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September 17, 2553 B.C. — Home news: Gujarati filmmaker Ashish Kakkad, whose film The Better Half, released in Mumbai and Gujarat, is betting on different tactics to stay.## # September 17, 2553 B.C. — Home news: Gujarati filmmaker Ashish Kakkad, whose film The Better Half, released in Mumbai and Gujarat, is betting on different tactics to stay.## #
September 17, 2553 BC
home news
Gujarati filmmaker Ashish Kakkad, whose film The Better Half, released in Mumbai and Gujarat, is betting on different tactics to stay.
Kakkad, 30, said he “doesn’t give up until an agreement is reached.”

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Gujarat Oct 4,  – Download MobiSave 4 7 Gujrat and Odisha Budget 2018 Calendar. Important dates, 10 Oct -The Gujarat Election Calendar 2018 2018 Karyyabandhu Charitable Trust,” Gujarat (India).
Q:

A string that contains a little-endian UTF-16 string

I need to make a string that contains a little-endian UTF-16 string, just like this one (which is in big-endian):
const char *tuple = “This is just a little-endian UTF-16 string\0”

In the above string, the “\0” is the NULL symbol (\u0000). I’m expecting to get this string by the following expression:
struct utf16string* utf16String = malloc(utf16string);
memcpy(utf16String->string, tuple, utf16string->length);

But this won’t work, because the “\0” is the 0 symbol and, of course, the 0 symbol is bigger than the “U” symbol (\u0080). I’d like to know if there’s a way to get my little-endian string.

A:

You can do this in two parts:

store the string in raw bytes without conversion, and
store it with all the null bytes converted to UTF-16 escape sequences (the byte order marks).

For the first, you can simply use strcpy(). For the second, you can use normal UTF-16 decoding to reverse this process.
This should suffice:
char *utf16String = malloc(utf16string);
char *p = strcpy(utf16String, tuple);
while(p length) {
char b = *p;
p++;
if(b & 0xC0 == 0x80)
b = *p;
p++;
*utf16String++ = (b >> 6) | 0xF4;
*utf16String++ = (b & 0x3F) | 0x80;
}
*utf16String = 0;

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