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Array of Bytes to Unicode chars (ISO-8859-1)

Hi ,
I want to convert an array of bytes like :

{79,104,-37,-66,24,123,30,-26,-99,-8,80,-38,19,14,-127,-3}

into Unicode character with ISO-8859-1 standard.

Can anyone help me .. how should I go about doing it ?

Thanks

Jul 25 '06
14 6441
In article <44************ **@cam.ac.ukIan Malone <ib***@cam.ac.u kwrites:
....
Not quite, ISO-8859-1 includes characters with the most significant
bit set (ie. 128-255). The low range matches (I believe, but could
be wrong...) ASCII, the printable characters of which do map directly
to Unicode. In general you need tables of corresponding characters
to do this, see P J Plauger's post.
Look at the table that P. J. Plaugher posted and see that it starts with:
0x00 0x0000
and goes on to
0xFF 0x00FF
or something like that. The first 256 codes of Unicode are identical to
ISO 8859-1.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Jul 25 '06 #11

Keith Thompson wrote:
"J. J. Farrell" <jj*@bcs.org.uk writes:
ab*****@gmail.c om wrote:
I want to convert an array of bytes like :

{79,104,-37,-66,24,123,30,-26,-99,-8,80,-38,19,14,-127,-3}

into Unicode character with ISO-8859-1 standard.

Can anyone help me .. how should I go about doing it ?
Your question is unclear. Are you saying that the current bit patterns
in the array are ISO 8859-1 encodings, presumably represented as signed
chars? (That is, that if I treat the bit patterns in the array as
unsigned chars, then the low-order 8 bits of each are ISO 8859-1
character values?)

Assuming that, you first need to pick a type with at least 21 value
bits to hold the Unicode characters. Life will usually be less
confusing if you make it an unsigned type. Unsigned long will be big
enough in all versions of Standard C. You then need to put the 8 low
order bits from the each entry in the source array into the 8 low order
bits in the destination entries, with zeroes in the higher bits. Use
pointers to iterate over the arrays doing something like

*targetp = *(unsigned char *)sourcep & 0xff;

The values of ISO 8859-1 characters are the same as the values of the
equivalent Unicode characters.

Assuming sourcep is a char*, I think converting the char value to
unsigned char is clearer than doing a pointer conversion:

*targetp = (unsigned char)*sourcep & 0xff;
Clearer perhaps, but also wrong in some environments given the
assumptions I stated.
The "& 0xff" may be unnecessary of CHAR_BIT==8.
Indeed. But why write it in a way that will be wrong if CHAR_BIT != 8,
when it takes so little extra effort to make it portable?

Jul 26 '06 #12
Simon Biber wrote:
Ian Malone wrote:
>Simon Biber wrote:
>>Simply copy them into the low byte of your Unicode characters!

Not quite, ISO-8859-1 includes characters with the most significant
bit set (ie. 128-255). The low range matches (I believe, but could
be wrong...) ASCII, the printable characters of which do map directly
to Unicode. In general you need tables of corresponding characters
to do this, see P J Plauger's post.

The Unicode character set was designed not only that 0-127 match ASCII,
but also that 128-255 match ISO-8859-1. This works for UTF-16 and UTF-32
but not, of course, UTF-8.
Sorry, my blood caffeine level was obviously dangerously low yesterday.
Hallucinations that occurred involved the fact it won't fit into UTF8
as you mention, and I somehow read low byte as 7 bits...

--
imalone
Jul 26 '06 #13
On Tue, 25 Jul 2006 16:15:10 GMT, Frederick Gotham
<fg*******@SPAM .comwrote:
<snip>
Perhaps something like:

#include <wchar.h>

void ToUnicode(wchar _t *pw, char const *pc)
{
while(*pw++ = *pc++);
}

int main(void)
{
char const str[] = "October is the tenth month of the year.";

wchar_t buf[sizeof str * sizeof(wchar_t)];
The *sizeof(wchar_t ) is unnecessary and wasteful. You need (only) the
same number of wchar_t as there were chars in the source.
ToUnicode(buf,s tr);
}
- David.Thompson1 at worldnet.att.ne t
Aug 7 '06 #14
Dave Thompson posted:
> char const str[] = "October is the tenth month of the year.";

wchar_t buf[sizeof str * sizeof(wchar_t)];
The *sizeof(wchar_t ) is unnecessary and wasteful. You need (only) the
same number of wchar_t as there were chars in the source.

You're correct, I wrote the code quickly and sloppily. I should have written:

char const str[] = "whatever";

wchar_t buf[sizeof str];

--

Frederick Gotham
Aug 7 '06 #15

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