CTF-practice-evening:2014-03-24

From Technologia Incognita
Revision as of 21:20, 24 March 2014 by MRieback (talk | contribs)
Jump to: navigation, search
CTF-practice-evening:2014-03-24
Date 2014/03/24
Time
Location ACTA
Type Workshop
Contact Melanie

Capture The Flag evening - Part 11

  • 24 March, 2014 - 7 PM
  • Please bring along a laptop with you!!!

General CTF Info

Walkthrough: Minibomb

  • Brainsmoke is explaining how he solved the challenge 'Minibomb' during the Codegate CTF
  • Minibomb is a small setuid binary
    • This is probably a handmade binary written in assembler, Linux ELF, 32 bit
    • You can see the ELF header if you use file or hexdump
      • For more information about the ELF header (including the binary entry point, memory pages being loaded, executable text, etc..), you can use readelf
    • Objdump allows us to disassemble the binary
    • It's a static binary - there's no dynamic loader
      • Dynamic binaries have an interpreter section, with more LD-* things that need to be resolved
      • The kernel needs to tell where the binary starts
    • You could also use IDA, but that's overkill for this binary
  • If you run it with strace, you see a list of signals and system calls
    • It starts, does an old_mmap call (you can get lots of information from the arguments, including the starting address), an unman (looks like a stack address - bfxx if usually on the stack in 32 bits)
    • It does a write and read
    • If you send lots of A's, you get a segfault - this gives away that you have a bug here
    • You can do this in gdb to get more information
      • You can see that a fault happens on the address 0x41414141 - our input!
      • It's easy to get arbitrary code execution here
  • Because it's a small file, we can take a look at the disassembled code
    • We can see the memory map
    • You can get system call information by typing 'syscall mmap' - we can see 0x5a, which is the syscall instruction in the disassembly!
    • We should read up to understand the meanings of: %eax, %ebx, etc…
      • We can give 6 arguments with a system call
    • Next command: %ebp is the frame pointer for the function call frames - this is also for the old_mmap system call
    • The one argument is an array of six arguments - that is a pointer to that argument
    • Next command: int 0x80 is the system call command on x86
  • readelf shows us that we have both data and text - the binary executable is loaded into 2 pages, starting from the start of the binary until 4096 bits later
    • The kernel loads it into the virtual address
    • You can also see another offset, used for page alignment with memory (in chunks of 4096 bytes)
    • The address we see in the disassembly is the same as the address in readelf
    • We can also visualise this by looking again in hex dump
      • We can see the protection bits (1-read, 2-write, 3-both)
      • This correlates to the arguments that we see in strafe
  • We can do the same with all of the other system calls
    • We can see the mmap, memunmap, write, read, etc…
  • This binary is so small that we can decode the whole thing
  • We can see a function call that allocated 16 bytes on the stack, by subtracting it from the stack pointer
    • It moves syscall 4 (write) to another address (look in hex-dump)
    • This writes passcode to the output
    • It prints 10 bytes to stdout - (0x1)
    • It does a write and a system call
    • It does a read and a system call
    • But then does something strange - it uses the stack pointer as the buffer it reads to
      • This gives a stack buffer overflow - you can write a page full of data to the stack - but there's only 16 bytes allocated to this purpose
  • If we run it again with strace, without having it crash, it also calls close
    • You can also see this in the disassembly
    • It closes stdin (this is a problem if you want to do shellcode, since you can't send data through it anymore)
    • It then does a write again, and then says BOOM!!, and then returns
    • (The binary doesn't really have any use)
  • The 4 bytes can be rewritten
    • We can test this by sending a bunch of A's again
    • i.e. echo -n 'AAAADDDDCCCCEEEEF' | strace ./minibomb
    • The read in strace now looks weird, because it's saying EFAULT (bad address)
    • We can look at this more carefully in gdb
      • We can print the stack pointer: x/40x $sp
      • (This behaves differently inside and outside of GDB since aslr is turned off in gdb)
      • We can write until the end of the page, and then it will give a fault
      • A complication: The address space is randomised, so we don't know exactly where the stack is
    • You have arbitrary code execution for free, but the problem is that you can't inject your shellcode directly and run it, since there's a special section that determines if the writeable address space is being protected as non-executable
      • This is tells the kernel if the stack should be non-executable - in this case, the stack is both readable and writeable
      • Since it's RW, (not-executable) you'll need to use Return Oriented Programming (ROP)
      • This explains why the binary is so tiny - in this case, the amount of addresses to return to is very small (actually impossible)
      • This makes it much more difficult
    • Run this in gdb with lots of A's, and look in the registers: info reg
      • eax is 0xc = the number of bytes written
      • ecx is the buffer
      • edx is the number of bytes that it wanted to write
      • ebx the first argument
      • esp is the stack pointer
    • Do 'info proc map' in gdb to see the memory mapping
