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Jorg hi I post too much
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Joined: 24 Dec 2007 Posts: 2276 Location: Minnesota
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Posted: Mon Mar 16, 2009 12:55 pm Post subject: [VB.net] How do you make a new integer type > than int64 |
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I'm using VB.net and the Int64 maxvalue is too small, I need one that can handle a maximum of int64^10, because I am making a brute forcer that will attempt to brute force server hash codes. Don't ask me why I just want it like that.
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iNoobHacker Advanced Cheater
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Posted: Mon Mar 16, 2009 1:57 pm Post subject: |
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You do know that int64, assuming you are talking about signed integer, can hold a maximum of 2^63-1 (2^63 numbers, 0 included), which is 9223372036854775807, which means the 10th power of it is 4.4555084156466750133735972424201e+189, which means there are:
4455508415646675013373597242420117818453694838130159772560668808816707086990958982033203334310070688731662890013605553436739351074980172000127431349940128178077122187317837794167991459381249! (factorial, which is n * (n-1) * ... * 3 * 2 * 1), which contains hundreds of thousands of digits, and that's only for signed int, as for unsigned int... (2^64-1)^20 contains 200 digits, imagine how many digits ((2^64-1)^10)! would contain.
Assuming you can calculate even 10^9 hashes a second, it'll take milions of bilions of years to calculate all possible values. which, by the way, I assume you calculate no more than 100 hashes a second.
So... good luck brute forcing.
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Last edited by iNoobHacker on Mon Mar 16, 2009 2:06 pm; edited 1 time in total |
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Jani Grandmaster Cheater
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Posted: Mon Mar 16, 2009 2:05 pm Post subject: |
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Get some external library which can handle "big digits."
I know nothing about VB, but it's the same for all the languages with limits like this.
Move to Python if you want to avoid this problem.
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Jorg hi I post too much
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Posted: Mon Mar 16, 2009 2:12 pm Post subject: |
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Ahh, but I am brute forcing a number to test with that is "49223372036854775807" and I can't test it with an int64...
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Dark Byte Site Admin
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Posted: Mon Mar 16, 2009 5:04 pm Post subject: |
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only one way to do it then, manually check if it's bigger
e.g split the value up into several pieces (e.g by using multiple 4 byte pointers that point to separate blocks of the large value)
and then check if the last part is bigger, if not, part before that, etc.. (also, if the parts before are smaller, stop checking, since it means it's smaller...)
with last part I mean the part that points to the last couple of bytes of the large value, assuming you're using the normal low-endian byte notation
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Jorg hi I post too much
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Joined: 24 Dec 2007 Posts: 2276 Location: Minnesota
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Posted: Mon Mar 16, 2009 7:03 pm Post subject: |
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| Dark Byte wrote: | only one way to do it then, manually check if it's bigger
e.g split the value up into several pieces (e.g by using multiple 4 byte pointers that point to separate blocks of the large value)
and then check if the last part is bigger, if not, part before that, etc.. (also, if the parts before are smaller, stop checking, since it means it's smaller...)
with last part I mean the part that points to the last couple of bytes of the large value, assuming you're using the normal low-endian byte notation |
Oh I get what you mean, I will make 2 int64 integers and and for the first one I'll start adding and once it is 999999999 or whatever then the second integer will be added by 1, and when I brute force it the forced value with be (int64 integer1 + int64 integer2).Then the first int64 will be reset to 0 thanks.
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VolatileAce Cheater
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Posted: Sat Mar 21, 2009 7:40 pm Post subject: |
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Don't know if this will be marked as spam but look up "BigInteger".
Also according to your goal, I would deem it impractical to "un-hash" hash functions because hashing is not encrypting, encryption would recover the original information after decryption. But the hash itself would "most likely not" contain any useful information of any means regarding the original information.
Hope it helps.
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