Used IP addresses
1
Name:
Anonymous
2009-03-02 18:05
Anyone got some C code that loops over all used IP addresses? Can you come up with any special optimizations? (I need to handle them in the "%u.%u.%u.%u"-form; computations are independant and can be run in parallel)
2
Name:
Anonymous
2009-03-02 18:08
for(i = 0; i < 256; i++)
for(h = 0; h < 256; h++)
for(j = 0; j < 256; j++)
for(k = 0; k < 256; k++)
printf("%d.%d.%d.%d",i,h,j,k);
3
Name:
Anonymous
2009-03-02 18:25
/*
EXPERT C CODING
*/
#include <stdio.h>
#include <stdint.h>
#include <arpa/inet.h>
#include <limits.h>
int main(int argc, char **argv) {
uint32_t ip = 0, ipn;
unsigned char *p = (unsigned char*)&ipn;
while(1) {
ipn = htonl(ip);
printf("%d.%d.%d.%d\n",p[0],p[1],p[2],p[3]);
if(ip++==UINT_MAX)
break;
}
return 0;
}
4
Name:
Anonymous
2009-03-02 18:37
>>2
yo dawg, we heard u like iteration, so we put a for loop in ur for loop so u can iterate while u iterate
5
Name:
Anonymous
2009-03-02 20:09
6
Name:
Anonymous
2009-03-02 20:22
[foldr1 ((++).(++".")) $ map show [a,b,c,d] | a <- [0..255], b <- [0..255], c <- [0..255], d <- [0..255]]
7
Name:
Anonymous
2009-03-02 20:23
for(i = 0; i < 256; i++)
for(h = 0; h < 256; h++)
for(j = 0; b < 256; j++)
for(k = 0; t < 256; k++)
printf("%d.%d.%d.%d",i,h,b,t);
8
Name:
Anonymous
2009-03-02 20:26
Have you read your poignant guide today?
256.times{|i|256.times{|j|256.times{|k|256.times{|l|puts "#{i}.#{j}.#{k}.#{l}"}}}}
THIS IS SO ELEGANT FUCK I NEED TO MASTURBATE.
I've run it for an hour, currently at 1.12.90.201
9
Name:
Anonymous
2009-03-02 20:52
[foldr1 ((++).(++".")) $ map show [a,b,c,d] | a <- [0..255], b <- [0..255], c <- [0..255], d <- [0..255]][/code
[code]map (intercalate "." . map show) . sequence . replicate 4 $ [0..255]
I came
10
Name:
Anonymous
2009-03-02 20:52
[/code]
11
Name:
Anonymous
2009-03-02 20:53
>>9
map (intercalate "." . map show) . sequence . replicate 4 $ [0..255]
12
Name:
Anonymous
2009-03-02 20:54
>>9,11
<interactive>:1:5: Not in scope: `intercalate'
13
Name:
Anonymous
2009-03-02 21:02
(2**32).times{|ip| printf("%d.%d.%d.%d\n",(ip>>24 )&0xff,(ip>>16 )&0xff,(ip>>8 )&0xff,ip&0xff)}
14
Name:
Anonymous
2009-03-02 21:14
(2**32).times{|ip|a=Array.new(4){|n|(ip>>(n*8))&0xff};puts a.reverse.join(".")}
15
Name:
Anonymous
2009-03-02 21:29
uint32_t n = 1;
while (n)
printf("%d\n", n++);
16
Name:
Anonymous
2009-03-03 5:15
print join '.',unpack 'C4',pack N,$_ foreach 0..4294967295
17
Name:
Anonymous
2009-03-03 5:34
Here's the code I already had, by the way:
for (a = 58; a < 223; ++a) {
if (a == 100) { a = 107; continue; }
if (a == 127) continue;
if (a == 175) { a = 177; continue; }
if (a == 179) { a = 183; continue; }
if (a == 185) continue;
for (b = 0; b < 256; ++b) {
for (c = 0; c < 256; ++c) {
for (d = 1; d < 255; ++d) {
SHA1_Init(&ctx);
sprintf(prefix1, "%u.%u.%u.%u", a, b, c, d);
SHA1_Update(&ctx, prefix1, strlen(prefix1));
SHA1_Update(&ctx, prefix2, 12);
SHA1_Update(&ctx, salt, 448);
SHA1_Final(md, &ctx);
if (memcmp(search, md, sizeof(search)) == 0)
goto done;
}
}
}
printf("%u/8 done.\n", a);
}
18
Name:
Anonymous
2009-03-03 6:27
>>17
Stop reading at
goto done;
19
Name:
Anonymous
2009-03-03 6:51
All right, I made some changes (most importantly I removed the overflow so it didn't loop forever on the inner loops, and I split the sprintf into four calls). But I'm not really sure about what IP addresses I should check. What are those LEGACY blocks?
