4.java多线程交替打印A和B?
3.java多线程 更优雅的实现线程同步：交替打印A、B LockSupport实现
2.Java实现两个线程交替打印问题
1.A、B两个线程交替打印1 -- 100
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4.java多线程交替打印A和B?
关键就是打印A和B的方法，synchronized的对象是同一个，即持有的锁是同一个，都是print这个类的实例
然后又一个标志，当不是自己的时候，就释放自己的锁，并让线程进入等待状态，就是this.wait（）
当while执行完后，说明到自己的回合，打印字符，设置标志，然后唤醒锁为this的线程。
class Print{
boolean nowA=true;
synchronized void printA(){
while(!nowA){
try {
this.wait();
} catch (Exception e) {
}
}
nowA=false;
System.out.println(Thread.currentThread()+" "+"A");
this.notify();
}
synchronized void printB(){
while(nowA){
try {
this.wait();
} catch (Exception e) {
}
}
nowA=true;
System.out.println(Thread.currentThread()+" "+"B");
this.notify();
}
}
public class Test3 {
public static void main(String[] args){
Print print=new Print();
new Thread(new Runnable() {
@Override
public void run() {
for(int i=0;i<10;i++){
print.printA();
}
}
}).start();
new Thread(new Runnable() {
@Override
public void run() {
for(int i=0;i<10;i++){
print.printB();
}
}
}).start();;
}
}
3.java多线程 更优雅的实现线程同步：交替打印A、B LockSupport实现
一 问题概述
线程或者进程之间有两种关系 同步和互斥，我们通常实现同步方法是使用线程的等待唤醒机制，而等待唤醒机制的使用是建立在互斥的继承上的。
但是同步线程并不一定是必须要实现互斥的。比如一个线程打印A，一个线程打印B。这两个线程就没有互斥关系，但是提出这么个需求：交替打印A、B 。
我们一般的解决方案，往往要使用wait()/notify机制。
二 LockSupport 介绍
LockSupport作为一个工具类，主要学习它的方法。
park():在线程内调用，表示当前线程自我阻塞，直到获得许可证
park(线程变量)：让指定的线程获得许可证。
一看这两个方法的定义，显然可以利用这两个方法实现线程的顺序调用（同步）
三 两种思路实现交替打印A/B
等待唤醒机制：
/**
* @program: test
* @description: 交替打印A/B 等待唤醒机制
* @author:
* @create: 2019-07-22 14:28
*/
public class Test3 {
static class MyRun implements Runnable {
static int i = 0;
@Override
public synchronized void run() {
for (int j = 0; j < 10; j++) {
if(i%2==0)
System.out.println(Thread.currentThread().getName()+":A");
else
System.out.println(Thread.currentThread().getName()+":B");
i++;
this.notifyAll();
try {
if(i>=19)
Thread.sleep(10);
else
this.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
public static void main(String[] args) {
MyRun myRun = new MyRun();
Thread a = new Thread(myRun);
Thread b = new Thread(myRun);
a.start();
b.start();
}
}
java.util.concurrent.locks.LockSupport 实现
/**
* @program: test
* @description:交替打印A,B LockSupport实现
* @author:
* @create: 2019-07-22 14:03
*/
public class Test2 {
static Thread a=null;
static Thread b=null;
public static void main(String[] args) {
a= new Thread(new Runnable() {
@Override
public void run() {
for (int i = 0; i < 10; i++) {
LockSupport.park();
System.out.println(Thread.currentThread().getName()+":B");
LockSupport.unpark(b);
}
}
});
b=new Thread((new Runnable() {
@Override
public void run() {
for (int i = 0; i < 10; i++) {
System.out.println(Thread.currentThread().getName()+":A");
LockSupport.unpark(a);
LockSupport.park();
}
}
}));
a.start();
b.start();
}
}
2.Java实现两个线程交替打印问题
线程1负责打印a,b,c,d
线程2负责打印1,2,3,4；要求控制台中输出的内容为 a1b2c3d4
public class TestMain {
static final Object object = new Object();
public static void main(String[] args) throws InterruptedException {
new Thread(new Runnable() {
String a[] = {"a","b","c","d"};
@Override
public void run() {
for(int i=0;i< 4 ;i++){
synchronized (object){
System.out.println("线程a 开始执行");
object.notify();
try {
System.out.println("线程a 开始等待");
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("线程a 继续执行");
System.out.println(a[i]);
System.out.println("线程a 执行结束");
object.notify();
}
}
}
}).start();
new Thread(new Runnable() {
int a[] = {1,2,3,4};
@Override
public void run() {
for(int i=0;i<4;i++){
synchronized (object){
System.out.println("线程1 开始执行");
object.notify();
try {
System.out.println("线程1 开始等待");
