日期:2014-05-16  浏览次数:20836 次

《coredump问题原理探究》Linux x86版4.5节函数的逆向之coredump例子

在了解了上面的规律后,现在开始来解决本章一开头的问题:

(gdb) bt
#0  0x4365b569 in vfprintf () from /lib/libc.so.6
#1  0x436629ff in printf () from /lib/libc.so.6
#2  0x080485b9 in main ()

看一下main函数的汇编:

(gdb) disassemble main
Dump of assembler code for function main:
   0x08048500 <+0>:     push   %ebp
   0x08048501 <+1>:     mov    %esp,%ebp
   0x08048503 <+3>:     and    $0xfffffff0,%esp
   0x08048506 <+6>:     sub    $0x20,%esp
   0x08048509 <+9>:     movl   $0x0,0x1c(%esp)

   0x08048511 <+17>:    jmp    0x80485bf <main+191>

   0x08048516 <+22>:    mov    0x1c(%esp),%eax
   0x0804851a <+26>:    lea    0x0(,%eax,4),%edx
   0x08048521 <+33>:    mov    0xc(%ebp),%eax
   0x08048524 <+36>:    add    %edx,%eax
   0x08048526 <+38>:    mov    (%eax),%eax
   0x08048528 <+40>:    mov    %eax,(%esp)

   0x0804852b <+43>:    call   0x80483d0 <strlen@plt>

   0x08048530 <+48>:    mov    %eax,0x18(%esp)
   0x08048534 <+52>:    mov    0x18(%esp),%eax
   0x08048538 <+56>:    cmp    $0x1,%eax

   0x0804853b <+59>:    je     0x8048570 <main+112>

   0x0804853d <+61>:    cmp    $0x2,%eax

   0x08048540 <+64>:    je     0x804858f <main+143>

   0x08048542 <+66>:    test   %eax,%eax

   0x08048544 <+68>:    jne    0x80485a5 <main+165>

   0x08048546 <+70>:    mov    0x1c(%esp),%eax
   0x0804854a <+74>:    lea    0x0(,%eax,4),%edx
   0x08048551 <+81>:    mov    0xc(%ebp),%eax
   0x08048554 <+84>:    add    %edx,%eax
   0x08048556 <+86>:    mov    (%eax),%eax
   0x08048558 <+88>:    movzbl (%eax),%eax
   0x0804855b <+91>:    movsbl %al,%eax
   0x0804855e <+94>:    mov    %eax,0x4(%esp)
   0x08048562 <+98>:    movl   $0x8048674,(%esp)

   0x08048569 <+105>:   call   0x80483e0 <printf@plt>
   0x0804856e <+110>:   jmp    0x80485ba <main+186>

   0x08048570 <+112>:   mov    0x1c(%esp),%eax
   0x08048574 <+116>:   add    $0x1,%eax
   0x08048577 <+119>:   lea    0x0(,%eax,4),%edx
   0x0804857e <+126>:   mov    0xc(%ebp),%eax
   0x08048581 <+129>:   add    %edx,%eax
   0x08048583 <+131>:   mov    (%eax),%eax
   0x08048585 <+133>:   mov    %eax,(%esp)

   0x08048588 <+136>:   call   0x80483f0 <puts@plt>
   0x0804858d <+141>:   jmp    0x80485ba <main+186>

   0x0804858f <+143>:   mov    0x1c(%esp),%eax
   0x08048593 <+147>:   mov    %eax,0x4(%esp)
   0x08048597 <+151>:   movl   $0x8048678,(%esp)

   0x0804859e <+158>:   call   0x80483e0 <printf@plt>
   0x080485a3 <+163>:   jmp    0x80485ba <main+186>

   0x080485a5 <+165>:   mov    0x1c(%esp),%eax
   0x080485a9 <+169>:   mov    %eax,0x4(%esp)
   0x080485ad <+173>:   movl   $0x804867c,(%esp)

   0x080485b4 <+180>:   call   0x80483e0 <printf@plt>

   0x080485b9 <+185>:   nop
   0x080485ba <+186>:   addl   $0x1,0x1c(%esp)
   0x080485bf <+191>:   mov    0x1c(%esp),%eax
   0x080485c3 <+195>:   cmp    0x8(%ebp),%eax
   0x080485c6 <+198>:   setl   %al
   0x080485c9 <+201>:   test   %al,%al

   0x080485cb <+203>:   jne    0x8048516 <main+22>

   0x080485d1 <+209>:   mov    $0x0,%eax
   0x080485d6 <+214>:   leave  
   0x080485d7 <+215>:   ret    
End of assembler dump.


   0x080485cb <+203>:   jne    0x8048516 <main+22>

可知,0x8048516到0x080485cb构成一个循环。

而0x080486cb的判断条件:

   0x080485bf <+191>:   mov    0x1c(%esp),%eax
   0x080485c3 <+195>:   cmp    0x8(%ebp),%eax
   0x080485c6 <+198>:   setl   %al
   0x080485c9 <+201>:   test