From: Jonathan V Hays (jhays@jtan.com)
Date: Tue May 27 2003 - 14:54:01 GMT-3
Since 'C' is 1100 in binary, the first and second (leftmost) bits of a
hex digit are "Don't Care" and can be 1 or 0. The last two (rightmost)
bits must be 0.
And note that 0x0C0C is actually a two-byte, 16-bit mask.
Hope that helps,
Jonathan
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On
> Behalf Of FATHALLAH
> Sent: Tuesday, May 27, 2003 11:09 AM
> To: lg01; ccielab@groupstudy.com
> Subject: RE: Filtering SNA
>
>
> 00 04 08 0B all are multiple of 4. so 0C0C mean that the
> first and second
> bits must be "0" and Third or/and fourth can be 1 or 0. witch
> give you 00,
> 04, 08,0B.
>
> Said FATHALLAH
>
> -----Message d'origine-----
> De : nobody@groupstudy.com [mailto:nobody@groupstudy.com]De la part de
> lg01
> Envoyi : mardi 27 mai 2003 11:01
> @ : ccielab@groupstudy.com
> Objet : Filtering SNA
>
>
> Hello group,
>
> If anyone can shed some lights on this, that would be greatly
> appreciated.
>
> In an exercise, it asked me to only allow SNA traffic from
> RTA. And it said
> that the SNA ports used will be 00, 04, 08 & 0B.
>
> But somehow... the answwer config gives:
>
> access-list 200 permit 0x0000 0x0C0C
>
> But I don't understand how they dervie / calculate this ACL.
>
> Thanks.
>
> H.
>
>
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