| ▲ | ggm a day ago | |||||||||||||||||||||||||||||||
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. | ||||||||||||||||||||||||||||||||
| ▲ | db48x a day ago | parent | next [-] | |||||||||||||||||||||||||||||||
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. | ||||||||||||||||||||||||||||||||
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| ▲ | doubletwoyou a day ago | parent | prev | next [-] | |||||||||||||||||||||||||||||||
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 | ||||||||||||||||||||||||||||||||
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| ▲ | ggm a day ago | parent | prev [-] | |||||||||||||||||||||||||||||||
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. | ||||||||||||||||||||||||||||||||
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