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C function to encode a Unicode code point as UTF-8 byte array
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#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; | |
} | |
} |
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
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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.