111  IOLI 0x01

This is the second IOLI crackme.

$ ./crackme0x01
IOLI Crackme Level 0x01
Password: test
Invalid Password!

Let’s check for strings with rz-bin.

$ rz-bin -z ./crackme0x01
nth paddr      vaddr      len size section type  string
0   0x00000528 0x08048528 24  25   .rodata ascii IOLI Crackme Level 0x01\n
1   0x00000541 0x08048541 10  11   .rodata ascii Password:
2   0x0000054f 0x0804854f 18  19   .rodata ascii Invalid Password!\n
3   0x00000562 0x08048562 15  16   .rodata ascii Password OK :)\n

This isn’t going to be as easy as 0x00. Let’s try disassembly with rizin.

$ rizin ./crackme0x01
-- Use `zoom.byte=printable` in zoom mode ('z' in Visual mode) to find strings
[0x08048330]> aa
[0x08048330]> pdf @ main
            ; DATA XREF from entry0 @ 0x8048347
/ 113: int main (int argc, char **argv, char **envp);
|           ; var int32_t var_4h @ ebp-0x4
|           ; var int32_t var_sp_4h @ esp+0x4
|           0x080483e4      55             push ebp
|           0x080483e5      89e5           mov ebp, esp
|           0x080483e7      83ec18         sub esp, 0x18
|           0x080483ea      83e4f0         and esp, 0xfffffff0
|           0x080483ed      b800000000     mov eax, 0
|           0x080483f2      83c00f         add eax, 0xf                ; 15
|           0x080483f5      83c00f         add eax, 0xf                ; 15
|           0x080483f8      c1e804         shr eax, 4
|           0x080483fb      c1e004         shl eax, 4
|           0x080483fe      29c4           sub esp, eax
|           0x08048400      c70424288504.  mov dword [esp], str.IOLI_Crackme_Level_0x01 ; [0x8048528:4]=0x494c4f49 ; "IOLI Crackme Level 0x01\n"
|           0x08048407      e810ffffff     call sym.imp.printf         ; int printf(const char *format)
|           0x0804840c      c70424418504.  mov dword [esp], str.Password: ; [0x8048541:4]=0x73736150 ; "Password: "
|           0x08048413      e804ffffff     call sym.imp.printf         ; int printf(const char *format)
|           0x08048418      8d45fc         lea eax, [var_4h]
|           0x0804841b      89442404       mov dword [var_sp_4h], eax
|           0x0804841f      c704244c8504.  mov dword [esp], 0x804854c  ; [0x804854c:4]=0x49006425
|           0x08048426      e8e1feffff     call sym.imp.scanf          ; int scanf(const char *format)
|           0x0804842b      817dfc9a1400.  cmp dword [var_4h], 0x149a
|       ,=< 0x08048432      740e           je 0x8048442
|       |   0x08048434      c704244f8504.  mov dword [esp], str.Invalid_Password ; [0x804854f:4]=0x61766e49 ; "Invalid Password!\n"
|       |   0x0804843b      e8dcfeffff     call sym.imp.printf         ; int printf(const char *format)
|      ,==< 0x08048440      eb0c           jmp 0x804844e
|      |`-> 0x08048442      c70424628504.  mov dword [esp], str.Password_OK_: ; [0x8048562:4]=0x73736150 ; "Password OK :)\n"
|      |    0x08048449      e8cefeffff     call sym.imp.printf         ; int printf(const char *format)
|      |    ; CODE XREF from main @ 0x8048440
|      `--> 0x0804844e      b800000000     mov eax, 0
|           0x08048453      c9             leave
\           0x08048454      c3             ret

“aa” tells rizin to analyze the whole binary, which gets you symbol names, among things.

“pdf” stands for

This will print the disassembly of the main function, or the main() that everyone knows. You can see several things as well: weird names, arrows, etc.

If you look carefully, you’ll see a cmp instruction, with a constant, 0x149a. cmp is an x86 compare instruction, and the 0x in front of it specifies it is in base 16, or hex (hexadecimal).

0x0804842b    817dfc9a140. cmp dword [ebp + 0xfffffffc], 0x149a

You can use rizin’s % command to display 0x149a in another numeric base.

[0x08048330]> % 0x149a
int32   5274
uint32  5274
hex     0x149a
octal   012232
unit    5.2K
segment 0000:049a
string  "\x9a\x14"
fvalue  5274.0
float   5274.000000f
double  5274.000000
binary  0b0001010010011010
trits   0t21020100

So now we know that 0x149a is 5274 in decimal. Let’s try this as a password.

$ ./crackme0x01
IOLI Crackme Level 0x01
Password: 5274
Password OK :)

Bingo, the password was 5274. In this case, the password function at 0x0804842b was comparing the input against the value, 0x149a in hex. Since user input is usually decimal, it was a safe bet that the input was intended to be in decimal, or 5274. Now, since we’re hackers, and curiosity drives us, let’s see what happens when we input in hex.

$ ./crackme0x01
IOLI Crackme Level 0x01
Password: 0x149a
Invalid Password!

It was worth a shot, but it doesn’t work. That’s because scanf() will take the 0 in 0x149a to be a zero, rather than accepting the input as actually being the hex value.

And this concludes IOLI 0x01.