CODEWITHSUNDEEP
assemblyx86additionx86-16
Ammar Alyousfi
Ammar Alyousfi

Reputation: 4382

Addition of different sized numbers in assembley

I'm asked to write a program to add a byte sized array elements. First 2 bytes are the number of elements in the array.

The array may contain thousands of numbers so I should put the sum in 32-bit register (e.g. ax:dx) but the problem is that I don't know how to take a byte from memory then add it to a 32-bit register (or a double variable in memory).

I have tried adding using a 16-bit register and a word variable in memory. Here is the code:

array db 07, 00, 30, 10, 77, 14, 9, 54, 100
sum dw ?


lea ax, data
mov ds, ax
mov es, ax

lea bx, array
mov cx, [bx] 
mov bx, 0002h
mov dx, 0000h
Addition:
mov dl, [bx]
mov dh, 00h
add sum, dx
add bx, 1
loop Addition 

mov ax, 4c00h
int 21h

It works correctly. But I want to know how to do that with a 32-bit register and a double variable.

I use emu8086

Upvotes: 0

Views: 610

Answers (1)

Michael
Michael

Reputation: 58507

To add an 8-bit value in BL to the 32-bit sum in DX:AX:

xor bh,bh  ; Clear BH (effectively zero-extends BL into BX)
add ax,bx  ; Add BX to the lower half of the sum
adc dx,0   ; If the lower half wrapped around (the above addition resulted
           ; in a carry), add 1 to the upper half, otherwise add 0.

Upvotes: 3

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