我最近在接受采访时被问到这个问题。
编写一个带有两个线程（A和B）的程序，其中A打印1，B打印2，依此类推，直到达到50。
我们如何做到这一点？
答案 0 :(得分：9)
赋值的本质是演示线程如何发出另一个信号。最常见的方法是使用阻塞队列，但这里信号不携带任何信息，因此信号量就足够了。
创建用2个信号量参数化的线程类：输入和输出：
class ThreadPrinter implements Runnable {
int counter;
Semaphore ins, outs;
ThreadPrinter(int counter, Semaphore ins, Semaphore outs) {
this.counter = counter;
this.ins = ins;
this.outs = outs;
}
@Override
public void run() {
for (int i = 0; i < 25; i++) {
ins.aquire(); // wait for permission to run
System.out.println("" + counter);
outs.release(); // allow another thread to run
counter += 2;
}
}
创建2个Semaphore并将它们传递给2个主题：
Semaphore a = new Semaphore(1); // first thread is allowed to run immediately
Semaphore b = new Semaphore(0); // second thread has to wait
ThreadPrinter tp1 = new ThreadPrinter(1, a, b);
ThreadPrinter tp2 = new ThreadPrinter(2, b, a);
注意信号量a和b以不同的顺序传递。
答案 1 :(得分：3)
public class Test {
private static int count = 0;
public static void main(String[] args) throws InterruptedException {
Thread t1 = new Thread(new Runnable() {
@Override
public void run() {
for (int i = 0; i < 25; i++) {
synchronized (CommonUtil.mLock) {
incrementCount();
CommonUtil.mLock.notify();
try {
CommonUtil.mLock.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
});
Thread t2 = new Thread(new Runnable() {
@Override
public void run() {
for (int i = 0; i < 25; i++) {
synchronized (CommonUtil.mLock) {
incrementCount();
CommonUtil.mLock.notify();
try {
CommonUtil.mLock.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
});
t1.start();
Thread.sleep(400);
t2.start();
t1.join();
t2.join();
}
private static void incrementCount() {
count++;
System.out.println("Count: " + count + " icnremented by: " + Thread.currentThread().getName());
}
}
class CommonUtil {
static final Object mLock = new Object();
}
答案 2 :(得分：2)
我遇到了同样的问题，并且预计只使用基础知识，所以我选择在线程之间的共享对象上等待通知
public class Message implements Runnable {
private static final int N = 10;
private Thread thread;
private static Object object = new Object();
public Message(String name){
thread = new Thread(this, name);
thread.start();
}
public void run(){
for(int i=0; i<N; i++){
synchronized (object) {
System.out.println(i + "--" + thread.getName());
object.notify();
try {
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
主要方法：
Message message1 = new Message("Ping");
Message message2 = new Message("Pong");
答案 3 :(得分：1)
嗨，请在这里找到答案...模式ABABABAB
package com.abhi.ThreadPractice;
public class Test {
public static void main(String[] args) throws InterruptedException {
final Object lock = new Object();
Thread t1 = new Thread(new Runnable() {
@Override
public void run() {
for (int i = 0; i < 10; i++) {
synchronized (lock) {
// count++;
System.out.println("A");
try {
lock.wait();
lock.notify();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
});
Thread t2 = new Thread(new Runnable() {
@Override
public void run() {
for (int i = 0; i < 10; i++) {
synchronized (lock) {
lock.notify();
//count++;
System.out.println("B");
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
});
t1.start();
t2.start();
t1.join();
t2.join();
}
}
答案 4 :(得分：0)
这是另一种解决方案：
Thread t1 = new Thread(new Runnable() {
@Override
public void run() {
synchronized (lock) {
for (int i = 1; i <= 50; i += 2) {
System.out.println("T1=" + i);
t1turn = false;
try {
lock.notifyAll();
lock.wait();
} catch (InterruptedException e) {
}
}
}
}
});
Thread t2 = new Thread(new Runnable() {
@Override
public void run() {
synchronized (lock) {
for (int i = 2; i <= 50; i += 2) {
if (t1turn)
try {
lock.wait();
} catch (InterruptedException e) {
}
System.out.println("T2=" + i);
t1turn = true;
lock.notify();
}
}
}
});
t1.start();
