java 值传递笔记_JAVA_编程开发_程序员俱乐部

package chunf.test;

public class Test2 {
//??? public static void doTest(int i, int j) {??
//??????? int a = i;??
//??????? i = j;??
//??????? j = a;??
//??? }??
//?
//??? public static void main(String[] args) {??
//??????? int a = 3;??
//??????? int b = 4;??
//??????? doTest(a, b);??
//?
//??????? System.out.println("a=" + a);??
//??????? System.out.println("b=" + b);??
//??? }
??? //以上输出的是 3 4 虽然走了doTest方法 但是，传递的只是对a b 的引用，并未改变a b 的值。
//-----------------------------------------------------------------------------------?
//??? public static void doTest(int[] counts) {??
//??????? counts[0] = 6;??
//??????? System.out.println(counts[0]);??
//??? }??
//?
//??? public static void main(String[] args) {??
//??????? int[] count = { 1, 2, 3};??
//??????? doTest(count);??
//??? }???
//以上输出为6，因为数组传递至方法内，方法将数组改变并在改变后打印改变后的值。
//----------------------------------------------------------------------------------
//??? public static void doTest(Test1 a) {??
//??????? a = new Test1();??
//??????? a.a++;??
//??? }??
//??????
//??? public static void main(String args[]) {??
//??? ??? Test1 a = new Test1();??
//??????? doTest(a);??
//??????? System.out.println(a.a);??
//??? }
/*
?* Test1{
?*? int a = 2;
?* }
?* */
//以上输出依然为TEST1类中的初始a值，原因是a并非静态的变量，每当new一个新的类，指针指向的地址是变化的。并非被方法处理的TEST1类
//------------------------------------------------------------------------------------
//??? public static void doTest(Test1 a) {??
//??????? a = new Test1();??
//??????? a.a++;??
//??? }??
//??????
//??? public static void main(String args[]) {??
//??? ??? Test1 a = new Test1();??
//??????? doTest(a);??
//??????? System.out.println(a.a);??
//??? }
/*
* Test1{
*? static int a = 2;
* }
* */
//以上输出为3，因为TEST1类中的变量a是静态的，因此无论new了多少个TEST1类，变量a的地址是不变的。所以操作的都是同一个a
//例如下面例子
//public static void doTest(Test1 a){
//??? a = new Test1();
//??? a.a++;
//}
//
//public static void main(String[] args) {
//??? Test1 tst1 = new Test1();
//??? doTest(tst1);
//??? Test1 tst2 = new Test1();
//??? doTest(tst2);
//??? System.out.println(tst1.a);
//??? System.out.println(tst2.a);
//}
//这两行输出都是4，因为无论是tst1还是tst2，变量a的地址是不变的，操作的都是同一个a
//----------------------------------------------------------------------------------
//??? String str = new String("haha");??
//??? ?
//??? char[] ch = { 'a', 'b', 'c' };??
?
//??? public static void main(String args[]) {??
//??? ??? Test2 tst = new Test2();??
//??????? ex.change(tst.str, tst.ch);??
//??????? System.out.print(tst.str + " and ");??
//??????? System.out.println(tst.ch);??
//??? }??
//?
//??? public void change(String str, char ch[]) {??
//??????? str = "lalala";??
//??????? ch[0] = 'd';??
//??? }?
//以上输出haha and dbc由于String是final的，那么tst.str是不被改变的。而数组并非如此，因此数组被改变
}??