About 10-11 months back I'd posted the question regarding BER encoding
of OID's, and now I am back, since some doubts are.
How do I decode the following octet string, using the explanation of the
encoding rules as specified in X.690.
06 0c 2a 86 3a 00 89 61 33 01 00 01 00 01
I'll write here, how far I manage to do it...
Tag = 0x06
Length = 0xc (12), count of folliowing value octets
Value = {
2a = 1.2.
86 = 6.
3a = 58.
00 = 0.
89 = (10001001)2 ??? (I thought that the MSbit could be one, only for
the terminal octet!!)
61 = (01100001)2 ? (.. so is this octet related to the previous one, the
one above in this list ? if so, how ?)
33 = 51. ??
01 = 1. ??
00 = 0. ??
01 = 1. ??
00 = 0. ??
01 = 1. ??
}
I think this list doesn't permit HTML formatting, which is sad, because
I'd colour coded the text to explain the issue in more clear terms. To
me the problem is in the octet's with value 89 onwards... since it has
the MSBit set. According to the explanation in X.690 & Olivier
Dubuisson's books, the MSBit = 1, would occur only once ??? (Or, that's
what I understood).
The difficulty is that you are misinterpreting the octets with the high
order bit set. Please bear in mind that the nodes of an OID can be very
large numbers and so there needs to be a way to specify the length. One
could have, of course, designed it such that each node would be preceded
by a length field, but that's a lot of wasted bits.
Instead since the numbers tend to be small, the method chosen is one where
the high bit specifies whether (0) this is the last octet or (1) more are
coming. That is, only the least significant 7 bits of each octet represent
the value. The high bit, so to speak, specifies the length. By way of
example (you might find the breakouts easier to read if you set this to
some fixed font)
01 0xxx xxxx this is the final octet
x000 0001 value is 1
863A 1xxx xxxx xxxx xxxx another octet follows
xxxx xxxx 0xxx xxxx this is the final octet
x000 0110 x011 1010 these 14 bits specify the value (826)
If you squeeze out the x's and use hex representation, you'll have
00 0011 0011 1010 033A which is 826.
You can therefore, by examining the high-order bits, see that you have the
following nodes
2a 1.2
863a 826
00 0
8961 1249
33 51
01 1
00 0
01 1
00 0
01 1
=====================================================================
Conrad Sigona Voice Mail : 1-732-302-9669 x400
OSS Nokalva Fax : 1-614-388-4156
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