3 ms·

whats the rule to say where the first 1 floats between the 2000: and the :1 at the end? the :: rule says "all zeros" but not how long.

• doubletwoyou 11mo ago
  the :1 is short for :0001 basically and then just put that bit of the address at the very end and put the first bit of the address at the front, and then just fill each missing group inbetween with 0000
  • webignition 11mo ago
    "just"
    • bigstrat2003 11mo ago
      Yes, in fact "just". This isn't remotely hard.
      • karlshea 11mo ago
        These types of complaints are how I know the objection to v6 is not serious.
      • Dagger2 11mo ago
        Well, okay, show us how to follow those instructions then. "the :1 is short for :0001 basically" is easy enough: you get 2001::0001::0001. Then "just put that bit at the very end" -- but which bit? If it means the ":0001", then there's two of them and they can't both go at the very end. If not, then it fails to specify which bit. Either way I don't see how these instructions are followable at all, let alone easily.
• db48x 11mo ago
  It’s a really complicated rule called “subtraction”. Addresses are always 128 bits long, or 8 groups of four hex digits. 2000::1 is two groups, so you need six groups in between to make 2000:0000:0000:0000:0000:0000:0000:1. But I don’t know why people always ask this, because it’s always the computer you are typing addresses in to that does the subtraction. You never ever have to type out the whole address. Just type the shortened version, because 2000::1 _is_ the whole address.
  • Dylan16807 11mo ago
    They were answering the question of why "2000::1::1" would be ambiguous if it was allowed.
• ggm 11mo ago
  My answer was too terse. IF there was two :: in the address, then the length of EACH :: denoted section is not known. It can be either longest left :: or longest right :: and that wasn't defined, because the rule is THERE IS ONLY ONE :: section. Posed as a question, disingenuously.
  • deleted 11mo ago
    [deleted]