How can i add two numbers with 12 bytes each-one?

I want to add two numbers that have 12 bytes and to store the result in a 16 bytes var. How can i do this?

section .data
    big_num1 dd 0x11111111, 0x22222222, 0x33333333
    big_num2 dd 0xffffffff, 0x22222222, 0x33333333
section .bss   
    result_4word resd 4

I think i can add the first 4 bytes number from number 1 with the other first 4 bytes from number 2 and so on.. but i don't know how to concatenate results in in my result variable. How should i do the carry,if it's needed? Is this solution the right one?

xor eax, eax
xor ebx, ebx
mov ecx, 3
loop1:
mov eax, dword[big_num1+4*(ecx-1)]
mov ebx, dword[big_num2+4*(ecx-1)]
mov [result_4word+4*(ecx-1)], eax
adc [result_4word+4*(ecx-1)], ebx
loop loop1

big_num1 dd 0x11111111, 0x22222222, 0x33333333
big_num2 dd 0xffffffff, 0x22222222, 0x33333333

What numbers are defined here?

Because x86 is a little-endian architecture, the lowest part of a number is stored in memory at the lowest addresses. For big_num1 the first defined dword (value is 0x11111111) is at the lowest address and thus is the lowest part of the number. In the normal number representation this is what goes at the right-handside.

big_num1 == 0x333333332222222211111111
big_num2 == 0x3333333322222222FFFFFFFF

Adding big numbers

You add corresponding digits going from right to left, just like everybody has learned at school.

In the hexadecimal representation of these numbers there are 24 digits to consider. However since the architecture is 32-bit, we can nicely make 3 groups of 8 digits.

For the 1st group we simply use ADD:

mov     eax, [big_num1]           ;   0x11111111
add     eax, [big_num2]           ; + 0xFFFFFFFF <-- This produces a carry
mov     [result_4dword], eax      ;   0x00000000

For the 2nd group we use ADC to pick up a possible carry from the previous addition:

mov     eax, [big_num1 + 4]       ;   0x22222222
adc     eax, [big_num2 + 4]       ; + 0x22222222 + CF=1  <-- No new carry
mov     [result_4dword + 4], eax  ;   0x44444445

For the 3rd group we use ADC to pick up a possible carry from the previous addition:

mov     eax, [big_num1 + 8]       ;   0x33333333
adc     eax, [big_num2 + 8]       ; + 0x33333333 + CF=0  <-- No new carry
mov     [result_4dword + 8], eax  ;   0x66666666

Turning this into a loop

Key here is that we can also use ADC for the 1st group if we expressly clear the carry flag beforehand:

clc
mov     eax, [big_num1]           ;   0x11111111
adc     eax, [big_num2]           ; + 0xFFFFFFFF + CF=0 <-- This produces a carry
mov     [result_4dword], eax      ;   0x00000000

Now we can write a loop with 3 iterations but we have to be careful about not changing the carry flag inadvertently. That's why I use LEA instead of ADD in order to advance the offset. DEC is also an instruction that does not destroy the carry flag. I've preferred the combo DEC ECX JNZ ... because it's better than LOOP ...:

    mov     ecx, 3
    xor     ebx, ebx              ; This additionally clears the carry flag
Again:
    mov     eax, [big_num1 + ebx]
    adc     eax, [big_num2 + ebx] ; Can produce a new carry flag
    mov     [result_4dword + ebx], eax
    lea     ebx, [ebx + 4]        ; This does not clobber the carry flag
    dec     ecx                   ; This does not clobber the carry flag
    jnz     Again

If after these 3 additions there's still a set carry, you'll have to write a 1 in the 4th dword of result_4dword, else you'll have to write a 0 here. Because result_4dword is in the .bss section, you should not count on any preset value like zero!

    setc    cl
    mov     [result_4dword + ebx], ecx  ; ECX=[0,1]

Please note that I've changed result_4word into result_4dword. Makes more sense...