Java and handling of UTF-8 codepoints in the supplementary range

UTF8.java

public class UTF8 {
  public static void main(String[] args) {
	  	// See http://www.fileformat.info/info/unicode/char/1f4a9/index.htm
		final String poo = "A pile of poo: 💩."; 
		
		System.out.println(poo);
		// Length of chars doesn't equals the "real" length, that is: the number of actual codepoints
		System.out.println(poo.length() + " vs " + poo.codePointCount(0, poo.length()));
		
		// Iterating over all chars
		for(int i=0; i<poo.length();++i) {
			char c = poo.charAt(i);
			// If there's a char left, we chan check if the current and the next char 
			// form a surrogate pair
			if(i<poo.length()-1 && Character.isSurrogatePair(c, poo.charAt(i+1))) {
				// if so, the codepoint must be stored on a 32bit int as char is only 16bit
				int codePoint = poo.codePointAt(i);
				// show the code point and the char
				System.out.println(String.format("%6d:%s", codePoint, new String(new int[]{codePoint}, 0, 1)));
				++i;
			}
			// else this can only be a "normal" char
			else 
				System.out.println(String.format("%6d:%s", (int)c, c));
		}
		
		// constructing a string constant with two \\u unicode escape sequences
		System.out.println("\ud83d\udca9".equals("💩"));
	}
}

Line #8 would print following in Java version "1.8.0_161":

System.out.println(poo.length() + " vs " + poo.codePointCount(0, poo.length()));
20 vs 20