Reputation: 4064
The question I have is quite simple, but I couldn't find a solution so far:
How can I convert a UTF8 encoded string
to a latin1 encoded string
in C++ without using any extra libs like libiconv?
Every example I could find so far is for latin1 to UTF8 conversion?
Upvotes: 6
Views: 10866
Reputation: 5426
Here is a version of filmor's answer that I wrote for my purposes. A bit more readable, probably a bit slower. I didn't need the template stuff since I was always dealing with char *
, and in my case I wanted to replace non-Latin1 character's with _. Just in case it helps someone:
int GetUtf8CharacterLength( unsigned char utf8Char )
{
if ( utf8Char < 0x80 ) return 1;
else if ( ( utf8Char & 0x20 ) == 0 ) return 2;
else if ( ( utf8Char & 0x10 ) == 0 ) return 3;
else if ( ( utf8Char & 0x08 ) == 0 ) return 4;
else if ( ( utf8Char & 0x04 ) == 0 ) return 5;
return 6;
}
char Utf8ToLatin1Character( char *s, int *readIndex )
{
int len = GetUtf8CharacterLength( static_cast<unsigned char>( s[ *readIndex ] ) );
if ( len == 1 )
{
char c = s[ *readIndex ];
(*readIndex)++;
return c;
}
unsigned int v = ( s[ *readIndex ] & ( 0xff >> ( len + 1 ) ) ) << ( ( len - 1 ) * 6 );
(*readIndex)++;
for ( len-- ; len > 0 ; len-- )
{
v |= ( static_cast<unsigned char>( s[ *readIndex ] ) - 0x80 ) << ( ( len - 1 ) * 6 );
(*readIndex)++;
}
return ( v > 0xff ) ? 0 : (char)v;
}
// overwrites s in place
char *Utf8ToLatin1String( char *s )
{
for ( int readIndex = 0, writeIndex = 0 ; ; writeIndex++ )
{
if ( s[ readIndex ] == 0 )
{
s[ writeIndex ] = 0;
break;
}
char c = Utf8ToLatin1Character( s, &readIndex );
if ( c == 0 )
{
c = '_';
}
s[ writeIndex ] = c;
}
return s;
}
Test code:
char s2[ 256 ] = "lif\xc3\xa9 is b\xc3\xa9tt\xc3\xa9r with acc\xc3\xa9nts";
Utf8ToLatin1String( s2 );
Upvotes: 1
Reputation: 32212
typedef unsigned value_type;
template <typename Iterator>
size_t get_length (Iterator p)
{
unsigned char c = static_cast<unsigned char> (*p);
if (c < 0x80) return 1;
else if (!(c & 0x20)) return 2;
else if (!(c & 0x10)) return 3;
else if (!(c & 0x08)) return 4;
else if (!(c & 0x04)) return 5;
else return 6;
}
template <typename Iterator>
value_type get_value (Iterator p)
{
size_t len = get_length (p);
if (len == 1)
return *p;
value_type res = static_cast<unsigned char> (
*p & (0xff >> (len + 1)))
<< ((len - 1) * 6);
for (--len; len; --len)
res |= (static_cast<unsigned char> (*(++p)) - 0x80) << ((len - 1) * 6);
return res;
}
This function will return the unicode code point at p
. You can now convert a string using
for (std::string::iterator p = s_utf8.begin(); p != s_utf8.end(); ++p)
{
value_type value = get_value<std::string::iterator&>(p));
if (value > 0xff)
throw "AAAAAH!";
s_latin1.append(static_cast<char>(value));
}
No guarantees, the code is quite old :)
Upvotes: 4
Reputation: 9211
latin1
(aka ISO-8859-1
) defines the first 256 code points of Unicode. Thus, in UTF-8
, if your character is 8 bits, then it will exactly map to the latin1
equivalent. If it's more than 8 bits in length, then there is no correspondent within latin1
and you should map it to some "unknown character" (e.g., \0
or ?).
Upvotes: -1