How much disk space would take to list every possible UEI-48 or MAC address?

Is it (2^48) possible addresses X 48 bits = 1536 terabytes ??

networking

ian

Posted 2011-02-19T21:54:04.073

Reputation: 11

10:1:2^48 takes about 9 bytes in my reckoning (including the terminating '\0' character). It's called utilizing entropy to compress your data - very useful for storing large amount of organized data. It's like storing 314159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706 as "first 100 digits of pi". – Floris – 2014-07-30T15:21:33.463

2Why would you want a list of all the numbers from 0 to 2^48 (281 474 976 710 656)? – martineau – 2011-02-20T08:41:33.650

Answers

There are 2⁴⁸ (281474976710656) possible EUI-48 addresses, which would occupy 2⁴⁸ × 48 bits or 1688849860263936 eight-bit bytes, which is 1536 terabytes or 1.5 petabytes.

user1686

Posted 2011-02-19T21:54:04.073

Reputation: 283 655

That sounds about right.

Plus some Bytes for the filesystem.

And a lot more bytes if you want to store everything as textfile. In this case we are talking about 2^48 (address range) * 48 bit (per address) * 8 bit (per char, assuming you store everything in ASCII) = about 12288 terabytes. And don't forget dividers between the single addresses - if required.

lajuette

Posted 2011-02-19T21:54:04.073

Reputation: 4 364

1If you're going to store in ASCII, you wouldn't store '0' or '1' for everything, you'd use hex. With 8-bit chars, you cut the expansion from 8:1 to 8:4, or 2. 48-bit address is 12 hex digits; 96 bits with 8bit characters or 84 with seven-bit ASCII. You could put two 48-bit addresses into 21 bytes of storage, even with character encoding. – mpez0 – 2011-02-20T01:28:53.083