29 of November 2011 was the date of public disclosure interesting vulnerability in lighttpd server. Xi Wang discovered that mod_auth for this server does not propely decode characters from the extended ASCII table. The vulnerable code is below:

"src/http_auth.c:67"
--- CUT ---
static const short base64_reverse_table[256] = ...;
static unsigned char * base64_decode(buffer *out, const char *in) {
    ...
    int ch, ...;
    size_t i;
    ...

        ch = in[i];
        ...
        ch = base64_reverse_table[ch];
    ...
}
--- CUT ---

Because variable ‘in’ is type ‘char’, characters above 0x80 lead to negative indices. This vulnerability may lead out-of-boud read and theoretically cause Segmentation Fault (Denial of Service attack). Unfortunately I couldn’t find any binaries where .rodata section before the base64_reverse_table table cause this situation.

I have added some extra debug in the lighttpd source code to see if this vulnerability is executed correctly. Here is output for one of the example:

--- CUT ---
ptr[0x9a92c48] size[0xc0] used[0x0]
127(. | 0 | 0)
-128(t | 1 | 0)
-127(e | 2 | 1)
-126(' | 3 | 2)
-125(e | 4 | 3)
-124(u | 5 | 3)
-123(r | 6 | 4)
-122(' | 7 | 5)
-121(s | 8 | 6)
-120(c | 9 | 6)
-119(i | 10 | 7)
-118(n | 11 | 8)
-117(i | 12 | 9)
-116(  | 13 | 9)
-115(a | 14 | 10)
-114(t | 15 | 11)
-113(. | 16 | 12)
-112(e | 17 | 12)
-111(u | 18 | 13)
-110(r | 19 | 14)
-109(' | 20 | 15)
-108(f | 21 | 15)
-107(i | 22 | 16)
-106(e | 23 | 17)
-105(: | 24 | 18)
-104(= | 25 | 18)
-103(o | 26 | 19)
-102(t | 27 | 20)
-101(o | 28 | 21)
-100(  | 29 | 21)
-99(a | 30 | 22)
-98(g | 31 | 23)
-97(. | 32 | 24)
-96(d | 33 | 24)
-95(g | 34 | 25)
-94(s | 35 | 26)
-93(: | 36 | 27)
-92(u | 37 | 27)
-91(s | 38 | 28)
-90(p | 39 | 29)
-89(o | 40 | 30)
-88(t | 41 | 30)
-87(d | 42 | 31)
-86(b | 43 | 32)
-85(c | 44 | 33)
-84(e | 45 | 33)
-83(d | 46 | 34)
-82(( | 47 | 35)
-81(n | 48 | 36)
-80(y | 49 | 36)
-79(h | 50 | 37)
-78(d | 51 | 38)
-77(g | 52 | 39)
-76(s | 53 | 39)
-75(  | 54 | 40)
-74(r | 55 | 41)
-73(p | 56 | 42)
-72(a | 57 | 42)
-71(n | 58 | 43)
-70(. | 59 | 44)
-69(. | 60 | 45)
-68(d | 61 | 45)
-67(g | 62 | 46)
-66(s | 63 | 47)
-65(: | 64 | 48)
-64(( | 65 | 48)
-63(d | 66 | 49)
-62(- | 67 | 50)
-61(e | 68 | 51)
-60(s | 69 | 51)
-59(  | 70 | 52)
-58(i | 71 | 53)
-57(s | 72 | 54)
-56(n | 73 | 54)
-55(  | 74 | 55)
-54(i | 75 | 56)
-53(l | 76 | 57)
-52(. | 77 | 57)
-51(. | 78 | 58)
-50(k | 79 | 59)
-49(0 | 80 | 60)
-48(% | 81 | 60)
-47(] | 82 | 61)
-46(p | 83 | 62)
-45(r | 84 | 63)
-44(0 | 85 | 63)
-43(% | 86 | 64)
-42(] | 87 | 65)
-41(s | 88 | 66)
-40(z | 89 | 66)
-39([ | 90 | 67)
-38(x | 91 | 68)
-37(x | 92 | 69)
-36(  | 93 | 69)
-35(s | 94 | 70)
-34(d | 95 | 71)
-33(0 | 96 | 72)
-32(% | 97 | 72)
-31(] | 98 | 73)
-30(. | 99 | 74)
-29(. | 100 | 75)
-28(d | 101 | 75)
-27(c | 102 | 76)
-26(d | 103 | 77)
-25(i | 104 | 78)
-24(g | 105 | 78)
-23(b | 106 | 79)
-22(s | 107 | 80)
-21(6 | 108 | 81)
-20(- | 109 | 81)
-19(t | 110 | 82)
-18(i | 111 | 83)
-17(g | 112 | 84)
-16(f | 113 | 84)
-15(i | 114 | 85)
-14(e | 115 | 86)
-13(. | 116 | 87)
-12(. | 117 | 87)
-11(. | 118 | 88)
-10(. | 119 | 89)
-9(. | 120 | 90)
-8(. | 121 | 90)
-7(. | 122 | 91)
-6(. | 123 | 92)
-5(. | 124 | 93)
-4(. | 125 | 93)
-3(. | 126 | 94)
-2(. | 127 | 95)
-1(. | 128 | 96)
k[0x60] ptr[0x9a92c48] size[0xc0] used[0x0]
ptr[0x9a92c48] size[0xc0] used[0x60]
string [.Yg.\...n.Xt.]r.ze.....g.Y..\..Yb.Y(..d..r.[..Y...-.xi..i.]
--- CUT ---

First column is the offset so vulnerability is executed like it should be (negative offsets). Second column is byte which is read out-of-bound.

How to run this very primitive Proof of Concept?

$ gcc p_cve-2011-4362.c -o p_cve-2011-4362
$ ./p_cve-2011-4362 

    ...::: -=[ Proof of Concept for CVE-2011-4362 (by Adam 'pi3' Zabrocki) ]=- :::...