[1] My research has shown that they might indeed be in use.
[1] http://www.iana.org/assignments/ipv4-address-space/
20
Name:
Anonymous
2009-03-03 6:58
>>19
Nevermind, it appears there is a section on that... I need my eyes checked I think.
Now then, I hope my algorithm is correct.
21
Name:
Anonymous
2009-03-03 7:06
>>18
}
}
}
printf("%u/8 done.\n", a);
}
22
Name:
Anonymous
2009-03-03 7:09
>>21
Thank you,
FrozenVoid .
23
Name:
Anonymous
2009-03-03 8:30
Turned out my algorithm was wrong after all, oh well, I changed it and got OpenMP working.
The input:
/* http://dis.4chan.org/read/sci/1172747610
* 1 Name: FrozenVoid : 2007-03-01 08:54 ID:fPuNCwa3 */
And the output:
80.230.22.238
Oh FrozenVoid, this explains everything.
24
Name:
Anonymous
2009-03-03 8:47
25
Name:
Anonymous
2009-03-03 8:55
Got 80.230.70.64 from the second day
[1] , I'd say this pretty much confirms it.
[1] http://dis.4chan.org/read/sci/1172747610/20
26
Name:
Anonymous
2009-03-03 12:27
27
Name:
Anonymous
2009-03-03 13:05
>>8
Interesting. I've been running it 25 minutes and its at 24.209.148.240
28
Name:
Anonymous
2009-03-03 14:12
>>23,25
What does it mean?
29
Name:
Anonymous
2009-03-03 14:37
>>28
80.230.22.238 = IGLD-80-230-22-238.inter.net.il.
80.230.70.64 = IGLD-80-230-70-64.inter.net.il.
il is the TLD of Israel.
30
Name:
Anonymous
2009-03-03 15:01
>>29
so FV is an arab?
JEWS don't internet.
31
Name:
Anonymous
2009-03-03 15:02
>>30
JEWS don't internet.
Yes they do, and they are often VERY annoying.
33
Name:
Anonymous
2009-03-03 15:51
>>29
I suddenly feel this burning desire to become a muslim and join Hamas.
34
Name:
Anonymous
2009-03-03 16:08
>>32
I suddenly feel this burning desire to purchase some Zion gold.
35
Name:
Anonymous
2009-03-03 22:22
>>33-34
I suddenly feel this burning sensation when I pee.
36
Name:
Anonymous
2009-03-03 23:35
>>35
I suddenly feel this burning vomit in small quantity in my mouth.
37
Name:
Anonymous
2009-03-04 0:39
failure
38
Name:
Anonymous
2009-03-04 1:03
};
39
Name:
Anonymous
2009-03-04 1:04
>>36
Go see you're local GP and get that checked out dude, that doesn't sound wright.
40
Name:
Anonymous
2009-03-04 2:40
Fuck used IP addresses, where can I get new ones?
41
Name:
Anonymous
2009-03-04 4:38
256.256.256.257
42
Name:
Anonymous
2009-03-04 4:47
>>41
FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU
43
Name:
Anonymous
2009-03-04 5:13
>>40
This problem was caused by the popularity of IBM's System/360 architecture, which had standardized on 8 bits per byte. This led to IP networking standards using these as a basic component in designing the structure of protocols.
Some computers at the time had 9 bits in a byte, and 36-bit words. Just think if our IP addresses were made up of four nonets instead - we'd have 68 billion addresses available and all this talk of IPv6 migration wouldn't be happening!
44
Name:
Anonymous
2009-03-04 6:08
>>43
But then it'd take 16 times longer to crack FrozenVoid's ID!
45
Name:
Anonymous
2009-03-04 6:37
>>44
How do you crack ID?
46
Name:
Anonymous
2009-03-04 6:39
>>45
With great difficulty.
47
Name:
Anonymous
2009-03-04 6:40
48
Name:
Anonymous
2009-03-04 6:43
>>47
You utilize great difficulty as a part of the cracking process.
49
Name:
Anonymous
2009-03-04 16:52
>>45-48
use great qw/difficulty/;
50
Name:
Anonymous
2009-03-04 16:57
>>23
JEWS DID FROZENVOID
Also, can you post the sauce?
51
Name:
Anonymous
2009-03-04 17:00
>>50
DISREGARD THAT, I SUCK COCKS AND CAN'T READ THREADS.
