  
      mov      ah, 02            
      mov      cx, 0204h        ; Count to display   
                                ; 2 digits  
      go:      rol dh, cl  
      mov      dl, dh  
      and      dl, 0fh  
      add      dl, 30h          ; display 2 msb digits       
      int      21h  
      dec      ch  
      jnz      go  
      mov      ch, 04h          ; Count of digits to be   
                                ; displayed  
      mov      cl, 04h          ; Count to roll by 4 bits  
 l12:     rol      bx, cl       ; roll bl so that msb   
                                ; comes to lsb                  
      mov      dl, bl           ; load dl with data to be   
                                ; displayed  
      and      dl, 0fH          ; get only lsb  
      cmp      dl, 09           ; check if digit is 0-9 or letter A-F      
      jbe      l14  
      add      dl, 07           ; if letter add 37H else only add 30H  
 l14:     add      dl, 30H  
      mov      ah, 02           ; Function 2 under INT 21H      (Display character)  
      int      21H  
      dec      ch               ; Decrement Count  
      jnz      l12  
      mov      ah, 4cH          ;  Terminate Program  
      int      21H  
 end

How to Run this Program

For Running this program you should have installed Tasm on you computer . If you have not installed Tasm  yet please install from Here .

 
 

C:\programs>tasm hex2bcd.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   hex2bcd.ASM
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  437k
C:\programs>tlink hex2bcd
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
C:\programs>hex2bcd
065535

 

Other Projects to Try:

  1. Add 8 Bit BCD Numbers
  2. Add Two 16 Bit BCD Numbers Code
  3. BCD to decimal Conversion in Assembly Language
  4. BCD to 7 Segment Code Conversion
  5. ASCII-Binary conversion in Assembly Language

BCD to decimal Conversion in Assembly Language

by 6 Comments

 Mobile payment system project

 

The project entitled Mobile payment system project is an application to assist mobile phone users to access the information of different bills, send and receive queries requested by user from anywhere and in anytime.

With the use of a single touch button the user is provided with the requested information as quickly as possible. Mobile payment system is focused for professionals as well as House wives and people who are busy can save their time to do their work in home.Mobile payment system project has lots of advanced accessibility features to facilitates users with a user-friendly as well as flexible interface.

When ever user chooses to launch Mobile payment system project, he will be asked for credentials and after that he is logged on to a window .Where he is asked to select which option or service he would like to access. Upon choosing we will be presented with a list of the particular service bill details. If the user selects the options provided on the screen given he is instantly provide with the information pertaining to that particular service.Some of the most unique features of Mobile payment system are accessing the data at the server and retrieving the data from the database.

Project Snapshots

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  5. BCD to 7 Segment Code Conversion

Program for Simulating AAA Instruction

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Objective 

Write a Program for Simulating AAA Instruction in assembly language.

Simulating AAA Instruction

AAA instruction stands for ASCII Adjust for Addition. It is used whenever we want to add two decimal digits which are Representative in ASCII code, without masking off “3” in upper nibble. In our program AAA instruction is not available in the instruction set of 8086. So we will first accept the two digits in AL and BL registers. Then we will mask the upper nibble i.e. “3” from AL and BL register by ANDing  with 0F H, so that we will get LSB of two numbers. Then the two numbers are added.
Result of addition is in AL. We will check if the result is valid BCD. If addition > 9, then result is invalid BCD and to make it valid we will add 6. If result is valid, display the result.

Algorithm for Simulating AAA Instruction

Step I          :   Initialize the data segment.
Step II        :   Load number 1 in AL.
Step III       :   Load number 2 in BL.
Step IV       :   Mask the number 1 and store result in AL.
Step V         :   Mask number 2 and store result in BL.
Step VI       :   Add = number 1 + number 2.
Step VII     :   Check if addition < 9. If yes go to step IX. else go to step VIII.
Step VIII    :   Add 6 to make the result valid.
Step IX       :   Display the result.
Step X        :   Stop.

 

Program for Simulating AAA Instruction 2

Program for Simulating AAA Instruction

 .model small  
 .data  
 a db 39H  
 b db 32H  
 .code  
      mov      ax, @data           ; Initialize data section  
      mov      ds, ax  
      mov      al, a               ; Load number1 in al  
      mov      bl, b               ; Load number2 in bl  
      and      al, 0fh             ; unmask numbers and result in al  
      and      bl, 0fh             ; unmask numbers and result in al  
      add      al, bl              ; add the numbers in al and bl   
      cmp      al, 09h             ; check if no is valid BCD   
      jb      next  
      add      al, 06h             ; for invalid BCD add 6 to        
                                   ; make it valid  
 next:      mov      ch, 02h       ; Count of digits to be        
                                   ; displayed  
      mov      cl, 04h             ; Count to roll by 4 bits  
      mov      bh, al              ; Result in reg bh  
 l2:     rol      bh, cl           ; roll bl so that msb comes   
                                   ; to lsb   
      mov      dl, bh              ; load dl with data to be        
                                   ; displayed  
      and      dl, 0fH             ; get only lsb  
      cmp      dl, 09              ; check if digit is 0-9 or        
                                   ; letter A-F  
      jbe      l4  
      add      dl, 07              ; if letter add 37H        
                                   ; else only add 30H  
 l4:     add      dl, 30H  
      mov      ah, 02              ; Function 2 under INT        
                                   ; 21H (Display character)  
      int      21H  
      dec      ch                  ; Decrement Count  
      jnz      l2                                
      mov      ah, 4cH             ; Terminate Program  
      int      21H  
 end
 
How to Run this Program

For Running this program you should have installed Tasm on you computer . If you have not installed Tasm  yet please install from Here .

C:\programs>tasm aaa.asm


Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International


Assembling file:   aaa.asm


Error messages:    None


Warning messages:  None


Passes:            1


Remaining memory:  438k


C:\programs>tlink aaa


Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International


Warning: No stack


C:\programs>aaa


11

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  1. Add 8 Bit BCD Numbers
  2. Add Two 16 Bit BCD Numbers Code
  3. Pack the Two Unpacked BCD Numbers Code
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  5. Assembly Language Codes