Question

Bomblab Phase 2

Asked by: Cosmonaut

Hello!
I am working on the second phase of the binary bomb/bomblab project, and I am a bit stuck.
Basically, we have to input 6 integers to keep the bomb from exploding - however they must be the 6 correct digits, and you have to find them in order.  I am using the gdb compiler.  
This is my code and the comments are how I understand it at the moment:

08048dc2 <phase_2>:
 8048dc2:      55                         push   %ebp
 8048dc3:      89 e5                      mov    %esp,%ebp
 8048dc5:      56                         push   %esi
 8048dc6:      53                         push   %ebx
 8048dc7:      83 ec 30                   sub    $0x30,%esp
 8048dca:      8d 45 e0                   lea    -0x20(%ebp),%eax
 8048dcd:      89 44 24 04                mov    %eax,0x4(%esp)
 8048dd1:      8b 45 08                   mov    0x8(%ebp),%eax
 8048dd4:      89 04 24                   mov    %eax,(%esp) //*esp = eax <- esp points to eax
 8048dd7:      e8 37 06 00 00             call   8049413 <read_six_numbers> //function call
 8048ddc:      bb 02 00 00 00             mov    $0x2,%ebx // ebx = 2
 8048de1:      8d 75 e0                   lea    -0x20(%ebp),%esi // loop -0x20(%ebp) = &esi
 8048de4:      8b 44 9e f8                mov    -0x8(%esi,%ebx,4),%eax // eax = esi[ebx-2]
 8048de8:      83 c0 05                   add    $0x5,%eax // eax += 5
 8048deb:      39 44 9e fc                cmp    %eax,-0x4(%esi,%ebx,4) // if (esi[ebx-1] == eax)
 8048def:      74 05                      je     8048df6 <phase_2+0x34> // safe
 8048df1:      e8 db 05 00 00             call   80493d1 <explode_bomb> // yuck
 8048df6:      83 c3 01                   add    $0x1,%ebx // ebx += 1
 8048df9:      83 fb 07                   cmp    $0x7,%ebx // if ebx != 7
 8048dfc:      75 e6                      jne    8048de4 <phase_2+0x22> // go to loop
 8048dfe:      83 c4 30                   add    $0x30,%esp
 8048e01:      5b                         pop    %ebx
 8048e02:      5e                         pop    %esi
 8048e03:      5d                         pop    %ebp
 8048e04:      c3                         ret  

08049413 <read_six_numbers>: //checks to make sure you have entered 6 integers & puts them in stack then returns to phase 2
 8049413:      55                         push   %ebp
 8049414:      89 e5                      mov    %esp,%ebp
 8049416:      83 ec 28                   sub    $0x28,%esp
 8049419:      8b 55 0c                   mov    0xc(%ebp),%edx
 804941c:      8d 42 14                   lea    0x14(%edx),%eax
 804941f:      89 44 24 1c                mov    %eax,0x1c(%esp)
 8049423:      8d 42 10                   lea    0x10(%edx),%eax
 8049426:      89 44 24 18                mov    %eax,0x18(%esp)
 804942a:      8d 42 0c                   lea    0xc(%edx),%eax
 804942d:      89 44 24 14                mov    %eax,0x14(%esp)
 8049431:      8d 42 08                   lea    0x8(%edx),%eax
 8049434:      89 44 24 10                mov    %eax,0x10(%esp)
 8049438:      8d 42 04                   lea    0x4(%edx),%eax
 804943b:      89 44 24 0c                mov    %eax,0xc(%esp)
 804943f:      89 54 24 08                mov    %edx,0x8(%esp)
 8049443:      c7 44 24 04 59 9c 04       movl   $0x8049c59,0x4(%esp)
 804944a:      08
 804944b:      8b 45 08                   mov    0x8(%ebp),%eax
 804944e:      89 04 24                   mov    %eax,(%esp)
 8049451:      e8 7e f5 ff ff             call   80489d4 <sscanf@plt>
 8049456:      83 f8 05                   cmp    $0x5,%
 8049459:      7f 05                      jg     8049460 <read_six_numbers+0x4d>
 804945b:      e8 71 ff ff ff             call   80493d1 <explode_bomb>
 8049460:      c9                         leave  
 8049461:      c3                         ret    

I have been using a few examples to try and walk through the code, and my main problem at the moment is that I feel I should be able to use "p $eax"  at this point "cmp    %eax,-0x4(%esi,%ebx,4)" to see what is being compared to the numbers I input.  It has been returning 6, however, if I use that for my first integer the bomb explodes.  

