TASM Program to Add Two 16 Bit Numbers

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Objective:
Write a menu based assembly language program to add  2 16 bit numbers.
Prerequisite:
TASM assembler
Algorithm
  1. Start
  2. Initialize data segment through AX register in the DS register.
  3. Display the 3 text message as “1. 16 bit addition 2 .16 bit subtraction 3. Exit Enter your choice”
  4. Compare accepted choice with 03h.
  5. If zero flag is set then goto step no.6  otherwise goto step no 7
  6. Exit the program
  7. Display the message as “Enter first 16 bit number”
  8. Read first digit in AL register through keyboard          (e.g. AL=32h)
  9. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number.AL=02h
  10. Move contents of AH with 00h.    (AHß 00h  so AX=0002h)
  11. Rotate AX contents in left directions by 12 bits. (AX=2000h)
  12. Move the contents of AX to BX(BXßAX so BX=2000h)
  13.   Read a second digit in AL register through keyboard AL=35h
  14. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=05h
  15. Move contents of AH with 00h.    (AHß 00h  so AX=0005h)
  16. Rotate AX contents in left directions by 8 bits. (AX=0500h)
  17. Add the contents of AX and BX (BX=BX+AX  so BX=2500h)
  18. Read a third digit in AL register through keyboard AL=31h
  19. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=01h
  20. Move contents of AH with 00h.    (AHß 00h  so AX=0001h)
  21. Rotate AX contents in left directions by 4 bits. (AX=0010h)
  22. Add the contents of AX and BX (BX=BX+AX  so BX=2510h)
  23. Read a fourth digit in AL register through keyboard AL=30h
  24. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=00h
  25. Move contents of AH with 00h.    (AHß 00h  so AX=0000h)
  26. Add the contents of AX and BX (BX=BX+AX  so BX=2510h)
  27. Display the message as “Enter second 16 bit number”
  28. Read first digit in AL register through keyboard          (e.g. AL=37h)
  29. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number.AL=07h
  30. Move contents of AH with 00h.    (AHß 00h  so AX=0007h)
  31. Rotate AX contents in left directions by 12 bits. (AX=7000h)
  32. Move the contents of AX to CX(CXßAX so CX=7000h)
  33.   Read a second digit in AL register through keyboard AL=35h
  34. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=05h
  35. Move contents of AH with 00h.    (AHß 00h  so AX=0005h)
  36. Rotate AX contents in left directions by 8 bits. (AX=0500h)
  37. Add the contents of AX and CX (CX=CX+AX  so CX=7500h)
  38. Read a third digit in AL register through keyboard AL=31h
  39. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=01h
  40. Move contents of AH with 00h.    (AHß 00h  so AX=0001h)
  41. Rotate AX contents in left directions by 4 bits. (AX=0010h)
  42. Add the contents of AX and CX (CX=CX+AX  so CX=7510h)
  43. Read a fourth digit in AL register through keyboard AL=34h
  44. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=04h
  45. Move contents of AH with 00h.    (AHß 00h  so AX=0004h)
  46. Add the contents of AX and BX (CX=CX+AX  so CX=7514h)
  47. Compare accepted choice from AL with 02h
  48. If zero flag is set then goto step no 69  otherwise goto step no  49
  49. Add the contents of BX and CX(BX=BX+CX  so BX=9A24h)
  50. Preserve the result in temporary variable as t of 16 bit (so t=9A24h)
  51. Mask the first nibble by AND operation with number F000h (AND BX,f000h so BX=9000h)
  52. Rotate the BX contents right by 12(in decimal so BX=0009h)
  53. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BX=0039h)
  54. Move the contents of BL to DL and display it on the screen
  55. Move result from temporary variable t to BX again (Now BX=9A24h)
  56. Mask the second nibble by AND operation with number 0F00h (AND BX,0F00h so BX=0A00h)
  57. Rotate the contents of  BX to right by 8(in decimal)
  58. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BX=0041h(ASCII hex value of ‘A’))
  59. Move the contents of BL to DL and display it on the screen.
  60. Move result from temporary variable to BX again (Now AX=9A24h)
  61. Mask the third nibble by AND operation with number 00F0h (AND BX,00F0h so BX=0020h)
  62. Rotate the contents of BX to right by 4(in decimal)
  63. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range      (BX=00032h)
  64. Move the contents of BL to DL and display it on the screen
  65. Move result back from temporary variable to BX again (Now BX=9A24h)
  66. Mask the fourth nibble by AND operation with number 000Fh (AND BX,000fh so BX=0004h)
  67. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BX=0004h)
  68. Move the contents of BL to DL and display it on the screen.
  69. Subtract  the contents of CX from BX(BXßBX-CX  so BX=AFFCh)
  70. Preserve the result in temporary variable as t of 16 bit (so t=AFFCh)
  71. Mask the first nibble by AND operation with number F000h (AND BX,F000h so BX=A000h)
  72. Rotate the BX contents right by 12(in decimal so BX=000Ah)
  73. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BX=0041h  (i.e. ASCII hex value for ‘A’ ))
  74. Move the contents of BL to DL and display it on the screen
  75. Move result from temporary variable t to BX again (Now BX=AFFCh)
  76. Mask the second nibble by AND operation with number 0f00h (AND BX,0F00h so BX=0F00h)
  77. Rotate the contents of  BX to right by 8(in decimal)
  78. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BX=0046h(i.e ASCII hex value of ‘F’))
  79. Move the contents of BL to DL and display it on the screen.
  80. Move result from temporary variable to BX again (Now BX=AFFCh)
  81. Mask the third nibble by AND operation with number 00F0h (AND BX,00F0h so BX=00F0h)
  82. Rotate the contents of BX to right by 4(in decimal)
  83. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range      (BX=0046h (i.e ASCII hex value of ‘F’))
  84. Move the contents of BL to DL and display it on the screen
  85. Move result back from temporary variable to BX again (Now BX=AFFCh)
  86. Mask the fourth nibble by AND operation with number 000fh (AND AX,000Fh so AX=000Ch)
  87. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BX=0043h(i.e ASCII hex value of ‘C))
  88. Move the contents of BL to DL and display it on the screen.
  89. Stop
Algorithm for Input procedure :( To accept input from 0 to f)
1.      Compare the contents of  AL with 41h(Small case)
2.      Jump to step no 4 if carry flag is set
3.      Sub  07h to AL register
4.      Sub 30h to AL register
5.      Return.
Algorithm for Output procedure:
1.      Compare the contents of  BL with 0Ah
2.      Jump to step no 4 if carry flag is set
3.      Add  07h to AL register
4.      Add 30h to AL register
5.      Return.