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("线程1 继续执行");
System.out.println(a[i]);
System.out.println("线程1 执行结束");
}
}
}
}).start();
}
}
方案一：
public class Test100_02 {
private static Object object = new Object();
private static boolean isFlag = false;
public static void main(String[] args) {
Thread t1 = new Thread(() -> {
for (int i = 1; i <= 999; i+=2) {
synchronized (object) {
if (!isFlag) {
System.out.println("current:" + i);
isFlag = true;
try {
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
object.notify();
}
}
}
});
Thread t2 = new Thread(() -> {
for (int i = 2; i <= 1000; i+=2) {
synchronized (object) {
if (isFlag) {
isFlag = false;
System.out.println("current:" + i);
object.notify();
try {
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
});
t1.setPriority(Thread.MAX_PRIORITY);
t1.start();
t2.start();
}
}
方案二：
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class Test100_01 {
private static Lock lock = new ReentrantLock();
private static Condition condition1 = lock.newCondition();
private static Condition condition2 = lock.newCondition();
public static void main(String[] args) {
// thread 1
Thread t1 = new Thread(() -> {
for (int i = 1; i <= 999; i += 2) {
lock.lock();
try {
System.out.println("current:" + i);
condition1.await();
condition2.signal();
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.unlock();
}
});
// thread 2
Thread t2 = new Thread(() -> {
for (int i = 2; i <= 1000; i += 2) {
lock.lock();
try {
System.out.println("current:" + i);
condition1.signal();
condition2.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.unlock();
}
});
t1.setPriority(Thread.MAX_PRIORITY);
t1.start();
t2.start();
}
}
你的代码我运行了貌似会在中途卡住，
我改进了一下顺序
condition2.signal();
System.out.println("current:" + i);
condition1.await();
condition1.signal();
System.out.println("current:" + i);
condition2.await();
方案一：
public class Test100_02 {
private static Object object = new Object();
private static boolean isFlag = false;
public static void main(String[] args) {
Thread t1 = new Thread(() -> {
for (int i = 1; i <= 999; i+=2) {
synchronized (object) {
if (!isFlag) {
System.out.println("current:" + i);
isFlag = true;
try {
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
object.notify();
}
}
}
});
Thread t2 = new Thread(() -> {
for (int i = 2; i <= 1000; i+=2) {
synchronized (object) {
if (isFlag) {
isFlag = false;
System.out.println("current:" + i);
object.notify();
try {
object.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
});
t1.setPriority(Thread.MAX_PRIORITY);
t1.start();
t2.start();
}
}
方案二：
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class Test100_01 {
private static Lock lock = new ReentrantLock();
private static Condition condition1 = lock.newCondition();
private static Condition condition2 = lock.newCondition();
public static void main(String[] args) {
// thread 1
Thread t1 = new Thread(() -> {
for (int i = 1; i <= 999; i += 2) {
lock.lock();
try {
System.out.println("current:" + i);
condition1.await();
condition2.signal();
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.unlock();
}
});
// thread 2
Thread t2 = new Thread(() -> {
for (int i = 2; i <= 1000; i += 2) {
lock.lock();
try {
System.out.println("current:" + i);
condition1.signal();
condition2.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.unlock();
}
});
t1.setPriority(Thread.MAX_PRIORITY);
t1.start();
t2.start();
}
}
你的代码我运行了貌似会在中途卡住，
我改进了一下顺序
condition2.signal();
System.out.println("current:" + i);
condition1.await();
condition1.signal();
System.out.println("current:" + i);
condition2.await();