三個線程，分別只用以打印a,b,c

用多線程實現(xiàn)打印abc重復10次

我想了一個比較丑的方法，就是一個命令者，用以發(fā)布命令，這次要打印那個字符，三個打印者就分別檢查這個指令是否是發(fā)給自己的

class ShId {private int value;public ShId(int id) {value = id;}public void set(int i) {value = i;}public boolean isEqual(Integer i) {return i.intValue() == value;}@Overridepublic boolean equals(Object o) {if (this == o) return true;return ((ShId)o).value == value;} }class Work implements Runnable {private String name;private Integer id;private ShId shId;public Work(String name, Integer id, ShId shId) {this.name = name;this.id = id;this.shId = shId;}@Overridepublic void run() {try {while (true) {synchronized (shId) {while(!shId.isEqual(id)) {shId.wait();}System.out.print(name);shId.set(-1);shId.notifyAll();if (name.equals("c")) System.out.println();}}} catch(InterruptedException e) {e.printStackTrace();}} }class Produce implements Runnable {private int count;private ShId shId;public Produce (int count, ShId shId) {this.count = count;this.shId = shId;}@Overridepublic void run() {try {while (count >0) {for (int i = 0; i < 3; ++i) {synchronized (shId) {while (!shId.isEqual(new Integer(-1))) {shId.wait();}shId.set(i);shId.notifyAll();}}--count;}} catch (InterruptedException e) {}} }public class testt {public static void main(String[] args) {ShId shId = new ShId(-1);Work w1 = new Work("a",0, shId);Work w2 = new Work("b",1, shId);Work w3 = new Work("c",2, shId);Produce p = new Produce(10, shId);new Thread(w1, "work1").start();new Thread(w2, "work2").start();new Thread(w3, "work3").start();new Thread(p, "produce").start();} }

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