Okay, I got it... but I think the text in X.690 definitely needs clarification, 
and text, in the otherwise excellent & very informative book by Olivier, needs 
correction, because IMHO, the following text:
 
"
                                                                                
   this form is similar
to the long-form encoding of the tag octets presented in Figure 18.2 on
page 396)
"
is somewhat misleading. The encoding is not like long-form encoding of tag 
octets, where the MSByte octet of the tag (in long form) contains the length of 
the remaining # of octets, forming the tag, which is not the case here.
 
So, the way to interpret the encoding is --
 
Value = { 
2a = 1.2.
86 = (10000110)2 (since the MSbit is 1, thus next octet also is part of this 
integer)
3a = (00111010)2 (this octet is to be contactenated to previous octet, as 
bit-string) => 110-0111010 = 826.
00 = 0.
89 = (10001001)2 (since the MSbit is 1, thus next octet also is part of this 
integer)
61 = (01100001)2 (this octet is to be contactenated to previous octet, as 
bit-string) => 1001-1100001 = 1249.
33 = 51.
01 = 1.
00 = 0.
01 = 1.
00 = 0.
01 = 1.
}
 
i.e.
{ 1.2.826.0.1249.51.1.0.1.0.1 }
 
Which is { iso(1).member-body(2).uk(826).0.1249.51.1.0.1.0.1 }, as I've found 
on the net via OID search. 
 
thanks,
Banibrata

________________________________

From: Dutta, Banibrata (STSD) 
Sent: Monday, March 05, 2007 8:05 PM
To: [email protected]
Subject: Confusion about OID BER encoding (again)



Hi,

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).
 
Thanks for clarifying this issue.
 
regards,
Banibrata Dutta
Hewlett-Packard | > [EMAIL PROTECTED] | Å +91 80 22051796 | ü GMT +5.5
 
 

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