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Re: Is 1111 1110 10 equal to 0xfe80 or 0x3fa?

Fred Baker <fredbaker.ietf@gmail.com> Fri, 07 June 2019 17:53 UTC

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From: Fred Baker <fredbaker.ietf@gmail.com>
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Subject: Re: Is 1111 1110 10 equal to 0xfe80 or 0x3fa?
Date: Fri, 07 Jun 2019 10:53:28 -0700
In-Reply-To: <c57bc938-d151-a96d-7ea6-efca7a48648b@gmail.com>
Cc: IPv6 <ipv6@ietf.org>
To: Alexandre Petrescu <alexandre.petrescu@gmail.com>
References: <a71b00b7-0e0c-242a-b3f7-147f4c6b2eb0@gmail.com> <D05C857F-42F6-4F17-8520-A0BF4C8FB775@steffann.nl> <c57bc938-d151-a96d-7ea6-efca7a48648b@gmail.com>
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> On Jun 6, 2019, at 10:55 PM, Alexandre Petrescu <alexandre.petrescu@gmail.com> wrote:
> 
> Is it wrong to write that 3fa::/10 is the link-local prefix?

Which is to say that the bit pattern is 0000 0011 ffff 1010, and the prefix is the first ten bits (0000 0011 11) of that? If that's what you intend, no, it's not wrong. If you are thinking in terms of https://tools.ietf.org/html/rfc4291#section-2.5.6, there is no way that would work. It doesn't conform to https://tools.ietf.org/html/rfc4632 (which only uses IPv4 examples, but IPv6 works the same way).

Here's a simple way to think about this that might help you get your head around it. Let's imagine I have a prefix 2001:db8::/32, and I want to divide it into two 33 bit prefixes. What are they? 2001:db8:0000::/33 and 2001:db8:8000::/33, of course. It turns around the next bit.

If I have prefix fe80::/10, as described in RFC 4291, the next bit is bit 11. Doing the same subdivision of the prefix is fe80::/11 and fea0::/11.

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