Difference between revisions of "CTF-practice-evening:2014-03-24"
Line 97: | Line 97: | ||
* PlaidCTF: https://ctftime.org/event/119 | * PlaidCTF: 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! |
Revision as of 20:34, 24 March 2014
CTF-practice-evening:2014-03-24 | |
---|---|
Date | 2014/03/24 |
Time | |
Location | ACTA |
Type | Workshop |
Contact | Melanie |
Contents
Capture The Flag evening - Part 11
- 24 March, 2014 - 7 PM
- Please bring along a laptop with you!!!
General CTF Info
- See the page for the Ctf-evenings
- Link to the Tech Inc Challenge Website Scoreboard
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]
Next CTF Competition
- PlaidCTF: 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!