一个21字节的UTF-8序列如何仅来自5个字符？

Question

After writing some basic code to count the number of characters in a String , I've found one example where the UTF-8 encoded output creates 21 bytes from a 5 "character" String .

Here's the output:

 String ==¦ อภิชาติ ¦== Code units 7 UTF8 Bytes 21 8859 Bytes 7 Characters 5 

I understand that Java's internal representation of a char is 2 bytes and there is a possibility that some characters may require two Unicode code units to display them.

As UTF-8 doesn't use any more than 4 bytes per character, how is a byte[] length of more than 20 possible for a 5 character String ?

Here's the source:

import java.io.UnsupportedEncodingException;

public class StringTest {

    public static void main(String[] args) {
        displayStringInfo("อภิชาติ");
    }

    public static void displayStringInfo(String s) {
        System.out.println("Code units " + s.length());     
        try {
            System.out.println("UTF8 Bytes " + s.getBytes("UTF-8").length);
        } catch (UnsupportedEncodingException e) { // not handled }
        System.out.println("Characters " + characterLength(s));
    }

    public static int characterLength(String s) {
        int count = 0;
        for(int i=0; i<s.length(); i++) {
            if(!isLeadingUnit(s.charAt(i)) && !isMark(s.charAt(i))) count++;
        }
        return count;
    }

    private static boolean isMark(char ch) {
        int type = Character.getType(ch);
        return (type == Character.NON_SPACING_MARK ||
               type == Character.ENCLOSING_MARK ||
               type == Character.COMBINING_SPACING_MARK);
    }

    private static boolean isLeadingUnit(char ch) {
        return Character.isHighSurrogate(ch);
    }
}

Answer 1

Your "5 character" string actually consists of 7 Unicode code points:

• U+0E2D THAI CHARACTER O ANG
• U+0E20 THAI CHARACTER PHO SAMPHAO
• U+0E34 THAI CHARACTER SARA I
• U+0E0A THAI CHARACTER CHO CHANG
• U+0E32 THAI CHARACTER SARA AA
• U+0E15 THAI CHARACTER TO TAO
• U+0E34 THAI CHARACTER SARA I

All of them are in the U+0800 to U+FFFF range that requires 3 bytes per character in UTF-8, hence a total length of 7×3 = 21 bytes.

Answer 2

There're 7 characters in the string:

 ' อ' (0x0e2d) encoded as {0xe0, 0xb8, 0xad}
  'ภ' (0x0e20) - / -      {0xe0, 0xb8, 0xa0}
  ' ิ' (0x0e34) - / -      {0xe0, 0xb8, 0xb4}
  'ช' (0x0e0a) - / -      {0xe0, 0xb8, 0x8a}
  'า' (0x0e32) - / -      {0xe0, 0xb8, 0xb2}
  'ต' (0x0e15) - / -      {0xe0, 0xb8, 0x95}
  ' ิ' (0x0e34) - / -      {0xe0, 0xb8, 0xb4}

each symbol is encoded by three bytes in UTF-8 and so you have 7 * 3 == 21 bytes altogeter