110 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
[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 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
Print
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 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.