Context

A couple of my internet fellas were working on a CTF that presented them a binary file, which had the flag inside they had to retrieve. I jumped on this without expecting much, but anyway.

Triage

I did a quick file bin to check what type of file it was:

The file was a non-stripped out linux binary file, which means debugging will be easier since we will be able to see original function names used in the binary.

Strings

I ran the file through strings strings bin to see if anything stood out:

We can notice some interesting things that we can make some assumptions about - notably the following strings:

• ACCESS GRANTED/ACCESS DENIED - possibly will need to enter a password somewhere in the binary and these messages will be printed to the user depending on if the provded password is correct/incorrect.

• some long strings - maybe something interesting encoded here or maybe those strings are used as part of the password decryption algorithm?

• a string %32s - maybe a C string output format (32 characters)?

Simply running the file prompted for a password and failed with an error message ACCESS DENIED:

Disassembly

Let's have a quick look at the disassembly of the file and look at its main function:

objdump -d bin | more

Note the following from the above screenshot:

• We can see that at offset b14 (cyan) there is a C function scanf called which reads from the standard input.

• instruction at b20 (orange) calls a check_pw routine - we can assume that the input captured from the instruction at b14 will be passed to check_pw function to decide if the string received from the standard input matches the password the binary is protected with or not

• instruction at b25 carries out a check against the eax register and based on if eax==0 or eax!=0, it will either take a jump to instructions at b27 (if eax==0) or continue executing instructions at b29 if eax!=0. Pressumably, the jumps are carried out based on if the provided password is correct or incorrect.

GDB

• Let's look at the file through GDB with Peda plugin

  • Let's set a break point on the main function

  • Do a quick disas of the main function to remind ourselves once again what the routine for password checking was

  • Let's set a breakpoint check_pw routine as well

gdb bin
b main
disas
b check_pw

Let's hit c to continue running the program until the scanf function is called and then provide it with some dummy password, say test:

Check_pw Routine: Round 1

Once the password is entered, the program breaks on check_pw:

If we skip through instructions one by one and keep observing how register values change over time and what instructions are executed, we will soon end up at check_pw+88:

Note this from the above screenshot:

• current instruction at check_pw+88: cmp dl, al - al and dl register values are being compared

• register rax and rdx values are b and t respectively (organge at the top). If you followed the register values whilst stepping through the code, you would notice that the value in the rdx is actually the first letter of our password test. Having said this, it looks like the binary is checking if the first character of the provided password is actually an ascii b

• If dl==al, the code should jump to check_pw+99 as seen at offset check_pw+90

However, stepping through the instructions further, we can see that the jump is NOT taken - the program continues executing instructions at offset check_pw+92 - suggesting the first character of the password does NOT start with a t:

Check_pw Routine: Round 2

What if we rerun the program and supply it with a password best this time (replacing the first t with b, since the binary seemed to be expecting to see in the dl register)?

Well, this time the cmp al,dl sets the zero flag to true and the jump at check_pw+90 is taken - suggesting that the first character of the password is indeed a b:

If we repeat this process 32 more times (remember the %32s string discussed previously?), we will eventually get the full password:

Going back to the long strings we saw earlier - they were indeed used in the password decryption routine, but going through the algorithm is out of scope for today:

Now, there is probably a better/automated way of solving this, so if you know a better way, I would like to hear about it!