【Java面试题】29 设计4个线程,其中两个线程每次对j增加1,另外两个线程对j每次减少1。写出程序。
本题并不难,实现方式有很多种,有很多种结构。
方法一:利用内部类实现,两个实现加减的类实现Runnable接口,然后再实现4个具体线程。
代码:
public class ManyThreads { private int j; public static void main(String[] args) { // TODO Auto-generated method stub ManyThreads many = new ManyThreads(); Inc inc = many.new Inc(); Dec dec = many.new Dec(); for (int i = 0; i < 2; i++) { Thread t = new Thread(inc); t.start(); t = new Thread(dec); t.start(); } } private synchronized void inc() { j++; System.out.println(Thread.currentThread().getName() + "inc" + j); } private synchronized void dec() { j--; System.out.println(Thread.currentThread().getName() + "dec" + j); } class Inc implements Runnable { @Override public void run() { // TODO Auto-generated method stub for (int i = 0; i < 20; i++) { inc(); } } } class Dec implements Runnable { @Override public void run() { // TODO Auto-generated method stub for (int i = 0; i < 20; i++) { dec(); } } } }
第二种方式:具体加减操作写在一个类的方法里,没有内部类,用另外一个类去调用。
代码:
public class MyTest { private ManyThreads2 many = new ManyThreads2(); public static void main(String[] args) { // TODO Auto-generated method stub MyTest myTest = new MyTest(); myTest.test(); } public void test() { for (int i = 0; i < 2; i++) { new Thread(new Runnable() { @Override public void run() { // TODO Auto-generated method stub for (int i = 0; i < 20; i++) { many.inc(); } } }).start(); new Thread(new Runnable() { @Override public void run() { // TODO Auto-generated method stub for (int i = 0; i < 20; i++) { many.dec(); } } }).start(); } } class ManyThreads2 { private int j = 0; public synchronized void inc() { j++; System.out.println(Thread.currentThread().getName() + "inc" + j); } public synchronized void dec() { j--; System.out.println(Thread.currentThread().getName() + "dec" + j); } } }
参考资料:
http://www.cnblogs.com/nannanITeye/p/3421943.html
http://blog.csdn.net/u014236541/article/details/51178905
http://aijuans.iteye.com/blog/1847835