Any help would be greatly appreciated!!!

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Asked On
2009-10-20 at 06:23:17ID24826743
Tags

Binary Bomb

,

Phase 2

,

Bomblab

Topics

Assembly Programming Language

,

Programming Languages

,

C Programming Language

Participating Experts
1
Points
500
Comments
8

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Answers

 

by: Infinity08Posted on 2009-10-20 at 07:24:49ID: 25614466

>>  8048dd4:      89 04 24                   mov    %eax,(%esp) //*esp = eax <- esp points to eax

Not exactly. This instruction places the value in eax at the top of the stack (esp is the stack pointer).


>>  8048de1:      8d 75 e0                   lea    -0x20(%ebp),%esi // loop -0x20(%ebp) = &esi

lea is the "load effective address" instruction. Also remember that the target is on the right, so :

        esi = ebp - 0x20


>>  8048de4:      8b 44 9e f8                mov    -0x8(%esi,%ebx,4),%eax // eax = esi[ebx-2]

ok, assuming that esi points to an array of 32bit values (which is the case).


>> It has been returning 6, however, if I use that for my first integer the bomb explodes.  

As long as you don't fill in the 6 correct numbers, the bomb will still explode. In your debugger however, you should see that you're getting further if some of the numbers are correct.

 

by: CosmonautPosted on 2009-10-20 at 07:40:26ID: 25614635

>> Not exactly. This instruction places the value in eax at the top of the stack (esp is the stack pointer).
Ok that makes much more sense now, thank you!

>>As long as you don't fill in the 6 correct numbers, the bomb will still explode. In your debugger
>>however, you should see that you're getting further if some of the numbers are correct.
So, would a good way to check my progress to be to have a breakpoint @ "cmp    %eax,-0x4(%esi,%ebx,4)" and use "p $eax"  then re-run the program (6 times) with the new information gained?

I'm sorry this is such a common question topic - I've been reading the old posts on this project, as well as reviewing stacks, but every version of the code seems to be slightly different and it's kind of throwing me off.

 

by: CosmonautPosted on 2009-10-20 at 08:13:48ID: 25615017

I went ahead and attempted the breakpoint solution I asked about in the last comment and it worked!!
Thank you so much for your help!
I am attaching a portion of the solution code, because I was wondering (just for a better knowledge of assembly) if there is a better way to do what I just did.

Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$1 = 11
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$2 = 16
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$3 = 21
 
(gdb) r input.txt
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Phase 1 defused. How about the next one?
6 11 16 21 99 98
 
Breakpoint 1, 0x08048dc7 in phase_2 ()
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$4 = 11
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$5 = 16
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$6 = 21
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$7 = 26
(gdb) r input.txt
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Phase 1 defused. How about the next one?
6 11 16 21 26 99
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$7 = 11
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$8 = 16
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$9 = 21
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$10 = 26
(gdb) c
Continuing.
 
Breakpoint 2, 0x08048deb in phase_2 ()
(gdb) p $eax
$16 = 31

                                              
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by: Infinity08Posted on 2009-10-20 at 08:26:09ID: 25615140

>> So, would a good way to check my progress to be to have a breakpoint @ "cmp    %eax,-0x4(%esi,%ebx,4)" and use "p $eax"  then re-run the program (6 times) with the new information gained?

That's one possible approach.
Another approach is to simply understand what's happening in the code, and you'll immediately know all 6 numbers.


>> but every version of the code seems to be slightly different and it's kind of throwing me off.

That's intentional ;)


>> I went ahead and attempted the breakpoint solution I asked about in the last comment and it worked!!

Great !


>> if there is a better way to do what I just did.

That's just fine. Personally I prefer reading the assembly code to understand what's happening, but that comes with practice :)

 

by: CosmonautPosted on 2009-10-20 at 08:36:22ID: 25615238

So, if you were just reading the code to understand what's going on could you just interpret "add    $0x5,%eax" to mean that each number will be the last number plus five?  If so, how would you go about getting the first number?

 

by: Infinity08Posted on 2009-10-20 at 09:59:01ID: 25616118

>> could you just interpret "add    $0x5,%eax" to mean that each number will be the last number plus five?

Indeed.


>> If so, how would you go about getting the first number?

Try using a different value as first number and see what happens ;)

 

by: Infinity08Posted on 2009-10-20 at 09:59:22ID: 25616122

>> Try using a different value as first number and see what happens ;)

With which I mean that it doesn't matter.

 

by: CosmonautPosted on 2009-10-20 at 10:43:08ID: 25616571

That's awesome!
Thanks again - you've been an amazing help!

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