    Usage: ./p_cve-2011-4362 <options>

        Options:
             -v <victim>
             -p <port>
             -d <remote_dir_for_auth>

$ ./p_cve-2011-4362 -h 127.0.0.1 -p 81 -d dupa

    ...::: -=[ Proof of Concept for CVE-2011-4362 (by Adam 'pi3' Zabrocki) ]=- :::...

        [+] Preparing arguments... OK
        [+] Creating socket... OK
        [+] Connecting to [127.0.0.1]... OK
        [+] Sending dirty packet... OK

        [+] Check the website!

$

Lighttpd will log this situation probably in error-log file like this:

--- CUT ---
...
...
2011-12-xx xx:xx:11: (http_auth.c.887) : is missing in ÇYg\§ÎúnöXt¾]rzeëÛô¾gYóï\ðÿYbîY(¿dßørÖ[YóúÙ-·xiþèi°kÂWpË    ]߶øò\äÂ×@VØä¦xóúÝize
--- CUT ---

Maybe you can find vulnerable binary?

Best regards,
Adam ‘pi3’ Zabrocki


http://pi3.com.pl
http://site.pi3.com.pl/exp/p_cve-2011-4362.c
http://blog.pi3.com.pl/?p=277

Second level of GCHQ ‘canyoucrackit’ challenge requires to implement own Virtual Machine(!). This VM must emulate segmented memory model with 16-byte segment size (notation seg:offset). For details please read this link:

http://www.canyoucrackit.co.uk/15b436de1f9107f3778aad525e5d0b20.js

I wrote quick overview about this challenge, how to solve it and some tips. It can be found here:

http://blog.pi3.com.pl/?p=213

Anyway, I am impressed how many people saw this post and how fast this link was shared in community 🙂 Of course I’m happy of that but also a bit terrified. Anyway, in this short post I didn’t put much details about how to implement this VM, if there is any difficulties, etc. This was one of the reason I received a few emails asking some help to solve it. This is the reason why I decide finally write this second post. I want to share with my VM which i wrote in pure C (I love this language). To be honest I didn’t implement it at the beginning like it is here. I found some implementation in the http://pastebin.com webpage in python language. Unfortunately it has some mistakes (in fact serious mistakes). This was the reason why this machine didn’t work properly and in fact after a few instruction put exceptions and of course whole VM stops. I spend some time to fix it and I did it. After rewriting this machine, python VM starts working. This machine had a few problems like doesn’t correct  implement the most important instructions (JMP and JMPE). Also there was mistakes in take care about MOD flag. Another bug was that CS and DS register can be used in operations like ADD via normal operand argument as register. Also operations which use addressing [seg:off] must especially take care if the arguments are inside of the SEGMENT, if not make them fix. In fact this was critical bugs.

Anyway because of that I rewrote almost whole program so after all I decide, OK let’s do that in my way and this was the point why I implement everything again in C. Here you have got my VM in pure C:

http://site.pi3.com.pl/exp/pi3_VM.c

 

Btw. In fact this challenge is NOT finished yet… Maybe it was mistake to publish solution BEFORE end of it? I feel a bit guilty.

 

Best regards,

Adam Zabrocki

Yesterday I read in one of the polish portal (with news) an  information about interesting challenge organized by the Government Communications Headquarters (GCHQ). This is a British intelligence agency responsible for providing signals intelligence (SIGINT) and information assurance to the UK government and armed forces. Based in Cheltenham, it operates under the guidance of the Joint Intelligence Committee. CESG (originally Communications-Electronics Security Group) is the branch of GCHQ which works to secure the communications and information systems of the government and critical parts of UK national infrastructure.

GCHQ, is aiming to attract the next generation of web-savvy spies by running an ad campaign that challenges computer hackers to crack a code to get an interview.

Ok so let’s look at it closer 🙂

 

First level

I will not analyze the security of webpage and server. I will try to discuss about pure challenge. OK so first question is what does this hexcode means? Let’s look it closer:

0xeb 0x04 0xaf 0xc2 0xbf 0xa3 0x81 0xec   0x00 0x01 0x00 0x00 0x31 0xc9 0x88 0x0c
0x0c 0xfe 0xc1 0x75 0xf9 0x31 0xc0 0xba   0xef 0xbe 0xad 0xde 0x02 0x04 0x0c 0x00
0xd0 0xc1 0xca 0x08 0x8a 0x1c 0x0c 0x8a   0x3c 0x04 0x88 0x1c 0x04 0x88 0x3c 0x0c
0xfe 0xc1 0x75 0xe8 0xe9 0x5c 0x00 0x00   0x00 0x89 0xe3 0x81 0xc3 0x04 0x00 0x00
0x00 0x5c 0x58 0x3d 0x41 0x41 0x41 0x41   0x75 0x43 0x58 0x3d 0x42 0x42 0x42 0x42
0x75 0x3b 0x5a 0x89 0xd1 0x89 0xe6 0x89   0xdf 0x29 0xcf 0xf3 0xa4 0x89 0xde 0x89
0xd1 0x89 0xdf 0x29 0xcf 0x31 0xc0 0x31   0xdb 0x31 0xd2 0xfe 0xc0 0x02 0x1c 0x06
0x8a 0x14 0x06 0x8a 0x34 0x1e 0x88 0x34   0x06 0x88 0x14 0x1e 0x00 0xf2 0x30 0xf6
0x8a 0x1c 0x16 0x8a 0x17 0x30 0xda 0x88   0x17 0x47 0x49 0x75 0xde 0x31 0xdb 0x89
0xd8 0xfe 0xc0 0xcd 0x80 0x90 0x90 0xe8   0x9d 0xff 0xff 0xff 0x41 0x41 0x41 0x41

For the first quick view we can say there is some interesting bytes. With red color I sign this bytes which can be a dump of x86 assembler instructions (of course there is more but this can be typical for shellcodes) and with the blue color I sign interesting bytes for me 🙂 – 0x41414141 and 0x424242 for me always will be connected with exploiting 😉

OK so let’s try to analyze this bytes as x86 instructions:

