I have to add two 3*3 arrays of words and store the result in another array. Here is my code:
.data
a1 WORD 1,2,3
WORD 4,2,3
WORD 1,4,3
a2 WORD 4, 3, 8
WORD 5, 6, 8
WORD 4, 8, 9
a3 WORD DUP 9(0)
.code
main PROC
mov eax,0;
mov ebx,0;
mov ecx,0;
mov edx,0;
mov edi,0;
mov esi,0;
mov edi,offset a1
mov esi,offset a2
mov ebx, offset a3
mov ecx,LENGTHOF a2
LOOP:
mov eax,[esi]
add eax,[edi]
mov [ebx], eax
inc ebx
inc esi
inc edi
call DumpRegs
loop LOOP
exit
main ENDP
END main
But this sums all elements of a2 and a1. How do I add them row by row and column by column? I want to display the result of sum of each row in another one dimensional array(Same for columns).
The
a1 WORD 1,2,3
WORD 4,2,3
WORD 1,4,3
will compile as bytes (in hexa):
01 00 02 00 03 00 04 00 02 00 03 00 01 00 04 00 03 00
Memory is addressable by bytes, so if you will find each element above, and count it's displacement from the first one (first one is displaced by 0 bytes, ie. it's address is a1+0), you should see a pattern, how to calculate the displacement of particular [y][x] element (x is column number 0-2, y is row number 0-2... if you decide so, it's up to you, what is column/row, but usually people tend to consider consecutive elements in memory to be "a row").
Pay attention to the basic types byte size, you are mixing it everywhere in every way, reread some lesson/tutorial about how qword/dword/word/byte differ and how you need to adjust your instructions to work with correct memory size, and how to calculate the address correctly (and what is the size of eax and how to use smaller parts of it).
If you have trouble to figure it on your own:
displacement = (y * 3 + x) * 2 => *2 because element is
word, each occupies two bytes. y * 3 because single row is 3 elements long.
In ASM instructions that may be achieved for example...
If [x,y] is [eax,ebx], this calculation can be done as
lea esi,[ebx+ebx*2] ; esi = y*3|lea esi,[esi+eax] ; esi = y*3+x|mov ax,[a1+esi*2] ; loads [x,y] element from a1.
Now if you know how to calculate address of particular element, you can do either loop doing all the calculation ahead of each element load, or just do the math in head how the addresses differ and write the address calculation for first element (start of row/column) and then mov + 2x add with hardcoded offsets for next two elements (making loop for 3 elements is sort of more trouble than writing the unrolled code without loop), and repeat this for all three columns/rows and store the results.
BTW, that call DumpRegs ... is not producing what you expected? And it's a bit tedious way to debug the code, may be worth to spend a moment to get some debugger working.
Couldn't help my self, but to write it, as it's such funny short piece of code, but you will regret it later, if you will just copy it, and not dissect it to atoms and understand fully how it works):
column_sums: DW 0, 0, 0
row_sums: DW 0, 0, 0
...
; columns sums
lea esi,[a3] ; already summed elements of a1 + a2
lea edi,[column_sums]
mov ecx,3 ; three columns to sum
sum_column:
mov ax,[esi] ; first element of column
add ax,[esi+6] ; 1 line under first
add ax,[esi+12] ; 2 lines under
mov [edi],ax ; store result
add esi,2 ; next column, first element
add edi,2 ; next result
dec ecx
jnz sum_column
; rows sums
lea esi,[a3] ; already summed elements of a1 + a2
lea edi,[row_sums]
mov ecx,3 ; three rows to sum
sum_row:
mov ax,[esi] ; first element of row
add ax,[esi+2] ; +1 column
add ax,[esi+4] ; +2 column
mov [edi],ax ; store result
add esi,6 ; next row, first element
add edi,2 ; next result
dec ecx
jnz sum_row
...
(didn't debug it, so bugs are possible, plus this expect a3 to contain correct element sums, which your original code will not produce, so you have to fix it first ... this code does contain lot of hints, how to fix each problem of original)
Now I feel guilty of taking the fun of writing this from you... nevermind, I'm sure you can find few more tasks to practice this. The question is, whether you got the principle of it. If not, ask which part is confusing and how you currently understand it.