This is the second IOLI crackme.
$ ./crackme0x01
IOLI Crackme Level 0x01
Password: test
Invalid Password!
Let's check for strings with rabin2.
$ rabin2 -z ./crackme0x01
[Strings]
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 r2.
$
r2 ./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 r2 to analyze the whole binary, which gets you symbol names, among things.
"pdf" stands for
• Disassemble
• Function
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.
• "imp." stands for imports. Those are imported symbols, like printf()
• "str." stands for strings. Those are strings (obviously).
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 radare2'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: 0.000000f
double: 0.000000
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.