c - 卡住拆卸

Question

需要一个功能的帮助，我认为这并不难，有人可以将它翻译成 C，以便我可以从那里获取逻辑吗？

0x004011cf mov al, byte [esi]

| : 0x004011d1 and eax, 0xff

| : 0x004011d6 mul ebx

| : 0x004011d8 inc esi

| : 0x004011d9 add edi, eax

| : 0x004011db inc ebx

| : 0x004011dc dec ecx

| `=< 0x004011dd jne 0x4011cf

Answer 1

干得好：

esi显然是指向某个长度缓冲区的指针ecx

LOOP:
      mov al, byte [esi]    ; read byte from memory pointed by esi into low bits of eax
      and eax, 0xff         ; mask eax with 0xff
      mul ebx               ; multiply eax with ebx (wherever ebx came from...)
                            ; put result in eax
      inc esi               ; increment buffer pointer
      add edi, eax          ; add eax to edi (whereever edi came from)
      inc ebx               ; increment ebx
      dec ecx               ; decrement ecx (which is probably some counter)
      jne LOOP              ; jump to LOOP if ecx is different from 0

但是如果没有任何上下文信息，很难判断这段代码实际上在做什么。

等效的 C 代码大致执行以下操作：

  char *esi;    // points to some buffer...
  int ebx;      // contains some value
  int edi;      // contains some value
  int ecx;      // some counter, presubably the length of the buffer pointed by esi
  ...
  do
  {  
    edi += *esi++ * ebx++;
  } while (--ecx != 0)

您需要学习 x86 汇编的基础知识。