On 12/11/2012 11:50 AM, vanis...@boil.afraid.org wrote:
From: James Lin <James_Lin_at_symantec.com>
Hi
Does anyone know why ill-form occurred on the UTF-8? besides it doesn't follow
> the pattern of UTF-8 byte-sequences, i just wondering how or why?
If i have a code point: U+4E8C or "二"
In UTF-8, it's "E4 BA 8C" while in UTF-16, it's "4E8C". Where is this "BA"
comes from?
thanks
-James
Each of the UTF encodings represents the binary data in different ways. So we
need to break the scalar value, U+4E8C, into its binary representation before
we proceed.
4E8C -> 0100 1110 1000 1100
Then, we need to look up the rules for UTF-8. It states that code points
between U+800 and U+FFFF are encoded with three bytes, in the form 1110xxxx
10xxxxxx 10xxxxxx. So plugging in our data, we get
4 E 8 C
0100 1110 10-00 1100
|||| ||||// \\||||
+ 1110xxxx 10xxxxxx 10xxxxxx
= 11100100 10111010 10001100
or E 4 B A 8 C
-Van Anderson
Nice!
A./
PS: I fixed a missing "\"
