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MaximuS
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 Posted: Tue Feb 17, 2015 2:52 am Post subject: Converting C++ algorithim to assembly language |
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The 4 loops is kinda confusing, I am a beginner at assembly. Please help. this is what I have so far
I am trying to convert this algorithim written in C++
into the assembly language.
Please help me if possible, and help me understand what codes to write as well please. Much appreciated.
Sorry for asking so much,
| Code: |
; h = 1;
; while ( (h*3+1) < a.length) {
; h = 3 * h + 1;
; }
mov dword [h], 1
whileLoop1:
mov eax, dword [dThree]
mul dword [h]
inc eax ;(h*3*1)
cmp eax, dword [length] ; while (x) < length
jge exitwhileLoop1
mov dword [h], eax ; h = 3 * h + 1
jmp whileLoop1
exitwhileLoop1:
; while( h > 0 )
; {
; for (i = h-1; i < a.length; i++) {
; tmp = a[i];
; j = i;
; for( j = i; (j >= h) && (a[j-h] > B); j -= h) {
; a[j] = a[j-h];
; }
; a[j] = tmp;
; }
; h = h / 3;
; }
whileLoop2:
cmp dword [h], 0
jg whileLoop2
forLoop1:
mov eax, dword [i]
forLoop2:
mov dword [j], eax ;j = i
cmp dword [j], dword [h]
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Dark Byte
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| Posted: Tue Feb 17, 2015 3:09 am Post subject: |
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it's just a for loop, inside a for loop, inside a while loop.
if you wish you can split up every loop into a subroutine called by the other loop to make it more clear to you
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MaximuS
I post too much
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Joined: 05 Apr 2007
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| Posted: Tue Feb 17, 2015 3:17 am Post subject: |
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; for( j = i; (j >= h) && (a[j-h] > B); j -= h) {
; a[j] = a[j-h];
; }
This part is really messing me up how to implement.
Can you show me how you would do that? |
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Dark Byte
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Joined: 09 May 2003
Posts: 25785
Location: The netherlands
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| Posted: Tue Feb 17, 2015 6:31 am Post subject: |
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for a general idea:
a for loop does a check before each iteration instead of after it. It runs as long as the statement in the middle entry equals true
let's assume you assign each variable a register
so: h=ebx, i=ecx, j=edx, a=esi(address of first element of a) a.length=edi
(eax=temp var)
and assuming that B is a constant variable (uppercase)
So, what this code would do then is:
| Code: |
mov edx,ecx ;initializer of the for loop (j=i )
loop:
;check the state ( (j >= h) && (a[j-h] > B) )
cmp edx,ebx ;cmp j,h
jl endloop ; if j<h go to endloop
mov eax,edx ;eax get the value of j
sub eax,ebx ;sub eax with h: eax is now j-h
cmp [esi+eax],B ;compare the value at a[j-h] with constant B
jle endloop ; if a[j-h] is smaller or equal to B exit the loop
;do stuff inside the loop body (a[j] = a[j-h])
mov eax,[esi+eax] ;eax already contains j-h, so now get the value a[j-h]
mov [esi+edx],eax ; a[j]([esi+edx])=a[j-h] (eax)
;next iteration (j -= h : decrease j with the value of h )
sub edx,ebx ; decrease j (edx) with h (ebx)
jmp loop
endloop:
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MaximuS
I post too much
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Posts: 3212
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| Posted: Tue Feb 17, 2015 12:07 pm Post subject: |
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Here's what I ended up doing, thanks for all your help Dark Byte.
| Code: |
whileLoop2:
cmp dword [h], 0 ; while( h > 0 )
jle exitwhileLoop2 ; if h is not > 0 then exit
;---
mov r8d, dword [h] ;;
dec r8d ;; h - 1
mov dword [i], r8d ;; (i = h - 1;
forLoop1: ;; for loop check
mov esi, dword [i] ;;
cmp esi, dword [len] ;; i < length;
jg exitforLoop1 ;;
mov r10d, dword [lst+esi*4] ; lst[i]
mov dword [tmp], r10d ; tmp = lst[i]
mov dword [j], esi ; j = i
;---
mov r8d, dword [i]
mov r9d, dword [h]
forLoop2: ;; for loop check
mov dword [j], r8d ;; j = i
cmp dword [j], r9d ;;
jl exitforLoop2 ;; for(j = i; (j >= h)
mov r10d, dword [j] ;;
sub r10d, dword [h] ;; j - h
mov r11d, dword [tmp] ;;
cmp dword [lst+r10d*4], r11d;; && (lst[j-h] > tmp);
jl exitforLoop2 ;;
mov dword [j], r10d ;; j = j-h)
mov ebx, dword [j] ;
sub ebx, dword [h] ; j - h
mov r12d, dword [lst+ebx*4] ;
mov r10d, dword [j] ;
mov dword [lst+r10d*4], r12d;lst[j] = lst[j-h]
jmp forLoop2
exitforLoop2:
mov eax, dword [tmp] ;
mov ebx, dword [j] ;
mov dword [lst+ebx*4], eax ; lst[j] = tmp
inc esi ; i++)
jmp forLoop1 ;
exitforLoop1:
mov eax, dword [h] ;
div dword [dThree] ; h / 3
mov dword [h],eax ; h = h /3
jmp whileLoop2
exitwhileLoop2:
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justa_dude
Grandmaster Cheater
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Joined: 29 Jun 2010
Posts: 891
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| Posted: Fri Feb 20, 2015 5:12 am Post subject: |
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BTW, pretty much every C/C++ compiler out there can compile to commented ASM. Hands-down, that's the easiest way to convert an algorithm from C/C++ to ASM.
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