我被要求在一次采访中编写一个双线程的Java程序。在这个程序中，一个线程应该打印偶数，另一个线程应该交替打印奇数。
示例输出：
线程1：1
线程2：2
线程1：3
线程2：4 ......等等
我写了以下程序。一个类Task，其中包含两种分别打印偶数和奇数的方法。从main方法，我创建了两个线程来调用这两个方法。面试官让我进一步改进，但我想不出任何进步。有没有更好的方法来编写相同的程序？
class Task
{
boolean flag;
public Task(boolean flag)
{
this.flag = flag;
}
public void printEven()
{
for( int i = 2; i <= 10; i+=2 )
{
synchronized (this)
{
try
{
while( !flag )
wait();
System.out.println(i);
flag = false;
notify();
}
catch (InterruptedException ex)
{
ex.printStackTrace();
}
}
}
}
public void printOdd()
{
for( int i = 1; i < 10; i+=2 )
{
synchronized (this)
{
try
{
while(flag )
wait();
System.out.println(i);
flag = true;
notify();
}
catch(InterruptedException ex)
{
ex.printStackTrace();
}
}
}
}
}
public class App {
public static void main(String [] args)
{
Task t = new Task(false);
Thread t1 = new Thread( new Runnable() {
public void run()
{
t.printOdd();
}
});
Thread t2 = new Thread( new Runnable() {
public void run()
{
t.printEven();
}
});
t1.start();
t2.start();
}
}
答案 0 :(得分：2)
我认为这应该正常而且非常简单。
package com.simple;
import java.util.concurrent.Semaphore;
/**
* @author Evgeny Zhuravlev
*/
public class ConcurrentPing
{
public static void main(String[] args) throws InterruptedException
{
Semaphore semaphore1 = new Semaphore(0, true);
Semaphore semaphore2 = new Semaphore(0, true);
new Thread(new Task("1", 1, semaphore1, semaphore2)).start();
new Thread(new Task("2", 2, semaphore2, semaphore1)).start();
semaphore1.release();
}
private static class Task implements Runnable
{
private String name;
private long value;
private Semaphore semaphore1;
private Semaphore semaphore2;
public Task(String name, long value, Semaphore semaphore1, Semaphore semaphore2)
{
this.name = name;
this.value = value;
this.semaphore1 = semaphore1;
this.semaphore2 = semaphore2;
}
@Override
public void run()
{
while (true)
{
try
{
semaphore1.acquire();
System.out.println(name + ": " + value);
value += 2;
semaphore2.release();
}
catch (InterruptedException e)
{
throw new RuntimeException(e);
}
}
}
}
}
答案 1 :(得分：1)
嗯，有很多选择。我可能会使用SynchronousQueue代替（我不喜欢低级wait / notify并尝试使用更高级别的并发原语代替）。此外，printOdd和printEven可以合并为单个方法，不需要其他标记：
public class App {
static class OddEven implements Runnable {
private final SynchronousQueue<Integer> queue = new SynchronousQueue<>();
public void start() throws InterruptedException {
Thread oddThread = new Thread(this);
Thread evenThread = new Thread(this);
oddThread.start();
queue.put(1);
evenThread.start();
}
@Override
public void run() {
try {
while (true) {
int i = queue.take();
System.out.println(i + " (" + Thread.currentThread() + ")");
if (i == 10)
break;
queue.put(++i);
if (i == 10)
break;
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
}
}
public static void main(String[] args) throws InterruptedException {
new OddEven().start();
}
}
答案 2 :(得分：0)
这样的短版本怎么样：
public class OddEven implements Runnable {
private static volatile int n = 1;
public static void main(String [] args) {
new Thread(new OddEven()).start();
new Thread(new OddEven()).start();
}
@Override
public void run() {
synchronized (this.getClass()) {
try {
while (n < 10) {
this.getClass().notify();
this.getClass().wait();
System.out.println(Thread.currentThread().getName() + ": " + (n++));
this.getClass().notify();
}
} catch (InterruptedException ex) {
ex.printStackTrace();
}
}
}
}
有一些技巧可以正确地启动线程 - 因此需要一个额外的notify()来启动整个事情（而不是让两个进程都等待，或者需要主线程调用一个通知）并且还处理线程启动的可能性，它是否正常工作并在第二个线程启动之前调用notify：）
答案 3 :(得分：0)
我的初步答案是无效的。编辑：
package test;
public final class App {
private static volatile int counter = 1;
private static final Object lock = new Object();
public static void main(String... args) {
for (int t = 0; t < 2; ++t) {
final int oddOrEven = t;
new Thread(new Runnable() {
@Override public void run() {
while (counter < 100) {
synchronized (lock) {
if (counter % 2 == oddOrEven) {
System.out.println(counter++);
}
}
}
}
}).start();
}
}
}
答案 4 :(得分：0)
有没有更好的方法来编写相同的程序？
嗯，问题是，编写程序的唯一好方法是使用单个线程。如果你想要一个程序按顺序执行X，Y和Z，那么编写一个执行X，然后是Y，然后执行Z的过程。没有比这更好的方法了。
这是我在与面试官讨论线索的适当性后我会写的内容。
import java.util.concurrent.SynchronousQueue;
import java.util.function.Consumer;
public class EvenOdd {
public static void main(String[] args) {
SynchronousQueue<Object> q1 = new SynchronousQueue<>();
SynchronousQueue<Object> q2 = new SynchronousQueue<>();
Consumer<Integer> consumer = (Integer count) -> System.out.println(count);
new Thread(new Counter(q1, q2, 2, 1, consumer)).start();
new Thread(new Counter(q2, q1, 2, 2, consumer)).start();
try {
q1.put(new Object());
} catch (InterruptedException ex) {
throw new RuntimeException(ex);
}
}
private static class Counter implements Runnable {
final SynchronousQueue<Object> qin;
final SynchronousQueue<Object> qout;
final int increment;
final Consumer<Integer> consumer;
int count;
Counter(SynchronousQueue<Object> qin, SynchronousQueue<Object> qout,
int increment, int initial_count,
Consumer<Integer> consumer) {
this.qin = qin;
this.qout = qout;
this.increment = increment;
this.count = initial_count;
this.consumer = consumer;
}
public void run() {
try {
while (true) {
Object token = qin.take();
consumer.accept(count);
qout.put(token);
count += increment;
}
} catch (InterruptedException ex) {
throw new RuntimeException(ex);
}
}
}
}