Note:

While masking F or f is not case sensitive. But in input procedure 41h number is considered for comparison because 41h is ASCII hex value for ‘A’. In output procedure ‘0A’ is considered not ‘a’ is considered as small case a has 61h ASCII hex value.So this input and output procedure are applicable for only capital ‘A’ to ‘F’ 
TASM Program :
Data Segment
 msg db 0dh,0ah,"Enter a 16-bit number: $"
 result db 0dh,0ah,"The Result is: $"
 newl db 0dh,0ah," $"
Data ends
Code Segment
assume CS:Code,DS:Data
Start:
 mov ax,Data
 mov DS,ax
 
 mov dx,offset msg;add16
 mov ah,09h
 int 21h
 
 call AcceptNum
 mov bh,bl
 
 call AcceptNum
 mov cx,bx

 mov dx,offset msg
 mov ah,09h
 int 21h
 
 call AcceptNum
 mov bh,bl
 
 call AcceptNum

 add cx,bx

 mov dx,offset result
 mov ah,09h
 int 21h
 
 mov bl,ch
 call DispNum

 mov bl,cl
 call DispNum

 mov ah,4ch
 int 21h
AcceptNum proc
 mov ah,01h
 int 21h
 
 call HexAccept
 
 mov bl,al
 rol bl,4
 
 mov ah,01h
 int 21h
 
 call HexAccept
 add bl,al
 ret
endp
DispNum proc
 mov al,bl
 and al,0f0h
 ror al,4

 mov dl,al
 call HexDisp
 mov ah,02h
 int 21h

 mov al,bl
 and al,0fh
 
 mov dl,al
 call HexDisp
 mov ah,02h
 int 21h
endp
HexAccept proc
 cmp al,41h
 jc norm
 sub al,07h
 norm: sub al,30h
 ret
endp
HexDisp proc
 cmp dl,0ah
 jc nothex
 add dl,07h
 nothex: add dl,30h
 ret
endp
Code ends
end Start
output:-
C:\TASM\BIN>tasm add16.asm
Turbo Assembler  Version 4.1  Copyright (c) 1988, 1996 Borland International

Assembling file:   add16.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  453k

C:\TASM\BIN>tlink add16.obj
Turbo Link  Version 7.1.30.1. Copyright (c) 1987, 1996 Borland International
Warning: No stack

C:\TASM\BIN>add16

Enter a 16-bit number: 1312
Enter a 16-bit number: 1205
The Result is: 2517
--------------------------------

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