   0x0804a040 <+0>:    jmp    0x804a046 <shellcode+6>
   0x0804a042 <+2>:    scas   %es:(%edi),%eax
   0x0804a043 <+3>:    ret    $0xa3bf
   0x0804a046 <+6>:    sub    $0x100,%esp
   0x0804a04c <+12>:    xor    %ecx,%ecx
   0x0804a04e <+14>:    mov    %cl,(%esp,%ecx,1)
   0x0804a051 <+17>:    inc    %cl
   0x0804a053 <+19>:    jne    0x804a04e <shellcode+14>
   0x0804a055 <+21>:    xor    %eax,%eax
   0x0804a057 <+23>:    mov    $0xdeadbeef,%edx
   0x0804a05c <+28>:    add    (%esp,%ecx,1),%al
   0x0804a05f <+31>:    add    %dl,%al
   0x0804a061 <+33>:    ror    $0x8,%edx
   0x0804a064 <+36>:    mov    (%esp,%ecx,1),%bl
   0x0804a067 <+39>:    mov    (%esp,%eax,1),%bh
   0x0804a06a <+42>:    mov    %bl,(%esp,%eax,1)
   0x0804a06d <+45>:    mov    %bh,(%esp,%ecx,1)
   0x0804a070 <+48>:    inc    %cl
   0x0804a072 <+50>:    jne    0x804a05c <shellcode+28>
   0x0804a074 <+52>:    jmp    0x804a0d5 <shellcode+149>
   0x0804a079 <+57>:    mov    %esp,%ebx
   0x0804a07b <+59>:    add    $0x4,%ebx
   0x0804a081 <+65>:    pop    %esp
   0x0804a082 <+66>:    pop    %eax
   0x0804a083 <+67>:    cmp    $0x41414141,%eax
   0x0804a088 <+72>:    jne    0x804a0cd <shellcode+141>
   0x0804a08a <+74>:    dec    %eax
   0x0804a08b <+75>:    cmp    $0x42424242,%eax
   0x0804a090 <+80>:    jne    0x804a0cd <shellcode+141>
   0x0804a092 <+82>:    pop    %edx
   0x0804a093 <+83>:    mov    %edx,%ecx
   0x0804a095 <+85>:    mov    %esp,%esi
   0x0804a097 <+87>:    mov    %ebx,%edi
   0x0804a099 <+89>:    sub    %ecx,%edi
   0x0804a09b <+91>:    rep movsb %ds:(%esi),%es:(%edi)
   0x0804a09d <+93>:    mov    %ebx,%esi
   0x0804a09f <+95>:    mov    %edx,%ecx
   0x0804a0a1 <+97>:    mov    %ebx,%edi
   0x0804a0a3 <+99>:    sub    %ecx,%edi
   0x0804a0a5 <+101>:    xor    %eax,%eax
   0x0804a0a7 <+103>:    xor    %ebx,%ebx
   0x0804a0a9 <+105>:    xor    %edx,%edx
   0x0804a0ab <+107>:    inc    %al
   0x0804a0ad <+109>:    add    (%esi,%eax,1),%bl
   0x0804a0b0 <+112>:    mov    (%esi,%eax,1),%dl
   0x0804a0b3 <+115>:    mov    (%esi,%ebx,1),%dh
   0x0804a0b6 <+118>:    mov    %dh,(%esi,%eax,1)
   0x0804a0b9 <+121>:    mov    %dl,(%esi,%ebx,1)
   0x0804a0bc <+124>:    add    %dh,%dl
   0x0804a0be <+126>:    xor    %dh,%dh
   0x0804a0c0 <+128>:    mov    (%esi,%edx,1),%bl
   0x0804a0c3 <+131>:    mov    (%edi),%dl
   0x0804a0c5 <+133>:    xor    %bl,%dl
   0x0804a0c7 <+135>:    mov    %dl,(%edi)
   0x0804a0c9 <+137>:    inc    %edi
   0x0804a0ca <+138>:    dec    %ecx
   0x0804a0cb <+139>:    jne    0x804a0ab <shellcode+107>
   0x0804a0cd <+141>:    xor    %ebx,%ebx
   0x0804a0cf <+143>:    mov    %ebx,%eax
   0x0804a0d1 <+145>:    inc    %al
   0x0804a0d3 <+147>:    int    $0x80
   0x0804a0d5 <+149>:    nop
   0x0804a0d6 <+150>:    nop
   0x0804a0d7 <+151>:    call   0x804a079 <shellcode+57>
   0x0804a0dc <+156>:    inc    %ecx
   0x0804a0dd <+157>:    inc    %ecx
   0x0804a0de <+158>:    inc    %ecx
   0x0804a0df <+159>:    inc    %ecx
   0x0804a0e0 <+160>:    add    %al,(%eax)

So this is it. This code make sense and this was good way of analyzing. I sign by read color this instruction which always cause exit() syscall and half of the code won’t be executed (as we will see further even more). But first, at the beginning this code jump over 2 next instruction (so they are never executed) and than allocate memory which is filled by natural numbers. Next they are converted to some more interested values and finally there is static jump to the code which cause syscall exit() – red colour. That’s all, so what next? As we can see directly after jump instruction, the program tries to get the new value for the stack pointer exactly from the stack. It gaves us an information that smth should be changed there 😉 Also as we can see further (blue colour) from the stack is popped also value for %%eax register and compared with the 0x41414141 value. Next this value is decremented by one and again compared but now with the 0x42424242 value. Logically it makes no sense. If first compare will be true than next will be bad – 0x41414141 – 1 = 0x41414140 so it will never be 0x42424242. If we want to pass all this checks and executed further code we must change a lot 😉

First change

Ok let’s come back to the syscall exit(). We don’t want to stop the execution flow but continue, and we know that further code expect the new stack pointer in the stack. Also we know that after this operation program tries to get new value for %%eax register from the new stack and compare with the value 0x414141. As we can see in the end of the shellcode we have instructions:

   0x0804a0dc <+156>:    inc    %ecx
   0x0804a0dd <+157>:    inc    %ecx
   0x0804a0de <+158>:    inc    %ecx
   0x0804a0df <+159>:    inc    %ecx

this is exactly the value 0x414141:

