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

Question

Please view the following code snippets:

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    $0x1,%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    

I need to figure out the meaning of every single line of it, kinda confused me.

Answer 1

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

This isn't the real destination of the call, it is something will be filled in by linker. If you run objdump -dr sum.o, you will find it is actually a call to __x86.get_pc_thunk.ax. Same for the following add, to set up a pointer to the GOT. (This function doesn't need one but you compiled without optimization, with -fpie on by default.)

For more details, take a look Why does gcc generates strange code without flag -fno-pie?

---

System V i386 ABI, Section 2.2.2 tells the structure of a stack frame.

So your stack frame looks like this:

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

For remaining instructions, here is a step-by-step analysis.

// 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)