Damn, Daniel, back it again with more CTFing. Yes. CTFs never end.
This level is Hanoi, and our message this time says some things about hardware:
Further down-screen, the message reads:
There is no default password on the LockIT Pro HSM-1. Upon receiving the LockIT Pro, a new password must be set by first connecting the LockitPRO HSM to output port two, connecting it to the LockIT Pro App, and entering a new password when prompted, and then restarting the LockIT Pro using the red button on the back. LockIT Pro Hardware Security Module 1 stores the login password, ensuring users can not access the password through other means. The LockIT Pro can send the LockIT Pro HSM-1 a password, and the HSM will return if the password is correct by setting a flag in memory. This is Hardware Version B. It contains the Bluetooth connector built in, and two available ports: the LockIT Pro Deadbolt should be connected to port 1, and the LockIT Pro HSM-1 should be connected to port 2.
As mentioned on the New Orleans level, I’ve already solved a bunch of these and am posting a cleaned-up and more sane version of the notes I took during the levels. After a certain point, multiple cities will open up after you solve a previous city. So, you might do the levels in a slightly different order than me.
Testing things out
If we start the level, and type c to continue program execution, we’re prompted for a password. Interesting that it limits the password length to 8-16 characters.
I type 16 “A"s and hit enter. If I scroll through the memory management window, I see that my password ended up in memory at 0x2400.
“AAAAAAAAAAAAAAAA” was not the correct password, of course.
If we look at main, we see:
4438 <main> 4438: b012 2045 call #0x4520 <login> 443c: 0f43 clr r15
In other words, virtually all the logic is in the
login function. Let’s see what’s going on there.
4520 <login> 4520: c243 1024 mov.b #0x0, &0x2410 4524: 3f40 7e44 mov #0x447e "Enter the password to continue.", r15 4528: b012 de45 call #0x45de <puts> 452c: 3f40 9e44 mov #0x449e "Remember: passwords are between 8 and 16 characters.", r15 4530: b012 de45 call #0x45de <puts> 4534: 3e40 1c00 mov #0x1c, r14 4538: 3f40 0024 mov #0x2400, r15 453c: b012 ce45 call #0x45ce <getsn> 4540: 3f40 0024 mov #0x2400, r15 4544: b012 5444 call #0x4454 <test_password_valid> 4548: 0f93 tst r15 454a: 0324 jz $+0x8 454c: f240 5800 1024 mov.b #0x58, &0x2410 4552: 3f40 d344 mov #0x44d3 "Testing if password is valid.", r15 4556: b012 de45 call #0x45de <puts> 455a: f290 0400 1024 cmp.b #0x4, &0x2410 4560: 0720 jne #0x4570 <login+0x50> 4562: 3f40 f144 mov #0x44f1 "Access granted.", r15 4566: b012 de45 call #0x45de <puts> 456a: b012 4844 call #0x4448 <unlock_door> 456e: 3041 ret 4570: 3f40 0145 mov #0x4501 "That password is not correct.", r15 4574: b012 de45 call #0x45de <puts> 4578: 3041 ret
In broad strokes, let’s describe what’s going on here.
In 4520 to 4540, we’re prompting the user for a password, and reminding them that there’s a 8-16 char limit. Then we’re reading in their input, and storing it in memory at 0x2400.
Next, we’re calling
test_password_valid, presumably to test if the password is valid.
Then, we tell the user we’re testing their password. Depending on the results, we say it’s invalid, or that we’re granting access, and display that message. Lots of
puts calls all over the place.
That function name looks intriguing, right? Let’s check it out.
After poking around with this function a bit, I’m still not sure how it’s useful to us. But, if you step through it, you can see how the interrupt-to-unlock functionality works. That’s the 0x7d line onwards. You can read more about it in the Embedded CTF manual on page 9.
So if that doesn’t help us, what next? If you keep stepping, you’ll eventually return back to the
This next part is interesting (for real this time!)
After we’ve returned from our
test_password_valid goose chase, we skip over the
mov.b call, and print out “Testing if password is valid.”
Oh, so NOW we’re testing it? If we hadn’t skipped over that
mov.b line, we’d have 0x58 in location 0x2410. But… we don’t, and it’s not entirely clear how we’d get to that line anyway.
It doesn’t matter though, since the next line (at 455a) is comparing the contents at address 0x2410 with “0x4”. And 0x2410 is pretty close to the location of our password, right? Our password is at 0x2400
I know the program said we have to enter 8-16 characters, but what happens if we don’t follow the rules?
If we reset the program, and this time enter in 16 A’s, followed by one B (just to make it easier to see):
Now we’ve got a 42 (“B”) at location 0x2410. We entered in “too many” characters, broke the rules, and it let us!
Let’s replace that 42 with a 0x4 instead, so we can pass our cmp.b call.
But 0x4 isn’t a readable ASCII character. We’ll have to use hex-encoded input instead.
Hanoi Level Solution
So, let’s craft our response using the hex-encoded input option.
If we check the hex input box, we need to send “41” instead of “A”, and so on.
That means we want 16 “41"s in a row, followed by “04”.
Does it matter that the first 16 bytes are 41? No, it doesn’t. Put whatever you want.
Hit send, and
continue through any breakpoints. We just buffer overflowed our way to success. : )
If we give a hex-encoded input of “4141414141414141414141414141414104”, and overwrite the comparison value, we can get in!
Type solve, and then rerun the program to complete the Hanoi level.