      • You can see the text, stack segments and the Virtual Dynamic Shared Object [vdso]
    • The mapped address spaces don't look randomised in gdb, so by default gdb turns aslr off
      • The first two are fixed addresses, the second two are usually randomised
      • We can jump to the fixed addresses - we just looked at this code with objdump
      • Example: x/20i <address>
    • The vdso is executable
      • If we look at that, (x/20s), we can see the ELF header
      • The data doesn't look very interesting at first
      • At a certain point, we can see symbols that the kernel needs to put into the address space (sigreturn and vsyscall)
      • If you support sysenter, it will use that - but we need to remember that we had code that we needed to return to
      • This provides us with code to return to!
      • We can perform system calls, pop values from the stack, do a return, etc…
      • This is usually randomized
    • If we do: cat /proc/self, we can look at the current process
      • Example: cat /proc/self/maps looks at the address space of cat
      • We can see how the addresses are randomised between executions - this makes exploitation difficult
    • If you set the maximum stack size to unlimited, the kernel will leave a gap where the stack can grow
      • You can see the executable and the libraries loading
      • You can see what the kernel does with the vdso
      • This will all grow from high memory to low memory
      • If you set the stack to unlimited, it will do it the other way around: ulimit -s unlimited
      • (This is a nice trick to know for 32-bit binaries - although it's sometimes disabled in challenges)
    • If we start our minibomb with lots of A's again, we can look at: info proc map
      • Our vdso is now an address - we can inspect using (x/25i <address> - we have enlarged the code that we can jump to!
      • We can now use vsyscall, sigreturn, vsyscall, sysenter, etc..
      • Sysenter is interesting - it has a weird habit of losing the stack pointer - the kernel has the convention of putting the stack pointer in the base pointer first
      • You push the base pointer on the stack - the kernel then returns to the value of the base pointer
      • If we jump here, the kernel will switch the stack pointer and base pointer - this is something that we can exploit!
    • There's 2 ways to exploit this: 1 way they wanted you to exploit it, and a 2nd way that Brainsmoke exploited it
      • When sigreturn comes in, program execution is suspended, the stack is saved, the signal handler is called, and the kernel uses a technique similar to ROP to jump to something
      • However there's a problem - on this stack, there's not much space - you can see this by looking at x/40x $sp
      • You can only write 60 bytes past the end of the instruction pointer - and the stack frame is a bit bigger
      • We need to pop %ebp, if we jump to that address
      • You also need to pop %edx (the 3rd argument to the system call) and %ecx (the 2nd argument)
      • These 3 values are now under your control - plus %eax (the number of bytes written) and %ebx (stdout)
    • It has written 12 bytes - we can then jump to a system call gadget
      • syscall 12 is chdir - not useful. Syscall 11 is execve (but we can't control the arguments - this won't work), 10 is unlink (not useful), etc..
      • syscall3 is read- this is useful
      • We want to do: read(1, <mypointer>, <mylength>)
      • They wanted us to look at the old_syscall memory map - the pointer is an array into the memory
      • If we overwrite this, we control all of the arguments to the memory map
      • In this case, then we can jump to <address> that allows us to perform an arbitrary memory map, and then continue where it was before
      • But there's a trick - with memory, you can have file-based memory maps
      • The read can be file backed, at a location we specify, and can be executable
      • We can map this page over the executable page - when the call returns the code will be replaced with your own code (i.e. starting a shell with shellcode)
      • However, we still need to do the read() system call
      • We need to make sure that read() returns less than 12 bytes - but ulimit can help us here
      •  %ebx is 1, and %ecx can point to the buffer of memory map arguments
      • The address of the code is the first argument, RWX is the 2nd argument, it's file backed, and we provide an offset into the file (set to 0) - this will overwrite the stack
      • The return will do a system call, subtract 16 from the stack, return again, and then jump to where we just loaded data into %ebx
      • From this point, our code is there
  • Brainsmoke wrote a C program to put arbitrary binary data on the stack
    • You just need to tokenize the variables on the null byte
    • You have to align it on the word boundary
    • He put the return gadget in there (kindof like a NOP sled), so we don't have to jump to the exact address, will repeat until we reach something interesting
    • We then call sigreturn with the correct structure on the stack, to be loaded in the CPU registers
    • He does an execve system call with a filename as the pointer


Next CTF Competition

  • PlaidCTF: http://plaidctf.com and https://ctftime.org/event/119
    • April 11, 2014, 7 p.m. — April 13, 2014, 7 p.m. (Pittsburgh time)
    • We could meet on Saturday April 12 (all day) to play!
      • Alex is volunteering to host our CTF game at his apartment!
    • TODO Melanie: sign up Team Knuffelhackers
Retrieved from "https://wiki.techinc.nl/index.php?title=CTF-practice-evening:2014-03-24&oldid=11455"