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Good Afternoon,

I'm a bit new to IPv6 and am curious on a couple of things. The loopback address is reserved as ::1 /128. If the mask is /128, wouldn't that indicate no available bits for hosts as all 128 are assigned to the network?

Also, I find the notation of link-local addresses a bit odd. The range indicates FE80 /10. But in practice, if you look at many assigned link-local addresses, they have other prefixes such as /12, /14, etc.

I'm sure I'm missing something simple, but can anyone help clear it up? Thank you.

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    Welcome to Server Fault. Our format doesn't work well with multiple unrelated questions. Please choose one, edit your question to include only it, and then post a second question with the other question. – Michael Hampton Oct 15 '15 at 20:04
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The /128 means that 128 bits are fixed and that means that no bits are flexible. So what this means in practice is that there is exactly one loopback address, not a range of addresses.

You are correct about the /10. That is the reserved range. In practice a /64 is used on LANs.

  • Thanks for your response. That part I'm hung up on is that the /128 would indicate that all 128 bits are used to represent the network. Is that correct? If that's the case, where does the host bit come from? Or in IPv6 does /128 just mean fixed but not necessarily network bits? My understanding was it indicated network bits. – Bob Oct 20 '15 at 14:27
  • 3 host bits means 2 to the power of 3 = 8 hosts. 2 bits: 2^2 is 4 hosts. 1 bit, 2^1 is 2 hosts. 0 bits: 2^0 is 1 host. I.e. the network is exactly one host big: the local host – Sander Steffann Oct 20 '15 at 15:27

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