3 ms·
In your scheme, we'll end up with multiple leading bytes, though. Also, how do you plan to handle 9 leading bits instead of the 10 from your example? That woul
by cygx 13y ago
In your scheme, we'll end up with multiple leading bytes, though.
Also, how do you plan to handle 9 leading bits instead of the 10 from your example? That would make the second byte start with 10, the marker for continuation bytes.
- drdaeman 13y agoOuch, you're right. Refreshed my memory of UTF-8 encoding and now it seems it's designed so it can't be made to express arbitrary values. Not sure why I imagined the opposite. My mistake.