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delphiassemblyx86basm

Translate Delphi style ASM to English?


A recent Delphi project i've inherited has a procedure in ASM. I'm a complete ASM newbie, so i dont understand it. I've read up on the various ASM instructions to try and decipher the procedures flow, but i still dont get it.

Could someone with ASM experience assist my understanding and translate the following procedure to English (then i can translate back to Delphi so the code is easier to read in the future!!!)

The declaration of Mem1 is an array [0..15] of Byte;. And Mem2 is a LongInt.

Here's the procedure:

 procedure TForm1.XorMem(var Mem1; const Mem2; Count : Cardinal); register;
 begin
 asm
   push esi
   push edi

   mov  esi, eax         //esi = Mem1
   mov  edi, edx         //edi = Mem2

   push ecx              //save byte count
   shr  ecx, 2           //convert to dwords
   jz   @Continue

   cld
 @Loop1:                 //xor dwords at a time
   mov  eax, [edi]
   xor  [esi], eax
   add  esi, 4
   add  edi, 4
   dec  ecx
   jnz  @Loop1

 @Continue:              //handle remaining bytes (3 or less)
   pop  ecx
   and  ecx, 3
   jz   @Done

 @Loop2:                 //xor remaining bytes
   mov  al, [edi]
   xor  [esi], al
   inc  esi
   inc  edi
   dec  ecx
   jnz  @Loop2

 @Done:
   pop  edi
   pop  esi
 end;

 end;

edit: Thanks to Roman R i've converted the ASM back to Delphi

procedure TForm1.XorMem2(var Mem1; const Mem2 :LongInt; Count : Cardinal);
var
  Key : array [0..3] of byte absolute Mem1;
  Num : array [0..3] of byte absolute Mem2;
  idx : byte;
begin
  for Idx := 0 to Count -1 do Key[idx] := Key[idx] Xor Num[idx];
end;

Solution

  • The function accepts two pointers (to arrays whatever) and their lengths in bytes. The function performs byte-to-byte XOR operation on first array bytes (Mem1) using second array bytes (Mem2). Pseudo-code:

    for Index = 0 to Count - 1
      (Mem1 as Byte Array) [Index] = (Mem1 as Byte Array) [Index] Xor (Mem2 as Byte Array) [Index]