Baby-Steps (50 points) [Security Mooc CTF]

2 minute read

This was part of a CTF conducted as an exercise during the SecurityMooc course organised by IIT Kanpur.

Here’s the file link.

This is a very simple gdb exercise given to us.

I have lost my password, entering which to the program I will be able to enter a bank
account which has billion dollars. I assure you million dollars if you get the key for
me :) 

Let’s run the file using ltrace:

$ ltrace ./baby_steps 
__libc_start_main(0x80485cb, 1, 0xffbc6ca4, 0x80486e0 <unfinished ...>
fopen("/dev/urandom", "r")                       = 0x9a2c008
__isoc99_fscanf(0x9a2c008, 0x804876f, 0xffbc6b88, 0xf7739c08) = 1
fclose(0x9a2c008)                                = 0
printf("Enter the password: ")                   = 20
__isoc99_scanf(0x8048789, 0xffbc6b24, 0xffbc6b88, 0xf7739c08Enter the password: ABCD
) = 1
strcmp("ABCD", "\352\272y\231$\320\277O\3262\247,\331/y\365\254j\274\255") = -1
printf("Wrong password!!")                       = 16
Wrong password!!+++ exited (status 0) +++

The binary asks for a password and after our input, prints that it’s wrong. Let’s disassemble main() and see what we can infer (key is correct password, input is our input):

gdb-peda $ pdisas main
Dump of assembler code for function main:
   0x080485cb <+0>:	lea    ecx,[esp+0x4]
   0x080485ed <+34>:	push   0x8048760
   0x080485f2 <+39>:	push   0x8048762
   0x080485f7 <+44>:	call   0x80484a0 <[email protected]>         "fopen("/dev/urandom", "r")  "
   0x080485fc <+49>:	add    esp,0x10
   0x080485ff <+52>:	mov    DWORD PTR [ebp-0xd8],eax      "bytes we get from urandom"
   0x08048605 <+58>:	sub    esp,0x4
   0x08048608 <+61>:	lea    eax,[ebp-0x70]
   0x0804860b <+64>:	push   eax
   0x0804860c <+65>:	push   0x804876f
   0x08048611 <+70>:	push   DWORD PTR [ebp-0xd8]
   0x08048617 <+76>:	call   0x8048450 <[email protected]>   "fscanf(bytes,"%s",&key)"
   0x08048643 <+120>:	lea    eax,[ebp-0xd4]                    
   0x08048649 <+126>:	push   eax
   0x0804864a <+127>:	push   0x8048789
   0x0804864f <+132>:	call   0x80484b0 <[email protected]>    "Like scanf("%s",&input)"
   0x08048654 <+137>:	add    esp,0x10
   0x08048657 <+140>:	sub    esp,0x8
   0x0804865a <+143>:	lea    eax,[ebp-0x70]
   0x0804865d <+146>:	push   eax
   0x0804865e <+147>:	lea    eax,[ebp-0xd4]
   0x08048664 <+153>:	push   eax
   0x08048665 <+154>:	call   0x8048430 <[email protected]>   "Comparing our input with the key"
   0x0804866a <+159>:	add    esp,0x10
   0x0804866d <+162>:	test   eax,eax                  "Checking for equality"
   0x0804866f <+164>:	jne    0x80486a3 <main+216>     
   0x08048671 <+166>:	sub    esp,0xc                  "If pass and key are equal"
   0x08048674 <+169>:	push   0x8048790
   0x08048679 <+174>:	call   0x8048440 <[email protected]>
   0x0804867e <+179>:	add    esp,0x10
   0x08048681 <+182>:	sub    esp,0xc
   0x08048684 <+185>:	push   0x804ad60
   0x08048689 <+190>:	call   0x8048480 <[email protected]>
   0x0804868e <+195>:	add    esp,0x10
   0x08048691 <+198>:	sub    esp,0xc
   0x08048694 <+201>:	push   0x80487b5
   0x08048699 <+206>:	call   0x8048480 <[email protected]>
   0x0804869e <+211>:	add    esp,0x10
   0x080486a1 <+214>:	jmp    0x80486b3 <main+232>
   0x080486a3 <+216>:	sub    esp,0xc                   "If pass and key unequal"
   0x080486a6 <+219>:	push   0x80487b8
   0x080486d0 <+261>:	ret    
End of assembler dump.

The program opens /dev/urandom and writes random bytes into they key.

If our input and the random bytes are equal, we fall through 0x0804866f <+164>: jne 0x80486a3.

Now this has 3 output statements:


We can see these values by simply issuing a x/s 0x8048440 and two others x/s’s for each address using gdb.

That subsequently leads to printing our flag:

Correct password!! The flag is flag{kH9h9skyxrgSZeN3oaqGnNo4amNJvhOa}