(gdb) x/x 0x0804a0dc
0x804a0dc <shellcode+156>:    0x41414141
(gdb)

so here we go with answer 😉 Like in the oldschool technique of getting current stack pointer used in viruses – let’s change syscall exit() to the call which gave us back the flow to the shellcode. Old code:

   0x0804a0d5 <+149>:    nop
   0x0804a0d6 <+150>:    nop
   0x0804a0d7 <+151>:    call   0x804a079 <shellcode+57> # value: 0x80cd

New code:

   0x804c095:    nop
   0x804c096:    nop
   0x804c097:    call   0x804c039 # value: 0x45eb

OK – works. First compare is passed but of course second is not and again game over. But If we think again about it, decrementing instruction can be also overwrite to ours and as we know the new stack pointer is now _after_ the shellcode. We are able to add new bytes after this shellcode and change the assembler instruction which decrements value in register %%eax to pop new value from the stack 🙂 This is in fact the answer 😉

Second change

Old instruction:

   0x0804a08a <+74>:    dec    %eax  # byte: 0x48

New instruction:

   0x804c04a:    pop    %eax  # byte: 0x58

Perfect. Now we have another problem – which data should be added in the end of shellcode? We can manually add 0x42424242 value to pass the compare check but what about further code? Maybe this 0x42424242  is a tip? In further code the value for %%edx register is also popped from the stack. And next bytes are used to copy in temporary place and manipulate them. So of course all of this is a tip. Lets come back again to the main page of the crack site. Bytes which we used to create shellcode are not in the site as text but as image. We were frustrating to rewrite them manually not just simply copying. But wait a minute why this is an image?

Steganography

Analyzing image can be hard 😉 But I’m lazy buster and usually before I move to real hard job like analyze or full RE I try to get as much information as I can in as simple way as it can be. So let’s run strings command 😉 There is 983 lines (not small image) but one line from the top is very interesting:

$ strings cyber.png |head
IHDR
sRGB
    pHYs
tIME
]iTXtComment
QkJCQjIAAACR2PFtcCA6q2eaC8SR+8dmD/zNzLQC+td3tFQ4qx8O447TDeuZw5P+0SsbEcYR
78jKLw==2
IDATx
.^cwuW
$

Here you go 😉 First impression is – this is base64. Let’s check in one of the online sites if there is logic in this string. After decoding to the ASCII we see that hexdump should be done _but_ look for the first 4 bytes in this string:

BBBB2∅∅∅Øñmp :«gÄûÇfüÍÌ´ú×w´T8
ëÃþÑ+ÆïÈÊ/?ÿØ∅∅

Yes, this is exactly 0x42424242 value.  So probably this is what we need 😉 After adding this hex in the end of the shellcode and all of our changes and adding the code for dumping the memory after whole process of executing shellcode (this dump function write by yourself) we will see this beautiful message from the memory:

GET /15b436de1f9107f3778aad525e5d0b20.js HTTP/1.1

Interesting, isn’t it? 🙂 Yep the first level is done.

 

Second level

What does this link have?

//--------------------------------------------------------------------------------------------------
//
// stage 2 of 3
//
// challenge:
//   reveal the solution within VM.mem
//
// disclaimer:
//   tested in ie 9, firefox 6, chrome 14 and v8 shell (http://code.google.com/apis/v8/build.html),
//   other javascript implementations may or may not work.
//
//--------------------------------------------------------------------------------------------------

var VM = {

  cpu: {
    ip: 0x00,

    r0: 0x00,
    r1: 0x00,
    r2: 0x00,
    r3: 0x00,

    cs: 0x00,
    ds: 0x10,

    fl: 0x00,

    firmware: [0xd2ab1f05, 0xda13f110]
  },

  mem: [
    0x31, 0x04, 0x33, 0xaa, 0x40, 0x02, 0x80, 0x03, 0x52, 0x00, 0x72, 0x01, 0x73, 0x01, 0xb2, 0x50,
    0x30, 0x14, 0xc0, 0x01, 0x80, 0x00, 0x10, 0x10, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,

    0x98, 0xab, 0xd9, 0xa1, 0x9f, 0xa7, 0x83, 0x83, 0xf2, 0xb1, 0x34, 0xb6, 0xe4, 0xb7, 0xca, 0xb8,
    0xc9, 0xb8, 0x0e, 0xbd, 0x7d, 0x0f, 0xc0, 0xf1, 0xd9, 0x03, 0xc5, 0x3a, 0xc6, 0xc7, 0xc8, 0xc9,
    0xca, 0xcb, 0xcc, 0xcd, 0xce, 0xcf, 0xd0, 0xd1, 0xd2, 0xd3, 0xd4, 0xd5, 0xd6, 0xd7, 0xd8, 0xd9,
    0xda, 0xdb, 0xa9, 0xcd, 0xdf, 0xdf, 0xe0, 0xe1, 0xe2, 0xe3, 0xe4, 0xe5, 0xe6, 0xe7, 0xe8, 0xe9,
    0x26, 0xeb, 0xec, 0xed, 0xee, 0xef, 0xf0, 0xf1, 0xf2, 0xf3, 0xf4, 0xf5, 0xf6, 0xf7, 0xf8, 0xf9,
    0x7d, 0x1f, 0x15, 0x60, 0x4d, 0x4d, 0x52, 0x7d, 0x0e, 0x27, 0x6d, 0x10, 0x6d, 0x5a, 0x06, 0x56,
    0x47, 0x14, 0x42, 0x0e, 0xb6, 0xb2, 0xb2, 0xe6, 0xeb, 0xb4, 0x83, 0x8e, 0xd7, 0xe5, 0xd4, 0xd9,
    0xc3, 0xf0, 0x80, 0x95, 0xf1, 0x82, 0x82, 0x9a, 0xbd, 0x95, 0xa4, 0x8d, 0x9a, 0x2b, 0x30, 0x69,
    0x4a, 0x69, 0x65, 0x55, 0x1c, 0x7b, 0x69, 0x1c, 0x6e, 0x04, 0x74, 0x35, 0x21, 0x26, 0x2f, 0x60,
    0x03, 0x4e, 0x37, 0x1e, 0x33, 0x54, 0x39, 0xe6, 0xba, 0xb4, 0xa2, 0xad, 0xa4, 0xc5, 0x95, 0xc8,
    0xc1, 0xe4, 0x8a, 0xec, 0xe7, 0x92, 0x8b, 0xe8, 0x81, 0xf0, 0xad, 0x98, 0xa4, 0xd0, 0xc0, 0x8d,
    0xac, 0x22, 0x52, 0x65, 0x7e, 0x27, 0x2b, 0x5a, 0x12, 0x61, 0x0a, 0x01, 0x7a, 0x6b, 0x1d, 0x67,
    0x75, 0x70, 0x6c, 0x1b, 0x11, 0x25, 0x25, 0x70, 0x7f, 0x7e, 0x67, 0x63, 0x30, 0x3c, 0x6d, 0x6a,
    0x01, 0x51, 0x59, 0x5f, 0x56, 0x13, 0x10, 0x43, 0x19, 0x18, 0xe5, 0xe0, 0xbe, 0xbf, 0xbd, 0xe9,
    0xf0, 0xf1, 0xf9, 0xfa, 0xab, 0x8f, 0xc1, 0xdf, 0xcf, 0x8d, 0xf8, 0xe7, 0xe2, 0xe9, 0x93, 0x8e,
    0xec, 0xf5, 0xc8, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,

