C function to encode a Unicode code point as UTF-8 byte array

utf8_encode.c

#include <stdint.h>

/**
 * Encode a code point using UTF-8
 * 
 * @author Ondřej Hruška <[email protected]>
 * @license MIT
 * 
 * @param out - output buffer (min 5 characters), will be 0-terminated
 * @param utf - code point 0-0x10FFFF
 * @return number of bytes on success, 0 on failure (also produces U+FFFD, which uses 3 bytes)
 */
int utf8_encode(char *out, uint32_t utf)
{
  if (utf <= 0x7F) {
    // Plain ASCII
    out[0] = (char) utf;
    out[1] = 0;
    return 1;
  }
  else if (utf <= 0x07FF) {
    // 2-byte unicode
    out[0] = (char) (((utf >> 6) & 0x1F) | 0xC0);
    out[1] = (char) (((utf >> 0) & 0x3F) | 0x80);
    out[2] = 0;
    return 2;
  }
  else if (utf <= 0xFFFF) {
    // 3-byte unicode
    out[0] = (char) (((utf >> 12) & 0x0F) | 0xE0);
    out[1] = (char) (((utf >>  6) & 0x3F) | 0x80);
    out[2] = (char) (((utf >>  0) & 0x3F) | 0x80);
    out[3] = 0;
    return 3;
  }
  else if (utf <= 0x10FFFF) {
    // 4-byte unicode
    out[0] = (char) (((utf >> 18) & 0x07) | 0xF0);
    out[1] = (char) (((utf >> 12) & 0x3F) | 0x80);
    out[2] = (char) (((utf >>  6) & 0x3F) | 0x80);
    out[3] = (char) (((utf >>  0) & 0x3F) | 0x80);
    out[4] = 0;
    return 4;
  }
  else { 
    // error - use replacement character
    out[0] = (char) 0xEF;  
    out[1] = (char) 0xBF;
    out[2] = (char) 0xBD;
    out[3] = 0;
    return 0;
  }
}

are you using linux?

Please stop posting to this gist, it is NOT related to the error you're getting and I don't have the time or the will to troubleshoot your program. StackOverflow is the place to get help.

Eugenetucha are you assuming this snippet is a standalone program? You need to provide your own main() or WinMain() that uses it, or compile it as library or linkable object.

how to convet a double to UTF-8 bytes?

@wangjunfeng90214 what do you mean by double?

@MightyPork just wanted to say thanks, due to this i dont have to rely on the icu library! <3