package object;

public class E08_StaticTest {
	  int petalCount = 0;
	  String s = "initial value";
//(1) 
	  E08_StaticTest(int petals) {
	    petalCount = petals;
	    System.out.println("Constructor w/ int arg only, petalCount= "
	      + petalCount);
	  }
//(2)
	  E08_StaticTest(String ss) {
	    System.out.println("Constructor w/ String arg only, s = " + ss);
	    s = ss;
	  }
//(3)
	  E08_StaticTest(String s, int petals) {
	    this(petals);
	//!    this(s); // Can't call two!
	    this.s = s; // Another use of "this"
	    System.out.println("String & int args");
	  }
//(4)
	  E08_StaticTest() {
	    this("hi", 47);
	    System.out.println("default constructor (no args)");
	  }
	  void printPetalCount() {
	//! this(11); // Not inside non-constructor!
	    System.out.println("petalCount = " + petalCount + " s = "+ s);
	  }
	  public static void main(String[] args) {
	    E08_StaticTest x = new E08_StaticTest();
	    x.printPetalCount();
	  }
	} /* Output:
	Constructor w/ int arg only, petalCount= 47
	String & int args
	default constructor (no args)
	petalCount = 47 s = hi
	*///:~
 

注意：this本身表示對當前物件的引用。

剛開始看這段程式碼時，沒怎麼看明白，本以為會直接輸出最後兩行，沒搞懂Constructor w/ int arg only, petalCount= 47；String & int args這兩行為什麼會輸出。後來才明白其原因，

其執行順序為： E08_StaticTest x = new E08_StaticTest();其後執行（4）的this語句->（3）的this語句->（1）確認int型別為47，輸出print->（3）的剩下語句->（4）的剩下語句。這樣輸出結果就和程式設計的一一對應了。原因：如果構造器的第一個語句形如this（...）這個構造器將呼叫同一個類的另一個構造器。