    0x37, 0x7a, 0x07, 0x11, 0x1f, 0x1d, 0x68, 0x25, 0x32, 0x77, 0x1e, 0x62, 0x23, 0x5b, 0x47, 0x55,
    0x53, 0x30, 0x11, 0x42, 0xf6, 0xf1, 0xb1, 0xe6, 0xc3, 0xcc, 0xf8, 0xc5, 0xe4, 0xcc, 0xc0, 0xd3,
    0x85, 0xfd, 0x9a, 0xe3, 0xe6, 0x81, 0xb5, 0xbb, 0xd7, 0xcd, 0x87, 0xa3, 0xd3, 0x6b, 0x36, 0x6f,
    0x6f, 0x66, 0x55, 0x30, 0x16, 0x45, 0x5e, 0x09, 0x74, 0x5c, 0x3f, 0x29, 0x2b, 0x66, 0x3d, 0x0d,
    0x02, 0x30, 0x28, 0x35, 0x15, 0x09, 0x15, 0xdd, 0xec, 0xb8, 0xe2, 0xfb, 0xd8, 0xcb, 0xd8, 0xd1,
    0x8b, 0xd5, 0x82, 0xd9, 0x9a, 0xf1, 0x92, 0xab, 0xe8, 0xa6, 0xd6, 0xd0, 0x8c, 0xaa, 0xd2, 0x94,
    0xcf, 0x45, 0x46, 0x67, 0x20, 0x7d, 0x44, 0x14, 0x6b, 0x45, 0x6d, 0x54, 0x03, 0x17, 0x60, 0x62,
    0x55, 0x5a, 0x4a, 0x66, 0x61, 0x11, 0x57, 0x68, 0x75, 0x05, 0x62, 0x36, 0x7d, 0x02, 0x10, 0x4b,
    0x08, 0x22, 0x42, 0x32, 0xba, 0xe2, 0xb9, 0xe2, 0xd6, 0xb9, 0xff, 0xc3, 0xe9, 0x8a, 0x8f, 0xc1,
    0x8f, 0xe1, 0xb8, 0xa4, 0x96, 0xf1, 0x8f, 0x81, 0xb1, 0x8d, 0x89, 0xcc, 0xd4, 0x78, 0x76, 0x61,
    0x72, 0x3e, 0x37, 0x23, 0x56, 0x73, 0x71, 0x79, 0x63, 0x7c, 0x08, 0x11, 0x20, 0x69, 0x7a, 0x14,
    0x68, 0x05, 0x21, 0x1e, 0x32, 0x27, 0x59, 0xb7, 0xcf, 0xab, 0xdd, 0xd5, 0xcc, 0x97, 0x93, 0xf2,
    0xe7, 0xc0, 0xeb, 0xff, 0xe9, 0xa3, 0xbf, 0xa1, 0xab, 0x8b, 0xbb, 0x9e, 0x9e, 0x8c, 0xa0, 0xc1,
    0x9b, 0x5a, 0x2f, 0x2f, 0x4e, 0x4e, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00
  ],

  exec: function()
  {
    // virtual machine architecture
    // ++++++++++++++++++++++++++++
    //
    // segmented memory model with 16-byte segment size (notation seg:offset)
    //
    // 4 general-purpose registers (r0-r3)
    // 2 segment registers (cs, ds equiv. to r4, r5)
    // 1 flags register (fl)
    //
    // instruction encoding
    // ++++++++++++++++++++
    //
    //           byte 1               byte 2 (optional)
    // bits      [ 7 6 5 4 3 2 1 0 ]  [ 7 6 5 4 3 2 1 0 ]
    // opcode      - - -
    // mod               -
    // operand1            - - - -
    // operand2                         - - - - - - - -
    //
    // operand1 is always a register index
    // operand2 is optional, depending upon the instruction set specified below
    // the value of mod alters the meaning of any operand2
    //   0: operand2 = reg ix
    //   1: operand2 = fixed immediate value or target segment (depending on instruction)
    //
    // instruction set
    // +++++++++++++++
    //
    // Notes:
    //   * r1, r2 => operand 1 is register 1, operand 2 is register 2
    //   * movr r1, r2 => move contents of register r2 into register r1
    //
    // opcode | instruction | operands (mod 0) | operands (mod 1)
    // -------+-------------+------------------+-----------------
    // 0x00   | jmp         | r1               | r2:r1
    // 0x01   | movr        | r1, r2           | rx,   imm
    // 0x02   | movm        | r1, [ds:r2]      | [ds:r1], r2
    // 0x03   | add         | r1, r2           | r1,   imm
    // 0x04   | xor         | r1, r2           | r1,   imm
    // 0x05   | cmp         | r1, r2           | r1,   imm
    // 0x06   | jmpe        | r1               | r2:r1
    // 0x07   | hlt         | N/A              | N/A
    //
    // flags
    // +++++
    //
    // cmp r1, r2 instruction results in:
    //   r1 == r2 => fl = 0
    //   r1 < r2  => fl = 0xff
    //   r1 > r2  => fl = 1
    //
    // jmpe r1
    //   => if (fl == 0) jmp r1
    //      else nop

    throw "VM.exec not yet implemented";
  }

};

