Question

diffusing binary bomb phase_2

Asked by: SirakMD

I tried several times but couldn't get it, perhaps i missed some point in the loop or keeping track of the stack, here is my disassembled code of phase_2. I don't think i understood clearly how it reads the numbers. I appreciate if anyone can help hep me. 

08048e99 <phase_2>:
 8048e99:	55                   	push   %ebp
 8048e9a:	89 e5                	mov    %esp,%ebp
 8048e9c:	56                   	push   %esi
 8048e9d:	53                   	push   %ebx
 8048e9e:	83 ec 30             	sub    $0x30,%esp
 8048ea1:	8d 45 e0             	lea    0xffffffe0(%ebp),%eax
 8048ea4:	89 44 24 04          	mov    %eax,0x4(%esp)
 8048ea8:	8b 45 08             	mov    0x8(%ebp),%eax
 8048eab:	89 04 24             	mov    %eax,(%esp)
 8048eae:	e8 1d 04 00 00       	call   80492d0 <read_six_numbers>
 8048eb3:	bb 02 00 00 00       	mov    $0x2,%ebx
 8048eb8:	8d 75 e0             	lea    0xffffffe0(%ebp),%esi
 8048ebb:	8b 44 9e f8          	mov    0xfffffff8(%esi,%ebx,4),%eax
 8048ebf:	83 c0 05             	add    $0x5,%eax
 8048ec2:	39 44 9e fc          	cmp    %eax,0xfffffffc(%esi,%ebx,4)
 8048ec6:	74 05                	je     8048ecd <phase_2+0x34>
 8048ec8:	e8 c1 03 00 00       	call   804928e <explode_bomb>
 8048ecd:	43                   	inc    %ebx
 8048ece:	83 fb 07             	cmp    $0x7,%ebx
 8048ed1:	75 e8                	jne    8048ebb <phase_2+0x22>
 8048ed3:	83 c4 30             	add    $0x30,%esp
 8048ed6:	5b                   	pop    %ebx
 8048ed7:	5e                   	pop    %esi
 8048ed8:	5d                   	pop    %ebp
 8048ed9:	c3                   	ret    
 
 
080492d0 <read_six_numbers>:
 80492d0:	55                   	push   %ebp
 80492d1:	89 e5                	mov    %esp,%ebp
 80492d3:	83 ec 28             	sub    $0x28,%esp
 80492d6:	8b 55 0c             	mov    0xc(%ebp),%edx
 80492d9:	8d 42 14             	lea    0x14(%edx),%eax
 80492dc:	89 44 24 1c          	mov    %eax,0x1c(%esp)
 80492e0:	8d 42 10             	lea    0x10(%edx),%eax
 80492e3:	89 44 24 18          	mov    %eax,0x18(%esp)
 80492e7:	8d 42 0c             	lea    0xc(%edx),%eax
 80492ea:	89 44 24 14          	mov    %eax,0x14(%esp)
 80492ee:	8d 42 08             	lea    0x8(%edx),%eax
 80492f1:	89 44 24 10          	mov    %eax,0x10(%esp)
 80492f5:	8d 42 04             	lea    0x4(%edx),%eax
 80492f8:	89 44 24 0c          	mov    %eax,0xc(%esp)
 80492fc:	89 54 24 08          	mov    %edx,0x8(%esp)
 8049300:	c7 44 24 04 b9 9a 04 	movl   $0x8049ab9,0x4(%esp)
 8049307:	08 
 8049308:	8b 45 08             	mov    0x8(%ebp),%eax
 804930b:	89 04 24             	mov    %eax,(%esp)
 804930e:	e8 d9 f5 ff ff       	call   80488ec <sscanf@plt>
 8049313:	83 f8 05             	cmp    $0x5,%eax
 8049316:	7f 05                	jg     804931d <read_six_numbers+0x4d>
 8049318:	e8 71 ff ff ff       	call   804928e <explode_bomb>
 804931d:	c9                   	leave  
 804931e:	89 f6                	mov    %esi,%esi
 8049320:	c3                   	ret

                                  
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Asked On
2009-10-29 at 19:05:50ID24856858
Topics

Assembly Programming Language

,

Unix Systems Programming

,

Algorithms

Participating Experts
1
Points
125
Comments
4

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Answers

 

by: Infinity08Posted on 2009-10-30 at 06:49:28ID: 25702894

What's your question ?

How far did you get, and where are you stuck ?

 

by: SirakMDPosted on 2009-10-31 at 14:35:26ID: 25711647

hi, I'm new to this site, and i didn't know the response was sent via my email, I figured out the first phase but i'm  struggling phase_2 and i've only few days for due. here are my questions
1) where the user inputs are stored or from which register does the function read_six_number read the inputs
2) when i run gdb, and check info registers, before the function read_six_numbers is called and right after the the function called, should not the value in the register *(%esp)  be the same as the input 6 numbers?
3) how do we check what values are in the registers at specific point to keep track of their values
for example if i want the values of the registers-- %eax and *(%esp) at this line
  8048eab:       89 04 24                mov    %eax,(%esp)
4) when i run info registers after the phase_1 through out the phase_2, to see what's goin' on there, i could get only the register values and when i check their values i'd get only garbage
5) in the following line is the cmp checking the value in the adresses that %eax and 0xfffffffc(%esi,%ebx,4) contain or just the adresses
 8048ec2:       39 44 9e fc             cmp    %eax,0xfffffffc(%esi,%ebx,4)
 8048ec6

thanks in advance,

 

by: Infinity08Posted on 2009-11-01 at 01:03:46ID: 25713071

>> 1) where the user inputs are stored or from which register does the function read_six_number read the inputs

It reads 6 numbers from the string that the user entered (passed as argument in 0x8(%ebp)). It uses sscanf for that :

        http://www.cplusplus.com/reference/clibrary/cstdio/sscanf/

>> 2) when i run gdb, and check info registers, before the function read_six_numbers is called and right after the the function called, should not the value in the register *(%esp)  be the same as the input 6 numbers?

No. The 6 numbers are stored on the stack, yes. But you'll have to read the read_six_numbers function to know where exactly they are stored. Hint : you can find the information before the sscanf call.


>> 3) how do we check what values are in the registers at specific point to keep track of their values

'info registers' shows you the content of all registers. You can use the gdb print command to show the contents of a specific register. Eg. :

>> for example if i want the values of the registers-- %eax and *(%esp) at this line
>>   8048eab:       89 04 24                mov    %eax,(%esp)

        print $eax
        print *($esp)

Refer to the gdb documentation for more information.

        http://sourceware.org/gdb/current/onlinedocs/gdb_11.html#SEC66


>> 4) when i run info registers after the phase_1 through out the phase_2, to see what's goin' on there, i could get only the register values

Well, yes, that's what the command is for. If you need to see more information (you probably want to look at the stack contents too), then you'll need to show that too.

>> and when i check their values i'd get only garbage

They might look like garbage, but they probably are not ;) Think about what the value is supposed to represent (and address ? A value ? ...), and then interpret it as such.


>> 5) in the following line is the cmp checking the value in the adresses that %eax and 0xfffffffc(%esi,%ebx,4) contain or just the adresses
>>  8048ec2:       39 44 9e fc             cmp    %eax,0xfffffffc(%esi,%ebx,4)

It is comparing the contents of the eax register with the contents of the memory location indicated by 0xfffffffc(%esi,%ebx,4).

 

by: SirakMDPosted on 2009-11-01 at 13:09:35ID: 25715488

yes!!! thanks a lot, I got it!!!
Now I'm working on phase_3

20120131-EE-VQP-002

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