Re: converting from windows wchar_t to linux wchar_t

From:
"Chris Becke" <chris.becke@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 15 Aug 2008 09:23:13 +0200
Message-ID:
<1218784677.587351@vasbyt.isdsl.net>

my Q is : is there a simple way to convert a 2 bytes wchar_t (windows
version ) to 4 bytes wchar_t ( linux version ).


wchar_t is a particularly useless type : Because its implementation =
defined it doesn't have (in protable code) any kind of assurance of what =
type of character encoding it may be using or capable of using.

The next point is that *unicode* characters are unsigned. so use an =
unsigned short for your UCS-2 / UTF-16 representation. =
http://en.wikipedia.org/wiki/UTF-16 has loads more information.

Finally, conversion for simple UCS-2 to UTF-32 is simple... Simply pad =
out the data by doing a direct characterwise copy:

  typedef ucs2char unsigned short;
  typedef utf32char unsigned long;

  void convert_ucs2_2_utf32(ucs2char const* src; utf32char* dest)
  {
    do {
      *dest++ = *src;
    } while(*src++);
  }

If you want to properly convert characters outside the basic =
multilingual plane, and the B.M.P covers all displayable characters from =
all modern languages that are in use :- european and eastern - then you =
need to be aware of surrogate pairs: Unicode codepoints in the range =
U+D800-U+DFFF are not assigned to valid characters, this range is used =
by UTF-16 to encode pairs of UTF-16 character each of which encodes 10 =
bits of the final codepoint.

So, something like this will do the translation of UTF-16 to UTF-32

  typedef utf16char unsigned short;
  void convert_utf16_to_utf32(ucs2char const* src; utf32char* dest)
  {
    do {
      if(*src & 0xD800 == 0xD800) {
        *dest++ = (*src++ & 0x07ff) << 10 + (*src & 0x7ff) + 0x10000;
      } else
         *dest++ = *src;
    } while(*src++);
  }

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