有线程T1
、T2
和T3
,我们如何确保线程T2
在T1
之后运行,线程T3
在T2
之后运行?
这个问题是我面试中被问到的,我没有回答。请详细解释一下。
有线程T1
、T2
和T3
,我们如何确保线程T2
在T1
之后运行,线程T3
在T2
之后运行?
这个问题是我面试中被问到的,我没有回答。请详细解释一下。
final Thread t1 = new Thread(new T1()); // assume T1 is a Runnable
t1.start();
t1.join();
final Thread t2 = new Thread(new T2());
t2.start();
t2.join();
final Thread t3 = new Thread(new T3());
t3.start();
t3.join();
java.util.concurrent.CyclicBarrier
类来完成。AtomicBoolean
类,我们仍需要synchronized
以便在其上wait()
和notify()
。begin()
方法在T2
和T3
中,这样我们就可以将锁隐藏在这些对象内部。final Object lock2 = new Object();
final Object lock3 = new Object();
boolean ready2;
boolean ready3;
...
public T1 implements Runnable {
public void run() {
...
synchronized (lock2) {
// notify the T2 class that it should start
ready2 = true;
lock2.notify();
}
}
}
...
public T2 implements Runnable {
public void run() {
// the while loop takes care of errant signals
synchronized (lock2) {
while (!ready2) {
lock2.wait();
}
}
...
// notify the T3 class that it should start
synchronized (lock3) {
ready3 = true;
lock3.notify();
}
}
}
...
public T3 implements Runnable {
public void run() {
// the while loop takes care of errant signals
synchronized (lock3) {
while (!ready3) {
lock3.wait();
}
}
...
}
}
T2
的 run()
方法中,你想要说的是 synchronized (lock2) {
而不是 synchronized (lock3) {
。 - Kuldeep JainAtomicBoolean
与Boolean
相比是否有优势? - LearnerBoolean
上进行同步,@John。这是一个常见问题。你可以有一个 final Object lock
和一个 boolean
,但没有太大的意义。此外,AtomicBoolean
允许您在不必同步的情况下测试锁定。 - Graypublic class Test1 {
public static void main(String[] args) {
final Thread t1 = new Thread(new Runnable() {
public void run() {
System.out.println("start 1");
System.out.println("end 1");
}//run
});
final Thread t2 = new Thread(new Runnable() {
public void run() {
System.out.println(" start 2 ");
try {
t1.join(2000);
} catch (Exception e) {
e.getStackTrace();
}
System.out.println(" end 2");
}
}) ;
final Thread t3 = new Thread( new Runnable() {
public void run() {
System.out.println(" start 3 ");
try {
t2.join(4000);
}catch(Exception e) {
e.getStackTrace();
}
System.out.println(" end 3 ");
}
});
// we are reversing the order of the start() method
t3.start();
t2.start();
t1.start();
}
}
从输出结果可以看出,由于我们不知道哪个线程会获得CPU,所以线程的启动顺序是不确定的。这是线程调度器的决定,我们无法干涉。但是,你可以看到线程的结束顺序是正确的,即T1、T2和T3。
还有一种方法可以实现相同的效果,伪代码如下:
t1.start();
t1.join(); // signals t2 to wait
if( !t1.isAlive()) {
t2.start();// if t1 is finished then t2 will start
}
t2.join();//signals t3 to wait
if (!t2.isAlive()) {
t3.start();
}
让我们来看一个完整的程序:
public class Tic implements Runnable{
public void run() {
try {
for (int i = 0; i < 2; i++) {
System.out.println("tic");
}
} catch (Exception e) {
// TODO: handle exception
e.getStackTrace();
}
}
}
public class Tac implements Runnable{
public void run() {
try {
for (int i = 0; i < 2; i++) {
System.out.println("tac");
}
} catch (Exception e) {
// TODO: handle exception
e.getStackTrace();
}
}
}
public class Toe implements Runnable{
public void run() {
try {
for (int i = 0; i < 2; i++) {
System.out.println("toe");
}
} catch (Exception e) {
// TODO: handle exception
e.getStackTrace();
}
}
}
public class RunThreads1 {
public static void main(String[] args) {
try {
Tic tic = new Tic();
Tac tac = new Tac();
Toe toe = new Toe();
Thread t1 = new Thread(tic);
Thread t2 = new Thread(tac);
Thread t3 = new Thread(toe);
t1.start();
t1.join(); // signals t2 to wait
if( !t1.isAlive()) {
t2.start();// if t1 is finished then t2 will start
}
t2.join();//signals t3 to wait
if (!t2.isAlive()) {
t3.start();
}
}catch(InterruptedException e) {
e.printStackTrace();
}
}
}
面试官问你的问题是三个线程按顺序完成工作。例如,如果一个线程打印1、4、5...第二个线程打印2、5、8和第三个线程打印3、6、9等,则输出应为1、2、3、4、5.....
