[英]Multi threaded java program to print even and odd numbers alternatively
我被要求在一次采訪中編寫一個雙線程的Java程序。 在這個程序中,一個線程應該打印偶數,另一個線程應該交替打印奇數。
樣本輸出:
線程1:1
線程2:2
線程1:3
線程2:4 ......依此類推
我寫了以下程序。 一類Task
,包含兩種分別打印偶數和奇數的方法。 從main方法,我創建了兩個線程來調用這兩個方法。 面試官讓我進一步改進,但我想不出任何進步。 有沒有更好的方法來編寫相同的程序?
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();
}
}
我認為這應該正常,非常簡單。
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);
}
}
}
}
}
嗯,還有很多選擇。 我可能會使用SynchronousQueue
(我不喜歡低級wait
/ notify
,而是嘗試使用更高級別的並發原語)。 printOdd
和printEven
也可以合並為單個方法,不需要額外的標志:
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();
}
}
這樣的短版本怎么樣:
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();
}
}
}
}
有一個技巧可以正確地啟動線程 - 因此需要一個額外的notify()
來啟動整個事情(而不是讓兩個進程都等待,或者要求主線程調用一個通知)並且還需要處理線程啟動的可能性,它是否正常工作並在第二個線程啟動之前調用notify :)
我的初步答案是無功能的。 編輯:
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();
}
}
}
有沒有更好的方法來編寫相同的程序?
嗯,問題是,編寫程序的唯一好方法是使用單個線程。 如果你想要一個程序按順序執行X,Y和Z,那么編寫一個執行X,然后是Y,然后是Z的過程。沒有比這更好的方法了。
這是我在與面試官討論線程的適當性之后我會寫的。
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);
}
}
}
}
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