//--------------------------------------------------------------------------------------------------

try
{
  VM.exec();
}
catch(e)
{
  alert('\nError: ' + e + '\n');
}

//--------------------------------------------------------------------------------------------------

As we can read this level is completely different from the previous 🙂 Short overview:

  • We must implement own Virtual Machine(!)
  • VM must emulate segmented memory model with 16-byte segment size (notation seg:offset)
  • There is defined own assembler with own simple architecture
  • CPU have 8 registers: 4 general-purpose registers (r0-r3), 2 segment registers (cs, ds equiv. to r4, r5), 1 flags register (fl), and of course IP register (Instruction Pointer)
  • We know how instruction encoding looks like.
  • We know how to manipulate flags register.
  • 8 instructions are defined
  • Memory dump is available which must be used to execute our emulator

To solve this level we must know that segment-offset architecture has always shifted memory etc. If we look closer we can find that emulated %%ds register has value 0x10 (because its 16 bits architecture).

I won’t post here my implementation of this VM, but after all again we must dump the memory how it is changed. Here it is:

1^D3<AA>@^B<80>^CR^@r^As^A<B2>P0^T<C0>^A<80>^@^P^P^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@2^@u^L32@^B<80>^CR^@r^As^C<B2>^@ð^@0C0>^A<FF>^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@u^P^A^@^@^@^@^@^@^@^@^@^@^@<CC>^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@}^_^U`MMR}^N'm^PmZ^FVG^TB^N<B6><B2><B2><E6>봃<8E><D7><E5><D4><D9><C3><F0><80><95><U+4209A><BD><95><A4><8D><9A>+0iJieU^\{i^\n^Dt5!&/`^CN7^^3T9溴<A2><AD><A4>ŕ<C8><C1><E4><8A><EC>璋<E8><81><U+2D624><D0><C0><8D><AC>"Re~'+Z^Ra
^Azk^]gGET /da75370fe15c4148bd4ceec861fbdaa5.exe HTTP/1.0^@^@^@^@^@^@^@^@^@^@^@^@^@^@7z^G^Q^_^]h%2w^^b#[GUS0^QB<F6><F1><B1><E6><C3><CC><F8><C5><E4><CC><C0>Ӆ<FD><9A><E3>恵<BB><D7>͇<A3><D3>k6oofU0^VE^   t\?)+f=^M
^M^B0(5^U        ^U<DD><EC><B8><E2><FB><D8><CB><D8>ыՂٚ<F1><92><AB><E8><A6><D6>Ќ<AA>Ҕ<CF>EFg }D^TkEmT^C^W`bUZJfa^QWhu^Eb6}^B^P"B2<BA><E2><B9><E2>ֹ<FF><C3>銏<C1><8F>Ḥ<96><U+4F071><8D><89><CC><D4>xvar>7#Vsqyc^Q iz^Th^E!^^2'Y<B7>ϫ<DD><D5>̗<93><F2><E7><C0><EB><FF>飿<A1><AB><8B><BB><9E><9E><8C><A0><C1><9B>Z//NN^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@

As we can see there is hidden link for the next level 😉 Now we must probably play with binary 😉

Btw. to be sure that our emulator is correct, last instruction must be ‘hlt’ – halt.

 

Third level (last)

First of all, this is Windows binary so I will use VM to play with it (I use VirtualBox with Win XP SP3 + IDA of course ;p). We need to install cygwin to run the binary because of the libcrypt which was used to compile the binary. After all when we run the binary we will be asked to give an argument

usage: keygen.exe hostname

Let’s look inside of the binary…

        mov    [esp+78h+var_74], offset aR ; "r"
        mov    [esp+78h+var_78], offset aLicense_txt ;    "license.txt"
        call    _fopen64
        mov    [ebp+var_4C], eax
        cmp    [ebp+var_4C], 0
        jnz    short loc_401120
        mov    [esp+78h+var_78], offset aErrorLicense_t ; "error: license.txt not found\n"
        call    printf
        mov    [ebp+var_50], 0FFFFFFFFh
        jmp    loc_401204

So program is looking for the license.txt file

        lea    eax, [ebp+var_38]
        mov    [esp+78h+var_70], eax
        mov    [esp+78h+var_74], offset aS ; "%s"
        mov    eax, [ebp+var_4C]
        mov    [esp+78h+var_78], eax
        call    fscanf

Get the string from this file (here is by the way stack overflow bug :D). Next:

        cmp    [ebp+var_38], 71686367h
        jnz    short loc_4011CF

program checks the first 4 bytes if they are equal to 0x71686367 which in fact is equal to the string “gchq” – so the pattern of The Government Communications Headquarters 🙂 Next:

        mov    eax, dword_402000
        mov    [esp+78h+var_74], eax
        lea    eax, [ebp+var_38]
        add    eax, 4
        mov    [esp+78h+var_78], eax
        call    crypt
        mov    edx, eax
        mov    eax, dword_402000
        mov    [esp+78h+var_74], eax
        mov    [esp+78h+var_78], edx
        call    strcmp
        test    eax, eax
        jnz    short loc_4011A5

Here is small trick 🙂 For me very clever, so we can see call for the crypt() function and next compare and jump. If we analyze it closer we can see that argument for the crypt() is the same like one argument for the strcmp() function. So crypt() must return exactly the same value like one which was passed as argument. Pseudocode can look like:

char *tmp="hqDTK7b8K2rvw";
if (strcmp(crypt(bufor+4, tmp), tmp)) {
   ...