第一个线程打印1并给第二个线程打印2的机会……以此类推。
我试着使用循环屏障来实现。一旦“one”打印了1,它会调用cb.wait并将机会交给two,当two运行时,它将以类似的方式调用three,并继续运行。请告诉我代码中是否存在任何错误。
import java.util.concurrent.BrokenBarrierException;
import java.util.concurrent.CyclicBarrier;
class one implements Runnable{
CyclicBarrier cb;
one(CyclicBarrier cb){this.cb=cb;}
public void run(){
int i=1;
while(true)
{
System.out.println(i);
try {
Thread.sleep(1000);
cb.await();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (BrokenBarrierException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
i=i+3;
}
}
}
class two implements Runnable{
CyclicBarrier cb;
int i=2;
two(CyclicBarrier cb){this.cb=cb;}
public void run(){
System.out.println(i);
try {
cb.await();
i=i+3;
Thread.sleep(1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (BrokenBarrierException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
public class oneTwoThree {
public static void main(String args[]){
Runnable threePrinter = new Runnable() {
int i=3;
public void run() {
System.out.println(i);
i=i+3;
}
};
CyclicBarrier bar2 =new CyclicBarrier(1,threePrinter);//, barrier1Action);
two twoPrinter =new two(bar2);
CyclicBarrier bar1 =new CyclicBarrier(1,twoPrinter);
Thread onePrinter=new Thread(new one(bar1));
onePrinter.start();
}
}
线程也是可运行的。您可以简单地按顺序运行它们:
t1.run();
t2.run();
t3.run();
这显然没有什么兴趣。
假设他们希望线程并行运行,一种解决方案是让每个线程启动下一个线程,因为 JMM保证:
对线程的start()调用发生在启动的线程中的任何操作之前。
NOTE: Assuming that it is not about scheduling the threads in the required order
public class State {
private int state ;
public State() {
this.state =3;
}
public synchronized int getState() {
return state;
}
public synchronized void setState(int state) {
this.state = state;
}
}
public class T1 implements Runnable {
State s;
public T1(State s) {
this.s =s;
}
@Override
public void run() {
int i =1;
while(i<50)
{
//System.out.println("s in t1 "+ s.getState());
while(s.getState() != 3)
{
synchronized(s)
{
try {
s.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
synchronized(s)
{
//if(s.getState() ==3)
if(s.getState()==3)
System.out.println("t1 "+i);
s.setState(1);
i = i +3 ;
s.notifyAll();
}
}
}
}
public class T2 implements Runnable {
State s;
public T2(State s) {
this.s =s;
}
@Override
public synchronized void run() {
int i =2;
while(i<50)
{
while(s.getState() != 1)
{
synchronized(s)
{
try {
s.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
synchronized(s)
{
//if(s.getState() ==3)
if(s.getState()==1)
System.out.println("t2 "+i);
s.setState(2);
i = i +3 ;
s.notifyAll();
}
}
}
}
public class T3 implements Runnable {
State s;
public T3(State s) {
this.s =s;
}
@Override
public synchronized void run() {
int i =3;
while(i<50)
{
while(s.getState() != 2)
{
synchronized(s)
{
try {
s.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
synchronized(s)
{
if(s.getState()==2)
System.out.println("t3 "+i);
i = i +3 ;
s.setState(3);
s.notifyAll();
}
}
}}
public class T1t2t3 {
public static void main(String[] args) {
State s = new State();
Thread t1 = new Thread(new T1(s));
Thread t2 = new Thread(new T2(s));
Thread t3 = new Thread(new T3(s));
t1.start();
t2.start();
t3.start();
}
}
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.LinkedBlockingQueue;
class Worker implements Runnable {
BlockingQueue<Integer> q = new LinkedBlockingQueue<>();
Worker next = null; // next worker in the chain
public void setNext(Worker t) {
this.next = t;
}
public void accept(int i) {
q.add(i);
}
@Override
public void run() {
while (true) {
int i;
try {
i = q.take(); // this blocks the queue to fill-up
System.out.println(Thread.currentThread().getName() + i);
if (next != null) {
next.accept(i + 1); // Pass the next number to the next worker
}
Thread.sleep(500); // Just sleep to notice the printing.
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
public class PrintNumbersSequentially {
public static void main(String[] as) {
Worker w1 = new Worker();
Worker w2 = new Worker();
Worker w3 = new Worker();
w1.setNext(w2);
w2.setNext(w3);
w3.setNext(w1);
new Thread(w1, "Thread-1: ").start();
new Thread(w2, "Thread-2: ").start();
new Thread(w3, "Thread-3: ").start();
//Till here all the threads have started, but no action takes place as the queue is not filled for any worker. So Just filling up one worker.
w1.accept(100);
}
}
我认为这可能会对你有所帮助。