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Asked: 2026-06-13T23:20:39+00:00

I’m having some problems trying to solve this expression in assembler. `$`z=(5*a-b/7)/(3/b+a*a) I would

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I’m having some problems trying to solve this expression in assembler.

`$`z=(5*a-b/7)/(3/b+a*a)

I would like to know how do you convert a word to a double word ( unsigned solution ) ,
do i have to use the cwb command or do i use AX:BX , if i do have to use those last registers ,
how do i properly write the command ?

I will be testing the code in Turbo Debugger under DosBox .
My full code 

assume cs:code,ds:data 
data segment
;
a db ?
b db ?
rez dw ?
;
data ends
;
code segment
;
mov ax,data
mov ds,ax
;
;
;#####prima paranteza
;
mov al,a    ;ah=a
mul 5       ;ax=a*5
mov cx,ax   ;cx=ax
mov ah,b    ; il mut pe ah in b  ( pregatire pt conversie fara semn )
mov al,0    ; l-am convertit pe b in word ( pe 2 octeti )
div 7       ; am impartit double word-ul b la 7 , catul a ramas in ah , restul a ramas in al 
sub cx,ax   ; (am tinut cont de faptul ca in ax a ramas rezultatul dupa impartire )  , cx=cx-ax, a*5 - b/7
;
; ####a 2 a paranteza
;
mov ah,3    
mov al,0    ; conversie de la b la w  ( fara semn )
div b       ; ax=3/b
mov bx,ax   ; bx = ax
mov al,a    
mul a       ; ax = a * a 
add ax,bx   ; ax = ax + bx
;
;
; #### calcul final 
mov bx,ax   ; bx = ax ( rezultatul celei de a 2 a paranteze )
mov ax,cx   ; ax = cx ( rezultatul primei paranteze )
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1 Answer

  1. Editorial Team

    word to double-word?

    Let’s see if I got you:

    word -> 8bit
    double-word -> 16bit

    AX, BX, CX and DX are 16 bit registers, and they are formed by two other 8-bit registers [ABCD]H and [ABCD]L, so, AX would be:

    AH                      AL
    |0|0|0|0|0|0|0|0| – |0|0|0|0|0|0|0|0|

    When you use AX, you’re using those two at the same time. So, if you want to convert a word to a double word, you just clear the whole [ABCD]X register, and then move your word to the [ABCD]L register, leaving [ABCD]X with the word value.

    Cheers

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