如何理解这段 AT&T i386 汇编代码片段？

Question

请查看以下代码片段：

int& sum(int& num1, int& num2) {
    num1++;
    num2++;
}

00000000 <_Z3sumRiS_>:
   0:   55                      push   %ebp
   1:   89 e5                   mov    %esp,%ebp
   3:   e8 fc ff ff ff          call   4 <_Z3sumRiS_+0x4>    // why here is a jump
   8:   05 01 00 00 00          add    [=11=]x1,%eax     

   // why 0x8, my understanding is there are in total 3 parameters
   // num2 -- 0xc(%ebp), num1 -- 0x8(%ebp), this -- 0x4(%ebp)
   // am I right????
   d:   8b 45 08                mov    0x8(%ebp),%eax

  10:   8b 00                   mov    (%eax),%eax
  12:   8d 50 01                lea    0x1(%eax),%edx        // what the heck is this?
  15:   8b 45 08                mov    0x8(%ebp),%eax
  18:   89 10                   mov    %edx,(%eax)
  1a:   8b 45 0c                mov    0xc(%ebp),%eax
  1d:   8b 00                   mov    (%eax),%eax
  1f:   8d 50 01                lea    0x1(%eax),%edx
  22:   8b 45 0c                mov    0xc(%ebp),%eax
  25:   89 10                   mov    %edx,(%eax)
  27:   90                      nop
  28:   5d                      pop    %ebp
  29:   c3                      ret

我需要弄清楚每一行的意思，我有点困惑。

Answer 1

   3:   e8 fc ff ff ff          call   4 <_Z3sumRiS_+0x4>

这不是调用的真正目的地，它是由链接器填充的。如果你运行objdump -dr sum.o，你会发现它实际上是对__x86.get_pc_thunk.ax的调用。以下add同理，设置指向GOT的指针。（这个函数不需要一个，但你编译时没有优化，默认情况下 -fpie 打开。）

详情请看Why does gcc generates strange code without flag -fno-pie?

System V i386 ABI, 2.2.2 节讲述了堆栈帧的结构。

所以你的堆栈框架看起来像这样：

0xc  |      num2      |
0x8  |      num1      |
0x4  | return address |
0x0  | previous %ebp  |  <-- %ebp

剩下的说明，这里一步步分析。

// as num1 and num2 are references, they represents address in assembly
   d:   8b 45 08                mov    0x8(%ebp),%eax        // load num1 to %eax
  10:   8b 00                   mov    (%eax),%eax           // load *num1 to %eax
  12:   8d 50 01                lea    0x1(%eax),%edx        // put *num1 + 1 into %edx
  15:   8b 45 08                mov    0x8(%ebp),%eax        // load num1 to %eax
  18:   89 10                   mov    %edx,(%eax)           // save *num1 + 1 at num1
  1a:   8b 45 0c                mov    0xc(%ebp),%eax        // same as above
  1d:   8b 00                   mov    (%eax),%eax
  1f:   8d 50 01                lea    0x1(%eax),%edx
  22:   8b 45 0c                mov    0xc(%ebp),%eax
  25:   89 10                   mov    %edx,(%eax)

如何理解这段 AT&T i386 汇编代码片段？

How to understand this AT&T i386 assembly code snippet?

assembly

32-bit

i386

att