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encoding OIDs using BER

nitelord
nitelord asked
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Last Modified: 2010-03-18
Can anybody give me an explanation on how to encode OID components using BER. I know the first two use the scheme 40x+y, but my problem is with components 3 onwards larger than 127. For example, how does 134 become 81 02 (hex).

Thanx
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You sure the BER for 134 is not 81 06?

Using 134;  binary is 10000110.  You start with the least sig bit of the least sig byte and grab the last 7 bits, 0000110.  Then look at the 8 th bit; if it was zero we would stop, but since it is 1, we need another octet, 10000001.  The most sig bit says there are more octets to follow; i.e. it is the continuation bit.  When it is 0 the decoder knows it has the complete component, when 1 it knows to grab the next octet.  Thus, if the 8th bit is significant, i.e. 1, for our data then it must be moved left into another octet to make room for the continuation bit.  Final encoding:

10000001 0000110
    81      06

Hope this helps
Rob
There are probably links out there that explain it better; I'll look.
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Commented:
when I calculated it I got 81 06 but the book I'm using (pg 589, Feit, Sidnie, SNMP a guide to network management) gave that value. I'll try a few more, thanx

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