This is very interesting situation because we must find string which after pass it to the crypt() function with the salt “hqDTK7b8K2rvw” will be encrypted to exactly the same value as salt (“hqDTK7b8K2rvw”) 🙂

Two ways can be used to find this string – bruteforce and rainbow tables. I’ve done both 😀 In fact bruteforcing won’t take as much time. Anyway the answer is….

cyberwin

🙂 Yes, we have it 🙂 Now we know that license key should looks like:

gchqcyberwin

Whatever we add after this string it won’t change the value of encrypted string (because of the salt). So let’s try to run the code with exactly this license.txt file:

_> da75370fe15c4148bd4ceec861fbdaa5.exe www.canyoucrackit.co.uk

keygen.exe

loading stage1 license key(s)...
loading stage2 license key(s)...

request:

GET /hqDTK7b8K2rvw/0/0/0/key.txt HTTP/1.0

response:

HTTP/1.1 404 Not Found
Content-Type: text/html; charset=us-ascii
Server: Microsoft-HTTPAPI/2.0
Date: Thu, 01 Dec 2011 23:55:05 GMT
Connection: close
Content-Length: 315

<HTML body>

So what we know now… File tries to connect to the server given as the argument in command line and tries to GET a key.txt file from the URL:

/hqDTK7b8K2rvw/0/0/0/key.txt

So again we have some question. Why there is three values of 0 (zero) number? Is the server given in command line is correct? When the program is running it prints:

loading stage1 license key(s)...
loading stage2 license key(s)...

Why he do that? If we look again to the java script code from the previous level we can read:

// stage 2 of 3

Is it somehow connected to this string? OK so… if we add anything after the magic string in license.txt file program tries to dump it as hex value and put in the URL, so:

_> da75370fe15c4148bd4ceec861fbdaa5.exe www.canyoucrackit.co.uk

keygen.exe

loading stage1 license key(s)...
loading stage2 license key(s)...

request:

GET /hqDTK7b8K2rvw/41414141/42424242/43434343/key.txt HTTP/1.0

response:

HTTP/1.1 404 Not Found
Content-Type: text/html; charset=us-ascii
Server: Microsoft-HTTPAPI/2.0
Date: Thu, 01 Dec 2011 23:58:05 GMT
Connection: close
Content-Length: 315

<HTML body>

Again we cannot put too long string because we can make stack overflow bug 😉 Of course we do not need to do that 😉 What is interesting we must find correct three 4bytes values and this is the main goal – find a correct path for the key.txt file – as I said in first level there is unused 4 bytes – use it + in 2 stage there was also 2 unused bytes but written directly in the .js file 😉

UPDATE: Some ppl didn’t believe I know what bytes should be used. But as I said in previous sentence – all bytes are in previous stages. Here is the correct link:

http://www.canyoucrackit.co.uk/hqDTK7b8K2rvw/a3bfc2af/d2ab1f05/da13f110/key.txt

So first value as I pointed in all previous sentence (when I described first level also I point it specially) is from the unused code from the shellcode. Two next bytes are the firmware from the second stage.

Ending

When we solve all problems and enter correct string in the main page which is:

Pr0t3ct!on#cyber_security@12*12.2011+

you will be redirect to this page:

http://www.canyoucrackit.co.uk/soyoudidit.asp

Here is screenshot:

After you click the button you are going to be redirected here:

http://www.gchq-careers.co.uk/cyber-jobs/

Here is screenshot:

and final redirection is here:

https://apply.gchq-careers.co.uk/fe/tpl_gchq01ssl.asp?newms=jj&id=35874

Here is some screenshot with salary:

Conclusion

This is very nice challenge and requires in fact huge skills. Very nice training for our brains 😉 Anyway few points which I must write now:

  • To apply for this job you must have English citizen – I don’t have so automagically I can’t apply even when I finished this challenge 😉
  • This challenge requires a lot of skills and the salary(!) is very low for the ppl with this skills so I understand why they cannot find good ppl 😉
  • I’m living and working now in London and I can say its quite nice place 😉

 

UPDATE 2: More information about second level + my implementation of VM can be found here:

http://blog.pi3.com.pl/?p=268

 

Best regards,

Adam Zabrocki

21

Jul

by pi3

Finally! 19 of July 2011 I had defence of my Master of Degree. I pass exam from whole study at mark 5.5 (the highest mark) and defence my thesis with mark 5.5 (the highest mark) and on the diploma I’m going to have final mark 5.0 (almost the highest mark ;)). My thesis was interesting not only for me but also for my University and they want to send it to the contest 😉 My topic was: “Elaboration of an automatic system of fuzz testing technique to use in the CERN grid applications”. To be honest now I have very powerful fuzzer ;>

At the beginning of March I second time moved to Switzerland (because of my work at CERN). Before that I was working in Wroclaw Center for Networking and Supercomputing in security team. In the middle of one pentesting work me and my friends (Bartek Balcerek and Maciej Kotowicz) discovered very nice vulnerability in the TORQUE server.

TORQUE (Terascale Open-Source Resource and Queue Manager) is very common in any GRID projects – including GRID in European Organization for Nuclear Research aka CERN 🙂 By using this bug attacker are able to create dirty job and put it to the queue and server responsible for executing this job will be hacked. This is very dangerous situation from the infrastructure – in the easiest way noone else will be able to use GRID resources. In worst situation we are able to overtake control on the edge machine which can manipulate any other machine – of course server is running with the root privileges 😉

Here is advisory in full-disclosure list.

Here is backup on my server.

