3 ms·
For a bijective base: 7 bits -> Decode 0xxxxxxx and add 0 (unchanged) 14 bits -> Decode 10xxxxxx xxxxxxxx and add 0x80 21 bits -> Decode 110xxxxx x
by KMag 11y ago
For a bijective base:
7 bits -> Decode 0xxxxxxx and add 0 (unchanged)
14 bits -> Decode 10xxxxxx xxxxxxxx and add 0x80
21 bits -> Decode 110xxxxx xxxxxxxx xxxxxxxx and add 0x4080
In a non-bijective base vs a bijective base:
7 bits encode 0 to 2&7 - 1 vs. 0 to 2^7 -1
14 bits encode 0 to 2^14 - 1 vs. 2^7 to 2^14 + 2^7 - 1
21 bits encode 0 to 2^21 - 1 vs. 2^14 + 2^7 to 2^21 + 2^14 + 2^7 - 1
...
In the bijective decoding routine, you need to special-case the maximum length case to check for numeric overflow.
- deleted 11y ago[deleted]
- deleted 11y ago[deleted]