TASM Program to subtract Two 8 Bit Numbers

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Objective:

Write an assembly language program for 8 bit subtraction.

Prerequisite:

TASM assembler

Algorithm for 8 bit subtraction:

1. Start
2. Initialize data segment through AX register in the DS register.
3. Display the message as “Enter the first number”
4. Read first digit in AL register through keyboard          (e.g. AL=31h)
5. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number.AL=01h
6. Move contents of AL register to a BL.    (BLß AL  so BL=01h)
7. Rotate the contents of BL register by 4 positions at left side. (BL=10h)
8. Read a second digit in AL register through keyboard AL=35h
9. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=05h
10. Add the contents of BL and AL store the result in BL   (BLßBL+AL so BL=15h)
11. Display the message as “Enter the second number”
12. Read first digit in AL register through keyboard          AL=32h
13. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number.AL=02h
14. Move contents of AL register to a CL.    (CLß AL  so CL=02h)
15. Rotate the contents of CL register by 4 positions at left side. (CL=20h)
16. Read a second digit in AL register through keyboard (AL=33h)
17. Call Input procedure to make a number from ASCII hexadecimal to a normal hexadecimal number. AL=03h
18. Add the contents of CL and AL store the result in CL   (CLßCL+AL so CL=23h) (Now both numbers are accepted as 15h and 23h)
19. Subtract the contents of CL from BL and result gets stored in BL (E.g SUB BL,CL so BL=F2h)
20. Preserve the result in some temporary variable say temp from BL.
21. Mask the first nibble by AND operation with number F0h (AND BL,F0h so BL=30h)
22. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BL=33h)
23. Move the contents of BL to DL and display it on the screen
24. Move result from temporary variable to BL again (So BL=38h)
25. Mask the second nibble by AND operation with number 0Fh (AND BL,0Fh so BL=08h)
26. Call Output procedure with BL register to make a digit back in ASCII hexadecimal range (BL=38h)
27. Move the contents of BL to DL and display it on the screen
28. Stop

Algorithm for Input procedure:(To accept input from 0 to f)

1.      Compare the contents of  AL with 41h

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 BL register

4.      Add 30h to BL 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 first number: $"
 msg1 db 0dh,0ah,"Enter second number: $"
 result db 0dh,0ah,"The Result is: $"

Data ends
Code segment
 assume CS:Code,DS:Data
start:
 mov ax,Data ; Move Data to Data Segment
 mov DS,ax

 mov dx,offset msg ; Display contents of variable msg sub8
 mov ah,09h
 int 21h

 mov ah,01h ; To accept input and store ASCII value into al
 int 21h

 sub al,30h ; Accept 10's place of the Number
 mov bl,al
 rol bl,4

 mov ah,01h ; To accept input and store ASCII value into al
 int 21h

 sub al,30h ; Accept unit's place of Number

 add bl,al ; Get the number by adding 10's and unit's place

 mov dx,offset msg1 ; Display contents of variable msg1
 mov ah,09h
 int 21h

 mov ah,01h ; To accept input and store ASCII value into al
 int 21h

 sub al,30h ; Accept 10's place of the Number
 mov cl,al
 rol cl,4

 mov ah,01h ; To accept input and store ASCII value into al
 int 21h

 sub al,30h ; Accept unit's place of Number

 add cl,al ; Get the number by adding 10's and unit's place

 sub bl,cl ; Subtract the two accepted Number's

 mov dx,offset result ; Display contents of string result
 mov ah,09h
 int 21h

 mov cl,bl ; Store the value of the Result

 and bl,0f0h ; Isolate 10's place of Result
 ror bl,4

 call AsciiConv ; Convert to ASCII to display

 mov dl,bl ; Display a Number/Alphabet
 mov ah,02h
 int 21h

 mov bl,cl ; Retrieve original Result

 and bl,0fh ; Isolate unit's place of Result

 call AsciiConv ; Convert to ASCII to display

 mov dl,bl ; Display a Number/Alphabet
 mov ah,02h
 int 21h

 mov ah,4ch ; Terminate the program
 int 21h

 AsciiConv proc ; Compare to 0a if it is less than A then we need to add only 30
  cmp bl,0ah ; If it is greater than or equal to 0a then we also need to add 07
  jc skip
  add bl,07h
  skip: add bl,30h
  ret
  endp
Code ends
end start

output:-

C:\TASM\BIN>tasm sub8.asm
Turbo Assembler  Version 4.1  Copyright (c) 1988, 1996 Borland International

Assembling file:   sub8.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  453k

C:\TASM\BIN>tlink sub8.obj
Turbo Link  Version 7.1.30.1. Copyright (c) 1987, 1996 Borland International
Warning: No stack

C:\TASM\BIN>sub8

Enter first number: 13
Enter second number: 12
The Result is: 01
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