 

Best regards,

Adam Zabrocki

 

17

Nov

by pi3

Finally! After few months of waiting we’ve got Phrack number 67! For me this is special release. Why? My article was accepted by Phrack staff and published at this release 🙂 I’m proud of that 😉 For me Phrack magazine is a legend. I grown on this magazine, so my connection with this magazine is even stronger 😉

At first I would like to thanks blackb1rd. He helps me very much with this article. If not blackb1rd, this article will never exists at this form like it is now. You’ve got beer from me, whenever we meet 😉

OK. So what about this release? Personally I think this is very good release. There is many interesting articles (for me). I love trick with exploiting user space vulnerabilities in the years of Non Executable Memory, ASLR, SSP, … and other shits 😉 There is Heap, there is Format Strings, and of course stack (my article), there is …. ah 😉

From the news the chapter ‘loopback’ comes back 😉 The full table of content is:

 0x01  Introduction ....................................... Phrack Staff

 0x02  Phrack Prophile on punk ............................ Phrack Staff

 0x03  Phrack World News .................................. EL ZILCHO

 0x04  Loopback (is back) ................................. Phrack Staff

 0x05  How to make it in Prison ........................... TAp

 0x06  Kernel instrumentation using kprobes ............... ElfMaster

 0x07  ProFTPD with mod_sql pre-authentication ............ FelineMenace

 0x08  The House Of Lore: Reloaded ........................ blackngel

 0x09  A Eulogy for Format Strings ........................ Captain Planet

 0x0a  Dynamic Program Analysis and Software Exploitation . BSDaemon

 0x0b  Exploiting memory corruptions in Fortran programs .. Magma
       under UNIX/VMS

 0x0c  PHRACKERZ: Two Tales ............................... Antipeace
                                                                &
                                                            The Analog Kid

 0x0d  Scraps of notes on remote stack overflow ........... pi3
       exploitation

 0x0e  Notes Concerning the Security, Design and .......... The Philosopher
       Administration of Siemens DCO-CS Digital
       Switching Systems                                                

 0x0f  Hacking the mind for fun and profit ................ lvxferis

 0x10  International Scenes ............................... various

Pure content looks very interesting, so what is inside? 🙂 Go and read! Go Go GO!!! 🙂

Btw. As you see, my article is:

Scraps of notes on remote stack overflow

available here and backup on my site here.

Btw2. I’m waiting for feedback 😉

.

Best regards,

Adam Zabrocki

Yesterday (30 of April) I gave a lecture in WA (White Area) at CERN. I was talking about my new project (in fact Master of Degree thesis topic). This is automated testing tool which uses fuzzing technique. It can be used for generate CLI, API, Unit, Functionally, Regression, … , tests – in fact we can use it for all types of tests. Generated programs are independent from language. It can generate output program in JAVA, C, C++, Assembler, Python, Perl, C#, … languages – we can simply add new modules for add new languages.  To be more flexible, framework used Aspect-Oriented Programming  (AOP). First beta version of framework is published on CERN svn servers. It is integrated with DPM CLI tests and works pretty well 😉

In the future maybe I will publish some more details.

Btw. This project can be simply adapted for search vulnerabilities in software 😉

Best regards,

Adam Zabrocki

18

Mar

by pi3

One day I was reviewing all bugs in bugtraq IDs (popular bids). I want to know which kind of bugs is it now popular and what is the trend of modern bugs. I came to two main conclusions:

1) The most popular are SQL/XSS bugs but in 60% this is found in software which nobody knows/uses (stupid kiddie)

2) We’ve got 2010 year and there is still possible to find stack overflow bugs! The most funny thing for me there is more remote stack overflow bugs than local 🙂

Stack overflow bugs is one of the oldest class of software bugs which still exists – more-less 10% of all bugs ! Of course it isn’t 199x year that you can find it using regexpression for ‘grep’ program. So what is conclusion? Exploit stack overflow bugs is still interesting from attackers point of view. The question is “Is it still possible to exploit this class of bugs in modern UNIX systems in 2010 year?”. The answer for this question isn’t simple. Let’s do simple review of modern defence systems. We’ve got:

*) Non-exec memory (not only stack – almost every region where it is NOT necessary)
*) W^X – “Write XOR Exec” memory. It forbids memory with Write and Exec bits in the same time.
*) AAAS – ASCII Armored Address Space
*) ASLR – Address Space Layout Randomization
*) mmap() and mprotect() protections
*) Heap protections – like safe-unlink(), safe malloc() implementation (OpenBSD)
*) Random canary of death protections                       ——————————-|
*) frame pointer protection by canary of death                                                |
*) move all pointers to the beginning of the frame                                           |==>  pro-police
*) move all local byte arrays to the end of the frame                                       |        protection (SSP)
*) Vulnerable arguments copied to the local variables and then reordered—-|
.

We can bypass most of this protection but if it isn’t connected. Is there any possibilities to exploit in modern UNIX systems REMOTE stack overflow bugs with enabled ALL of this protections?! It sounds crazy… but STILL we CAN DO IT 🙂 I wrote simple server with remote stack overflow bug and EXPLOIT it. Proof Of Concept of course is private but I created a movie of exploiting. You can find it here:

http://site.pi3.com.pl/priv/bypass-all-protections.flv

We’ve got 2010 year and we can still exploit remote stack overflow bugs in modern UNIX systems 🙂 Amazing… but it could be that this techniques (yes it isn’t one technique which is used to exploit this bug) is the last opportunity to exploiting remote stack overflow bugs… OK so… have a nice watching 🙂

Best regards,

Adam Zabrocki

15

Dec

by pi3

More than year ago I was publish advisory in ‘mtr’ software. I think, personally, it is great bug because it can’t exist without unspecified situation in  libresolv library 🙂 The question is why have I written information about it on blog?

I forgot add this advisory in my site (sic!) 🙂 Now it’s ok and you can find this advisory here.

I attached to this advisory details and Proof Of Concept. If you haven’t read it yet i strongly recommend you to do it because it shows that sometimes if  we read source code we think bug doesn’t exists but sometimes other external stuff/bugs/unspecified situation help us to trigger and exploit unexisting bug 🙂

Here is link – once again:
http://site.pi3.com.pl/adv/advisory-libresolv-mtr.txt

Btw. In future I want to continue research about CPU bugs and probably it will cause news posts in this topic 🙂

Best regards,

Adam Zabrocki

Is it a dream? Impossible? Bugs in CPU? No… it’s reality! CPU is only a piece of hardware. Everything have bugs… CPU too. I will give here only a piece of information about bugs in INTEL products…

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