Java中this的用法及在构造器中调用构造器



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= 47String & int argsdefault constructor (no args)petalCount = 47 s = hi*///:~

注意：this本身表示对当前对象的引用。

刚开始看这段代码时，没怎么看明白，本以为会直接输出最后两行，没搞懂Constructor w/ int arg only, petalCount= 47；String & int args这两行为什么会输出。后来才明白其原因，

其执行顺序为： E08_StaticTest x = new E08_StaticTest();其后执行（4）的this语句->（3）的this语句->（1）确认int类型为47，输出print->（3）的剩下语句->（4）的剩下语句。这样输出结果就和程序设计的一一对应了。原因：如果构造器的第一个语句形如this（...）这个构造器将调用同一个类的另一个构造器。