CSAW 2019 beleaf
When we take a look at the binary, we see this:
$ file beleaf
beleaf: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 3.2.0, BuildID[sha1]=6d305eed7c9bebbaa60b67403a6c6f2b36de3ca4, stripped
$ pwn checksec beleaf
[*] '/Hackery/pod/modules/3-beginner_re/csaw19_beleaf/beleaf'
Arch: amd64-64-little
RELRO: Full RELRO
Stack: Canary found
NX: NX enabled
PIE: PIE enabled
$ ./beleaf
Enter the flag
>>> 15935728
Incorrect!
So we can see that we are dealing with a 64 bit binary. When we run the binary, it prompts us for input. This is probably a crackme challenge. This is a type of challenge that scans in input, and checks it. The goal is to pass it the correct input, to pass whatever check it does.
Reversing
Looking at the main function at 0x1008a1, we see this:
undefined8 main(void)
{
size_t inputLen;
long transformedInput;
long in_FS_OFFSET;
ulong i;
char input [136];
long stackCanary;
stackCanary = *(long *)(in_FS_OFFSET + 0x28);
printf("Enter the flag\n>>> ");
__isoc99_scanf(&DAT_00100a78,input);
inputLen = strlen(input);
if (inputLen < 0x21) {
puts("Incorrect!");
/* WARNING: Subroutine does not return */
exit(1);
}
i = 0;
while (i < inputLen) {
transformedInput = transformFunc(input[i]);
if (transformedInput != *(long *)(&desiredOutput + i * 8)) {
puts("Incorrect!");
/* WARNING: Subroutine does not return */
exit(1);
}
i = i + 1;
}
puts("Correct!");
if (stackCanary != *(long *)(in_FS_OFFSET + 0x28)) {
/* WARNING: Subroutine does not return */
__stack_chk_fail();
}
return 0;
}
So we can see, it starts off by scanning in input. If our input is less than 0x21 (33) bytes, the code exits (so our input probably has to be 33) bytes, Looking at it later, we see that it enters into a for loop. It will run each character through the transformFunc (or at least until the code calls exit). It will then compare the output of that functions (stored in transformedInput) against the corresponding character in the bss array desiredOutput (characters are stored at offsets of 8) bytes. If the two are not equivalent, exit is called and we fail the challenge. We can see the contents of desiredOutput by double clicking on it. When we look at desiredOutput, we see this:
desiredOutput XREF[2]: main:0010096b(*),
main:00100972(R)
003014e0 01 ?? 01h
003014e1 00 ?? 00h
003014e2 00 ?? 00h
003014e3 00 ?? 00h
003014e4 00 ?? 00h
003014e5 00 ?? 00h
003014e6 00 ?? 00h
003014e7 00 ?? 00h
003014e8 09 ?? 09h
003014e9 00 ?? 00h
003014ea 00 ?? 00h
003014eb 00 ?? 00h
003014ec 00 ?? 00h
003014ed 00 ?? 00h
003014ee 00 ?? 00h
003014ef 00 ?? 00h
003014f0 11 ?? 11h
003014f1 00 ?? 00h
003014f2 00 ?? 00h
003014f3 00 ?? 00h
003014f4 00 ?? 00h
003014f5 00 ?? 00h
003014f6 00 ?? 00h
003014f7 00 ?? 00h
003014f8 27 ?? 27h '
003014f9 00 ?? 00h
003014fa 00 ?? 00h
003014fb 00 ?? 00h
003014fc 00 ?? 00h
003014fd 00 ?? 00h
003014fe 00 ?? 00h
003014ff 00 ?? 00h
00301500 02 ?? 02h