t2.start();
答案 5 :(得分：0)
可能这仍然是相关的：
public class MyRunnable implements Runnable {
public static int counter = 0;
public static int turn = 0;
public static Object lock = new Object();
@Override
public void run() {
while (counter < 50) {
synchronized (lock) {
if (turn == 0) {
System.out.println(counter + " from thread "
+ Thread.currentThread().getName());
turn = 1;
try {
lock.wait();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
} else {
turn = 0;
lock.notify();
}
}
}
}
}
然后是主要功能
public static void main(String[] args) {
Thread threadA = new Thread(new MyRunnable());
Thread threadB = new Thread(new MyRunnable ());
threadA.start();
threadB.start();
}
答案 6 :(得分：0)
public class PingPong extends Thread {
static StringBuilder object = new StringBuilder("");
public static void main(String[] args) throws InterruptedException {
Thread t1 = new PingPong();
Thread t2 = new PingPong();
t1.setName("\nping");
t2.setName(" pong");
t1.start();
t2.start();
}
@Override
public void run() {
working();
}
void working() {
while (true) {
synchronized (object) {
try {
System.out.print(Thread.currentThread().getName());
object.notify();
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
答案 7 :(得分：0)
public class ThreadCounter implements Runnable {
private static int count = 0;
private Thread t;
public ThreadCounter(String tName){
t= new Thread(this, tName);
t.start();
}
@Override
public void run() {
for(int i=1; i<=5; i++){
synchronized (CommonUtil.mLock) {
incrementCount(t.getName());
CommonUtil.mLock.notify();
try {
CommonUtil.mLock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
private void incrementCount(String tName){
System.out.println(tName+": "+(++ThreadCounter.count));
}
public static void main(String[] args) throws InterruptedException {
new ThreadCounter("Thread1");
Thread.sleep(500);
new ThreadCounter("Thread2");
}
}
class CommonUtil{
public static Object mLock = new Object();
}
答案 8 :(得分：0)
这是最简单的解决方案，我能够想到。它使用synchronized方法并使用notify（）和wait（）来交替打印数字。希望能帮助到你。 :)
public class program implements Runnable
{
static int count =1;
private static final int MAX_COUNT = 50;
public synchronized void print ()
{
System.out.println(Thread.currentThread().getName() + " is printing " + count);
count++;
notify();
try{
if(count>MAX_COUNT)
return;
wait();
}catch (InterruptedException e){
e.printStackTrace();
}
}
public void run()
{
for(int i=0;i<MAX_COUNT/2;i++)
{
print();
}
}
public static void main(String[] args) {
program x= new program();
Thread t0= new Thread(x);
Thread t1= new Thread(x);
t0.start();
try
{
Thread.sleep(1);
} catch (InterruptedException e){
e.printStackTrace();
}
t1.start();
}
}
答案 9 :(得分：0)
//simply use wait and notify and and set a counter and it will do
public class ThreadalternatePrint implements Runnable {
static int counter =0;
@Override
public synchronized void run() {
try {
Thread.sleep(10);
} catch (InterruptedException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
while(counter<51)
{ ++counter;
notify();
System.out.println(Thread.currentThread().getName());
try {
wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
public static void main(String[] args) {
ThreadalternatePrint obj1 = new ThreadalternatePrint();
Thread Th1 = new Thread(obj1);
Thread Th2 = new Thread(obj1);
Th1.setName("Thread1");
Th2.setName("Thread2");
Th1.start();
Th2.start();
}
}
答案 10 :(得分：0)
public class Testing implements Runnable {
private static int counter = 1;
private static final Object lock = new Object();
public static void main(String[] args) {
Thread t1 = new Thread(new Testing(), "1");
t1.start();
Thread t2 = new Thread(new Testing(), "2");
t2.start();
}
@Override
