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

"Manfredi (US), Albert E" <albert.e.manfredi@boeing.com> Sat, 15 June 2019 23:30 UTC

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From: "Manfredi (US), Albert E" <albert.e.manfredi@boeing.com>
To: Alexandre Petrescu <alexandre.petrescu@gmail.com>, "ipv6@ietf.org" <ipv6@ietf.org>
Subject: RE: Is 1111 1110 10 equal to 0xfe80 or 0x3fa?
Thread-Topic: Is 1111 1110 10 equal to 0xfe80 or 0x3fa?
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Date: Sat, 15 Jun 2019 23:30:29 +0000
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-----Original Message-----
From: ipv6 <ipv6-bounces@ietf.org> On Behalf Of Alexandre Petrescu

> YEs, but if you only use the /10 to define a range, you introduce 
ambiguity in notation.

Alex, honestly, I never thought there was any ambiguity, and am surprised this issue lingers.

Consideration 1. The notation fe80::/10 means that the 10 bits represent a *prefix*. A prefix, by definition, the left-most 10 bits. That definition precludes the existence of any leading zeroes. If the 10-bit prefix included leading zeroes, it would have been written quite differently.

Consideration 2: Hex notation represents binary nibbles - groups of 4 bits. Therefore, however the 10-bit prefix is represented in hex, the only bits that matter are the left-most 10 bits.

Done. Now, the number 0xfe81 and the number 0xfe80 are identical, in terms of the left-most 10 bits. But someone might wonder, if you use fe81::/10, why on earth you bothered to set that 16th bit to 1 in your definition, when it was irrelevant to the 10bit prefix!

Bert

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