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16-January 12
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User is offline Mar 13 2013 09:16 PM
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Country:
PH
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Windows
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Favorite Processor:
Intel
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PC
Your Car:
Honda
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Posts I've Made

  1. In Topic: Simple Calculator

    Posted 12 Mar 2013

    thanks master..Ive learn alot.

  2. In Topic: Simple Calculator

    Posted 11 Mar 2013

    View PostGunnerInc, on 12 March 2013 - 12:58 AM, said:

    Ok, so what is your problem with division? Did you look up the div mnemonic? What don't you understand?

    storing input to registers is my problem that's why I prefer to use sub in looping to get the answer of division.

    this is what I've done.

    .model small
.stack 10h
.data
	prom db "Enter the 1st no.:$"
	promp db "Enter the 2nd no.:$"
	result db "The answer is$"
;	num db 2 dup(?),'$'
	stop_num db,10,13,'$'
.code
	mov ax,@data
	mov ds,ax
	
	mov ah,9
	lea dx,prom
	int 21h
	
	mov ah,1
	int 21h
	
;	add al,30h		
	mov bl,al

;	mov cl,bl
	
	mov ah,2
	mov dl,10
	int 21h
	
	mov dl,13
	int 21h
	
	mov ah,9
	lea dx,promp
	int 21h	
	
	mov ah,1
	int 21h
	
	mov bl,al	
	mov al,0
	mov al,cl	
	
	mov ch,bl	

	try:
;	add bl,num[0]
	sub al,bl
	
	loop try
	
	mov bl,al
	
	mov ah,2
	mov dl,10
	int 21h
	
	mov dl,13
	int 21h
	
	mov ah,9
	lea dx,result
	int 21h
	
	mov dl,13
	int 21h	
	
	mov ah,2
	mov dl,bl
	int 21h
	
	mov ah,4ch
	int 21h
	
end


    what can you say??

  3. In Topic: Simple Calculator

    Posted 10 Mar 2013

    my problem now is division.

  4. In Topic: Simple Calculator

    Posted 10 Mar 2013

    View PostGunnerInc, on 10 March 2013 - 05:22 AM, said:

    Quote

    Difficult to understand.

    What is difficult to understand???

        mov     eax, 15
    mov     ecx, 3
    mul     ecx
    ; eax now contains 45


    and you can use 32 bit registers in 16bit DOS code.

    ok..I will explore more.

  5. In Topic: Simple Calculator

    Posted 8 Mar 2013

    Anyone....help!Whats wrong with this code for multiplication? 
    
.model small
.stack 10h
.data
	prom db "Enter the 1st no.:$"
	promp db "Enter the 2nd no.:$"
	result db "The answer is$"
	num db 2 dup(?),'$'
	stop_num db,10,13,'$'
.code
	mov ax,@data
	mov ds,ax
	
	mov ah,9
	lea dx,prom
	int 21h
	
	mov ah,1
	int 21h
	
	mov num[0],al
	
	mov ah,2
	mov dl,10
	int 21h
	
	mov dl,13
	int 21h
	
	mov ah,9
	lea dx,promp
	int 21h	
	
	mov ah,1
	int 21h
	
	mov num[1],bl
	
	mov bl,0
	
;	mov ax,bx
	mov cx,ax
	ulit:
	add bl,num[0]
;	add bl,num[1]
	
	loop ulit
	
;	sub bl,30h
	
	mov ah,2
	mov dl,10
	int 21h
	
	mov dl,13
	int 21h
	
	mov ah,9
	lea dx,result
	int 21h
	
	mov dl,13
	int 21h	
	
	mov ah,2
	mov dl,bl
	int 21h
	
	mov ah,4ch
	int 21h
	
end

    what should I add to bl in the loop?is it bl itself???thanks for the help again.

My Information

Member Title:
D.I.C Head
Age:
19 years old
Birthday:
August 15, 1993
Gender:
Location:
Romblon
Interests:
Programming,Sports
Full Name:
Fredex Rainin
Years Programming:
3
Programming Languages:
C,C++,VB6,Assembly language

Contact Information

E-mail:
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Yahoo:
Yahoo  dexter_lab_0803@yahoo.com
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Comments

Page 1 of 1
  1. Photo

    Fredex Icon

    24 Jan 2012 - 04:21
    Hello to everyone..I want to learn more from you guys,share more useful tricks and idea that can improve our programming skills.
Page 1 of 1