Question

void function(int a, int b, int c) {
   char buffer1[5];
   char buffer2[10];
   int *ret;

   ret = buffer1 + 12;
   (*ret) += 8;//why is it 8??
}

void main() {
  int x;

  x = 0;
  function(1,2,3);
  x = 1;
  printf("%d\n",x);
}

The above demo is from here:

http://insecure.org/stf/smashstack.html

But it’s not working here:

D:\test>gcc -Wall -Wextra hw.cpp && a.exe
hw.cpp: In function `void function(int, int, int)':
hw.cpp:6: warning: unused variable 'buffer2'
hw.cpp: At global scope:
hw.cpp:4: warning: unused parameter 'a'
hw.cpp:4: warning: unused parameter 'b'
hw.cpp:4: warning: unused parameter 'c'
1

And I don’t understand why it’s 8 though the author thinks:

A little math tells us the distance is
8 bytes.

My gdb dump as called:

Dump of assembler code for function main:
0x004012ee <main+0>:    push   %ebp
0x004012ef <main+1>:    mov    %esp,%ebp
0x004012f1 <main+3>:    sub    $0x18,%esp
0x004012f4 <main+6>:    and    $0xfffffff0,%esp
0x004012f7 <main+9>:    mov    $0x0,%eax
0x004012fc <main+14>:   add    $0xf,%eax
0x004012ff <main+17>:   add    $0xf,%eax
0x00401302 <main+20>:   shr    $0x4,%eax
0x00401305 <main+23>:   shl    $0x4,%eax
0x00401308 <main+26>:   mov    %eax,0xfffffff8(%ebp)
0x0040130b <main+29>:   mov    0xfffffff8(%ebp),%eax
0x0040130e <main+32>:   call   0x401b00 <_alloca>
0x00401313 <main+37>:   call   0x4017b0 <__main>
0x00401318 <main+42>:   movl   $0x0,0xfffffffc(%ebp)
0x0040131f <main+49>:   movl   $0x3,0x8(%esp)
0x00401327 <main+57>:   movl   $0x2,0x4(%esp)
0x0040132f <main+65>:   movl   $0x1,(%esp)
0x00401336 <main+72>:   call   0x4012d0 <function>
0x0040133b <main+77>:   movl   $0x1,0xfffffffc(%ebp)
0x00401342 <main+84>:   mov    0xfffffffc(%ebp),%eax
0x00401345 <main+87>:   mov    %eax,0x4(%esp)
0x00401349 <main+91>:   movl   $0x403000,(%esp)
0x00401350 <main+98>:   call   0x401b60 <printf>
0x00401355 <main+103>:  leave
0x00401356 <main+104>:  ret
0x00401357 <main+105>:  nop
0x00401358 <main+106>:  add    %al,(%eax)
0x0040135a <main+108>:  add    %al,(%eax)
0x0040135c <main+110>:  add    %al,(%eax)
0x0040135e <main+112>:  add    %al,(%eax)
End of assembler dump.

Dump of assembler code for function function:
0x004012d0 <function+0>:        push   %ebp
0x004012d1 <function+1>:        mov    %esp,%ebp
0x004012d3 <function+3>:        sub    $0x38,%esp
0x004012d6 <function+6>:        lea    0xffffffe8(%ebp),%eax
0x004012d9 <function+9>:        add    $0xc,%eax
0x004012dc <function+12>:       mov    %eax,0xffffffd4(%ebp)
0x004012df <function+15>:       mov    0xffffffd4(%ebp),%edx
0x004012e2 <function+18>:       mov    0xffffffd4(%ebp),%eax
0x004012e5 <function+21>:       movzbl (%eax),%eax
0x004012e8 <function+24>:       add    $0x5,%al
0x004012ea <function+26>:       mov    %al,(%edx)
0x004012ec <function+28>:       leave
0x004012ed <function+29>:       ret

In my case the distance should be – = 5,right?But it seems not working..

Why function needs 56 bytes for local variables?( sub $0x38,%esp )

Answer 1

As joveha pointed out, the value of EIP saved on the stack (return address) by the call instruction needs to be incremented by 7 bytes (0x00401342 – 0x0040133b = 7) in order to skip the x = 1; instruction (movl $0x1,0xfffffffc(%ebp)).

You are correct that 56 bytes are being reserved for local variables (sub $0x38,%esp), so the missing piece is how many bytes past buffer1 on the stack is the saved EIP.

---

A bit of test code and inline assembly tells me that the magic value is 28 for my test. I cannot provide a definitive answer as to why it is 28, but I would assume the compiler is adding padding and/or stack canaries.

The following code was compiled using GCC 3.4.5 (MinGW) and tested on Windows XP SP3 (x86).

unsigned long get_ebp() {
   __asm__("pop %ebp\n\t"
           "movl %ebp,%eax\n\t"
           "push %ebp\n\t");
}

void function(int a, int b, int c) {
   char buffer1[5];
   char buffer2[10];
   int *ret;

   /* distance in bytes from buffer1 to return address on the stack */
   printf("test %d\n", ((get_ebp() + 4) - (unsigned long)&buffer1));

   ret = (int *)(buffer1 + 28);

   (*ret) += 7;
}

void main() {
   int x;

   x = 0;
   function(1,2,3);
   x = 1;
   printf("%d\n",x);
}

I could have just as easily used gdb to determine this value.

(compiled w/ -g to include debug symbols)

(gdb) break function
...
(gdb) run
...
(gdb) p $ebp
$1 = (void *) 0x22ff28
(gdb) p &buffer1
$2 = (char (*)[5]) 0x22ff10
(gdb) quit

(0x22ff28 + 4) – 0x22ff10 = 28

(ebp value + size of word) – address of buffer1 = number of bytes

---

In addition to Smashing The Stack For Fun And Profit, I would suggest reading some of the articles I mentioned in my answer to a previous question of yours and/or other material on the subject. Having a good understanding of exactly how this type of exploit works should help you write more secure code.