public void run() {
while (counter<=100) {
synchronized (lock) {
if (counter % 2 == 0) {
System.out.println(counter +" Written By Thread-"+ Thread.currentThread().getName());
counter++;
try {
lock.notifyAll();
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
} else if (counter % 2 == 1) {
System.out.println(counter +" Written By Thread-"+ Thread.currentThread().getName());
counter++;
try {
lock.notifyAll();
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
}
答案 11 :(得分：0)
我使用可重入锁为它创建了一个非常基本的解决方案。
package com.multithreding.trylock;
import java.util.concurrent.locks.ReentrantLock;
public class TryLock extends Thread {
static int intitialCount = 50; //Value till which you want to print
int valueToSubtract = 0; //Value by which difference you want to print the series like 1,2,3
static ReentrantLock alternate = new ReentrantLock();
public TryLock(String name) {
this.setName(name);
}
public void run() {
while (intitialCount > 1) {
if (valueToSubtract > 0) {
alternate.lock();
intitialCount = intitialCount - valueToSubtract;
valueToSubtract = 0;
try {
Thread.sleep(200);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("value Subtracted " + intitialCount + " by the Thread" + this.getName());
alternate.unlock();
} else {
try {
Thread.sleep(100);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
valueToSubtract++;
}
}
}
}
package com.multithreding.trylock;
public class AlternatePrint {
public static void main(String[] args) throws InterruptedException{
//You can add as many thread to print then in different number of series
TryLock t1 = new TryLock("Odd One");
TryLock t2 = new TryLock("Even Value");
t1.start();
t2.start();
}
}
该解决方案也是模块化的，
您可以添加“ n”个线程来打印备用系列。即一次使用3个线程
您还可以打印“差值大于1”的系列，即1,3,5等
答案 12 :(得分：0)
package thread;
public class Pingpong extends Thread {
static StringBuilder object = new StringBuilder("");
static int i=1;
@Override
public void run() {
working();
}
void working() {
while (i<=10) {
synchronized (object) {
try {
System.out.println(Thread.currentThread().getName() +" "+ i);
i++;
object.notify();
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
public static void main(String[] args) throws InterruptedException {
Thread t1 = new Pingpong();
Thread t2 = new Pingpong();
t1.setName("Thread1");
t2.setName("Thread2");
t1.start();
t2.start();
}
}
Thread1 1
Thread2 2
Thread1 3
Thread2 4
Thread1 5
Thread2 6
Thread1 7
Thread2 8
Thread1 9
Thread2 10
答案 13 :(得分：-1)
我想这可能会有所帮助。 虽然它不是标准的，但我希望它提供了一种更简单的方法。
public class ThreadDemo
{
public static void main (String [] args)
{
PrintDemo pd=new PrintDemo();
MyThread1 mt1 = new MyThread1 ("T1",pd);
MyThread2 mt2 = new MyThread2 ("T2",pd);
mt1.start ();
mt2.start();
}
}
class PrintDemo {
private boolean oddFlag=true;
public synchronized void printOdd(int i,String tName){
if(oddFlag==false){
try {
wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}else{
System.out.println("\nThread "+tName+" count:"+i);
oddFlag=false;
notify();
}
}
public synchronized void printEven(int i,String tName){
if(oddFlag==true){
try {
wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}else{
System.out.println("\nThread "+tName+" count:"+i);
oddFlag=true;
notify();
}
}
}
class MyThread1 extends Thread
{
private PrintDemo pd;
private String name;
MyThread1(String threadName,PrintDemo pd){
this.name=threadName;
this.pd=pd;
}
public void run ()
{
for(int i=1;i<=50;i+=2){
pd.printOdd(i,name);
}
}
}
class MyThread2 extends Thread
{
private PrintDemo pd;
private String name;
MyThread2(String threadName,PrintDemo pd){
this.name=threadName;
this.pd=pd;
}
public void run ()
{
for(int i=2;i<=50;i+=2){
pd.printEven(i,name);
}
}
}