So here we see that our first output has to be equal to 0x1, our second has to be 0x9, our third has to be 0x11, and so on and so forth. Looking at the transformFunc, we see this:
long transformFunc(char input)
{
long i;
i = 0;
while ((i != -1 && ((int)input != *(int *)(&lookup + i * 4)))) {
if ((int)input < *(int *)(&lookup + i * 4)) {
i = i * 2 + 1;
}
else {
if (*(int *)(&lookup + i * 4) < (int)input) {
i = (i + 1) * 2;
}
}
}
return i;
}
Here we can see that it essentially just takes a character, and looks at what it's index is in the lookup bss array. The characters are stored at offsets of 4 bytes. Let's take a look at the array:
lookup XREF[6]: transformFunc:00100820(*),
transformFunc:00100827(R),
transformFunc:00100844(*),
transformFunc:0010084b(R),
transformFunc:00100873(*),
transformFunc:0010087a(R)
00301020 77 ?? 77h w
00301021 00 ?? 00h
00301022 00 ?? 00h
00301023 00 ?? 00h
00301024 66 ?? 66h f
00301025 00 ?? 00h
00301026 00 ?? 00h
00301027 00 ?? 00h
00301028 7b ?? 7Bh {
00301029 00 ?? 00h
0030102a 00 ?? 00h
0030102b 00 ?? 00h
0030102c 5f ?? 5Fh _
0030102d 00 ?? 00h
0030102e 00 ?? 00h
0030102f 00 ?? 00h
00301030 6e ?? 6Eh n
00301031 00 ?? 00h
00301032 00 ?? 00h
00301033 00 ?? 00h
00301034 79 ?? 79h y
00301035 00 ?? 00h
00301036 00 ?? 00h
00301037 00 ?? 00h
00301038 7d ?? 7Dh }
00301039 00 ?? 00h
0030103a 00 ?? 00h
0030103b 00 ?? 00h
0030103c ff ?? FFh
0030103d ff ?? FFh
0030103e ff ?? FFh
0030103f ff ?? FFh
00301040 62 ?? 62h b
00301041 00 ?? 00h
00301042 00 ?? 00h
00301043 00 ?? 00h
00301044 6c ?? 6Ch l
00301045 00 ?? 00h
00301046 00 ?? 00h
00301047 00 ?? 00h
00301048 72 ?? 72h r
00301049 00 ?? 00h
0030104a 00 ?? 00h
0030104b 00 ?? 00h
0030104c ff ?? FFh
0030104d ff ?? FFh
0030104e ff ?? FFh
0030104f ff ?? FFh
00301050 ff ?? FFh
00301051 ff ?? FFh
00301052 ff ?? FFh
00301053 ff ?? FFh
00301054 ff ?? FFh
00301055 ff ?? FFh
00301056 ff ?? FFh
00301057 ff ?? FFh
00301058 ff ?? FFh
00301059 ff ?? FFh
0030105a ff ?? FFh
0030105b ff ?? FFh
0030105c ff ?? FFh
0030105d ff ?? FFh
0030105e ff ?? FFh
0030105f ff ?? FFh
00301060 ff ?? FFh
00301061 ff ?? FFh
00301062 ff ?? FFh
00301063 ff ?? FFh
00301064 61 ?? 61h a
00301065 00 ?? 00h
00301066 00 ?? 00h
00301067 00 ?? 00h
00301068 65 ?? 65h e
00301069 00 ?? 00h
0030106a 00 ?? 00h
0030106b 00 ?? 00h
0030106c 69 ?? 69h i
Here we can see that the character f is stored at 00301024. This will output 1 since ((0x00301024 - 0x00301020) / 4) = 1 (0x00301020 is the start of the array). This also corresponds to the first byte of the desiredOutput array, since it is 1. The second byte is 0x9, so the character that should correspond to it is (0x00301020 + (4*9)) = 0x301044, and we can see that the character there is l:
00301044 6c ?? 6Ch l
00301045 00 ?? 00h
00301046 00 ?? 00h
00301047 00 ?? 00h
00301048 72 ?? 72h r
So the second character is l. Moving on through the rest of the list, we can find the full string flag{we_beleaf_in_your_re_future}:
$ ./beleaf
Enter the flag
>>> flag{we_beleaf_in_your_re_future}
Correct!
Just like that, we solved the challenge!