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Hotel California

This is a shellcode challenge developed for the Defcon Quals 2019.

In a Nutshell

In a classic shellcode-like scenario, the service simply asks the user to provide a X64 shellcode, shorter than 1K. It then executes the code inside a TSX Hardware Lock Elision (HLE) transaction. If the transaction fails, the service prints an error message and asks the user to try again.

The catch is that there is really nothing useful you can do in a TSX transaction, as invoking syscalls would make the transaction fail and all read/write memory and register operations are cached by the CPU and reverted in case of failure.

Intended Exploit

The idea is to first close the transaction, and then execute a traditional shellcode to read and print the flag. However, a HLE transaction can only be closed by restoring the lock value to its original content - which unfortunately is not possible as the service acquires the lock by XORing two random 32bit numbers (properly deleted afterwards).

At a closer look, before jumping to the shellcode, one of the random value was momentarily spilled on the stack, just to be immediately overwritten by the frame of a function responsible to copy the received shellcode on the heap. Which brings us to the exploit procedure (a bit simplified here, but it gives the idea):

  1. Send a shellcode, padded to be at least 1K long. The shellcode is copied on the heap and executed.. but it fails as it has no way to close the transaction properly.
  2. The service prints the error message using printf, which internally also allocates a buffer on the heap.
  3. The buffer is de-allocated, causing libc to leak on the heap the address of a symbol on the libc (related to the allocator arena).
  4. When the service asks for a shellcode again, close the transmission side of the socket. This makes the read in the service fail. Since no bytes are received, the service does not invoke the function to copy the buffer to the heap. Therefore, the secret value is now available somewhere on the stack. Sadly, we have no reference to where the stack is.
  5. The service now tries to execute the buffer again. But since it did not copy anything new there, it ends up executing bytes that were put there by the first shellcode (modulo some that got overwritten and that therefore the shellcode needs to be careful to jump over).
  6. Use the shellcode to retrieve the pointer to the libc, from there move to the environ global variable which points to the stack. From the base of the environment, move down till you find the secret value, use it to close the transaction, and you are done.

(Instead of using the allocator to leak the libc position, it is also possible to use the cs register.)