52
Name:
Anonymous
2009-03-04 17:02
>>51
but post the new algorithm
anyways
53
Name:
Anonymous
2009-03-04 17:03
>>51
Oh, ok. Consider it disregarded.
54
Name:
Anonymous
2009-03-04 17:03
>>52
Heh heh hehhh shut up beavis
55
Name:
Anonymous
2009-03-04 17:05
>>52
You'll probably have to read the Shiichan source to understand everything.
% cc -fopenmp -O3 -Wall -s -fomit-frame-pointer -march=native idcrack.c -lssl -lroken
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <openssl/sha.h>
#include <base64.h>
#include <omp.h>
/* http://dis.4chan.org/read/sci/1172747610/1
* 1 Name: FrozenVoid : 2007-03-01 08:54 ID:fPuNCwa3 */
/* ?fPu NCwa 3???
* = 5df3ee 342c1a dd75d7 */
/* static const unsigned char search[] = { 0xf3, 0xee, 0x34, 0x2c, 0x1a }; */
/* http://dis.4chan.org/read/sci/1172747610/20
* 20 Name: FrozenVoid : 2007-03-02 01:35 ID:+sP42Nst
* _+sP 42Ns t___
* 5feb0f e3636c b575d7 */
static const unsigned char search[] = { 0xeb, 0x0f, 0xe3, 0x63, 0x6c };
/* http://www.iana.org/assignments/ipv4-address-space/
* Using 2009-01-28 revision. */
static const unsigned char ablocks[] = {
80, 85, 213, 212, 62, 84, 195, 78, 217, 79, 94, 81, 82,
3, 4, 6, 7, 8, 9, 11, 12, 13, 15, 16, 17, 18, 19, 20,
21, 22, 24, 25, 26, 28, 29, 30, 32, 33, 34, 35, 38, 40,
41, 43, 44, 45, 47, 48, 51, 52, 53, 54, 55, 56, 57, 58,
59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72,
73, 74, 75, 76, 77, 86,
87, 88, 89, 90, 91, 92, 93, 95, 96, 97, 98, 99, 108,
109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119,
120, 121, 122, 123, 124, 125, 126, 128, 129, 130, 131,
132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142,
143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153,
154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164,
165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 178,
184, 186, 187, 188, 189, 190, 191, 192, 193, 194,
196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206,
207, 208, 209, 210, 211, 214, 215, 216,
218, 219, 220, 221, 222 };
int
main(int argc, char* argv[])
{
unsigned char salt[448];
unsigned char search[5];
FILE* fp;
int a;
fp = fopen("salt.cgi", "rb");
fread(salt, 448, 1, fp);
fclose(fp);
#pragma omp parallel
{
#pragma omp for schedule(static, 1)
for (a = 0; a < sizeof(ablocks); ++a) {
SHA_CTX ctx;
char prefix[16] = { '\0' };
unsigned char md[SHA_DIGEST_LENGTH] = { '\0' };
unsigned int b, c, d;
char* bpos;
char* cpos;
char* dpos;
bpos = prefix + sprintf(prefix, "%u.", ablocks[a]);
for (b = 0; b < 256; ++b) {
cpos = bpos + sprintf(bpos, "%u.", b);
for (c = 0; c < 256; ++c) {
dpos = cpos + sprintf(cpos, "%u.", c);
for (d = 0; d < 256; ++d) {
sprintf(dpos, "%u", d);
SHA1_Init(&ctx);
SHA1_Update(&ctx, prefix, strlen(prefix));
SHA1_Update(&ctx, "021172747610", 12);
SHA1_Update(&ctx, salt, 448);
SHA1_Final(md, &ctx);
if (memcmp(search, md + 1, 5) == 0) {
char* base64 = NULL;
base64_encode(md, SHA_DIGEST_LENGTH, &base64);
base64[9] = '\0';
printf("%u.%u.%u.%u: %s\n", ablocks[a], b, c, d, base64 + 1);
free(base64);
exit(0);
}
}
}
}
printf("%u/8 (%u) done. [%u]\n", ablocks[a], a, omp_get_thread_num());
}
}
printf("Couldn't find it...\n");
exit(1);
}
56
Name:
Anonymous
2009-03-04 17:07
>>55
unsigned char search[5];
I think I left this line from something else... remove it.
57
Name:
Anonymous
2009-03-04 19:56
>>55
Hmm, clever. And also rather worrying.
58
Name:
Anonymous
2009-03-05 8:55
FrozenVoid is a TERRORIST!
59
Name:
Anonymous
2009-03-05 10:47
JEWS