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https://github.com/carlospolop/hacktricks
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121 lines
6.7 KiB
Markdown
121 lines
6.7 KiB
Markdown
---
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description: 'https://pentesterlab.com/'
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---
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# Padding Oracle
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**Post from** [**https://pentesterlab.com/**](https://pentesterlab.com/)\*\*\*\*
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## Cipher Block Chaining
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CBC is an encryption mode in which the message is split into blocks of X bytes length and each block is XORed with the previous encrypted block. The result is then encrypted.
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The following schema \(source: [Wikipedia](http://en.wikipedia.org/wiki/Block_cipher_mode_of_operation)\) explains this method:
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During the decryption, the reverse operation is used. The encrypted data is split in block of X bytes. Then the block is decrypted and XORed with the previous encrypted block to get the cleartext. The following schema \(source: [Wikipedia](http://en.wikipedia.org/wiki/Block_cipher_mode_of_operation)\) highlights this behavior:
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Since the first block does not have a previous block, an initialization vector \(IV\) is used.
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## Padding
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As we saw, the encryption is done by blocks of fixed size. To ensure that the cleartext exactly fit in one or multiple blocks, padding is often used. Padding can be done in multiple ways. A common way is to use PKCS7. With PKCS7, the padding will be composed of the same number: the number of bytes missing. For example, if the cleartext is missing 2 bytes, the padding will be `\x02\x02`.
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Let's look at more examples with a 2 blocks:
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| Block \#0 | Block \#1 | | | | | | | | | | | | | | |
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| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
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| byte \#0 | byte \#1 | byte \#2 | byte \#3 | byte \#4 | byte \#5 | byte \#6 | byte \#7 | byte \#0 | byte \#1 | byte \#2 | byte \#3 | byte \#4 | byte \#5 | byte \#6 | byte \#7 |
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| 'S' | 'U' | 'P' | 'E' | 'R' | 'S' | 'E' | 'C' | 'R' | 'E' | 'T' | '1' | '2' | '3' | **0x02** | **0x02** |
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| 'S' | 'U' | 'P' | 'E' | 'R' | 'S' | 'E' | 'C' | 'R' | 'E' | 'T' | '1' | '2' | **0x03** | **0x03** | **0x03** |
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| 'S' | 'U' | 'P' | 'E' | 'R' | 'S' | 'E' | 'C' | 'R' | 'E' | 'T' | **0x05** | **0x05** | **0x05** | **0x05** | **0x05** |
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| 'S' | 'U' | 'P' | 'E' | 'R' | 'S' | 'E' | 'C' | **0x08** | **0x08** | **0x08** | **0x08** | **0x08** | **0x08** | **0x08** | **0x08** |
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## Padding Oracle
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When an application decrypts encrypted data, it will first decrypt the data; then it will remove the padding. During the cleanup of the padding, **if** an **invalid** **padding** triggers a detectable **behaviour**, you have a **padding oracle vulnerability**. The detectable behaviour can be an **error**, a **lack** of **results**, or a **slower response**.
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If you detect this behaviour, you can **decrypt the encrypted data** and even **encrypt any cleartext**.
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### How to exploit
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You could use [https://github.com/AonCyberLabs/PadBuster](https://github.com/AonCyberLabs/PadBuster) to exploit this kind of vulnerability.
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In order to test if the cookie of a site is vulnerable you could try:
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```bash
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perl ./padBuster.pl http://10.10.181.45/index.php "Nl0OpaQYeGPMJeWSih2iiQ==" 8 -encoding 0 -cookies "auth=Nl0OpaQYeGPMJeWSih2iiQ=="
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```
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**Encoding 0** means that **base64** is used \(but others are available, check the help menu\).
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You could also **abuse** this **vulnerability** to **encrypt new data**. For example, imagine that the content of the cookie is "_user=MyUsername_", then you may change it to "_**user=administrator**_" and escalate privileges inside the application. You could also do it using `paduster`specifying the **-plaintext** parameter:
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```bash
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perl ./padBuster.pl http://10.10.181.45/index.php "Nl0OpaQYeGPMJeWSih2iiQ==" 8 -encoding 0 -cookies "auth=Nl0OpaQYeGPMJeWSih2iiQ==" -plaintext "user=administrator"
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```
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If the site is vulnerable `padbuster`will automatically try to find when the padding error occurs, but you can also indicating the error message it using the **-error** parameter.
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```bash
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perl ./padBuster.pl http://10.10.181.45/index.php "Nl0OpaQYeGPMJeWSih2iiQ==" 8 -encoding 0 -cookies "hcon=Nl0OpaQYeGPMJeWSih2iiQ==" -error "Invalid padding"
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```
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### The theory
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In **resume**, you can start decrypting the encrypted data by **guessing** the correct **values** that can be used to **create** all the **different paddings**. Then, the padding oracle attack will start **decrypting** bytes **from** the **end** to the start by **guessing** which will be the correct **value** that **creates a padding of 1, 2, 3, etc**.
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If we zoom in, we can see that the cleartext byte `C15` is just a XOR between the encrypted byte `E7` from the previous block, and byte `I15` which came out of the block decryption step:
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This is also valid for all other bytes:
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* `C14 = I14 ^ E6`
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* `C13 = I13 ^ E5`
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* `C12 = I12 ^ E4`
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* ...
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Now if we modify `E7` and keep changing its value, we will keep getting an invalid padding. Since we need `C15` to be `\x01`. However, there is one value of `E7` that will give us a valid padding. Let's call it `E'7`. With `E'7`, we get a valid padding. And since we know we get a valid padding we know that `C'15` \(as in `C15` for `E'7`\) is `\x01`.
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`\x01 = I15 ^ E'7`
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The gives us:
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`I15 = \x01 ^ E'7`
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So we are able to compute `I15`.
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Since we know `I15`, we can now compute `C15`
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`C15 = E7 ^ I15 = E7 ^ \x01 ^ E'7`
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Now that we have `C15`, we can move to brute-forcing `C14`. First we need to compute another `E7` \(let's call it `E''7`\) that gives us `C15 = \x02`. We need to do that since we want the padding to be `\x02\x02` now. It's really simple to compute using the property above and by replacing the value of `C15` we want \(`\x02`\) and `I15` we now know:
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`E''7 = \x02 ^ I15`
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After brute force `E6`, to find the value that gives us a valid padding `E''6`, we can re-use the formula:
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`C14 = I14 ^ E6`
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to get
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`I14 = \x02 ^ E''6`
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Once we get `I14`, we can compute `C14`:
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`C14 = E6 ^ I14 = E6 ^ \x02 ^ E''6`
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Using this method, we can keep going until we get all the ciphertext decrypted.
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### Detection of the vulnerability
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To get started, you can register an account and log in with this account \(to make things easier, you get automatically logged in when you register\).
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If you create an account and log in two times with this account, you can see that the cookie sent by the application didn't change.If you log in many times and always get the same cookie, there is probably something wrong in the application. The cookie sent back should be unique each time you log in. If the cookie is always the same, it will probably always be valid and there won't be anyway to invalidate it.
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Now, if you try to modify the cookie, you can see that you get an error from the application.
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