我有3个帖子 第一次印刷A. 第二次印刷B. 第三次印刷C
我想按顺序打印A B C A B C A B C等......
所以我写了下面的程序,但我无法实现同样的目标。 我知道当status = 1时,例如B1和C1线程正在等待,当我执行notifyAll()时,两个等待线程都被唤醒,并且根据CPU分配,它可能会打印B或C.
在这种情况下,我只希望在A之后打印B.
我需要做哪些修改。
public class NotifyAllExample {
int status=1;
public static void main(String[] args) {
NotifyAllExample notifyAllExample = new NotifyAllExample();
A1 a=new A1(notifyAllExample);
B1 b=new B1(notifyAllExample);
C1 c=new C1(notifyAllExample);
a.start();
b.start();
c.start();
}
}
class A1 extends Thread{
NotifyAllExample notifyAllExample;
A1(NotifyAllExample notifyAllExample){
this.notifyAllExample = notifyAllExample;
}
@Override
public void run() {
try{
synchronized (notifyAllExample) {
for (int i = 0; i < 100; i++) {
if(notifyAllExample.status!=1){
notifyAllExample.wait();
}
System.out.print("A ");
notifyAllExample.status = 2;
notifyAllExample.notifyAll();
}
}
}catch (Exception e) {
System.out.println("Exception 1 :"+e.getMessage());
}
}
}
class B1 extends Thread{
NotifyAllExample notifyAllExample;
B1(NotifyAllExample notifyAllExample){
this.notifyAllExample = notifyAllExample;
}
@Override
public void run() {
try{
synchronized (notifyAllExample) {
for (int i = 0; i < 100; i++) {
if(notifyAllExample.status!=2){
notifyAllExample.wait();
}
System.out.print("B ");
notifyAllExample.status = 3;
notifyAllExample.notifyAll();
}
}
}catch (Exception e) {
System.out.println("Exception 2 :"+e.getMessage());
}
}
}
class C1 extends Thread{
NotifyAllExample notifyAllExample;
C1(NotifyAllExample notifyAllExample){
this.notifyAllExample = notifyAllExample;
}
@Override
public void run() {
try{
synchronized (notifyAllExample) {
for (int i = 0; i < 100; i++) {
if(notifyAllExample.status!=3){
notifyAllExample.wait();
}
System.out.print("C ");
notifyAllExample.status = 1;
notifyAllExample.notifyAll();
}
}
}catch (Exception e) {
System.out.println("Exception 3 :"+e.getMessage());
}
}
}
答案 0 :(得分:8)
将这些IF语句转换为WHILE语句以获得所需的行为:
if (notifyAllExample.status != 2){
notifyAllExample.wait();
}
到
while (notifyAllExample.status != 2){
notifyAllExample.wait();
}
这将确保如果一个线程被通知,它将不会退出while循环,直到状态值达到预期值。
另外,将status
标记为volatile,以便线程不会有本地副本。
答案 1 :(得分:7)
public class RunThreadsInOrder implements Runnable {
static int numThread = 1;
static int threadAllowedToRun = 1;
int myThreadID;
private static Object myLock = new Object();
public RunThreadsInOrder() {
this.myThreadID = numThread++;
System.out.println("Thread ID:" + myThreadID);
}
@Override
public void run() {
synchronized (myLock) {
while (myThreadID != threadAllowedToRun) {
try {
myLock.wait();
} catch (InterruptedException e) {
} catch (Exception e) {}
}
try {
Thread.sleep(2000);
} catch (InterruptedException e) {
}
System.out.println("myThreadID is running: " + myThreadID);
myLock.notifyAll();
threadAllowedToRun++;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Thread t1 = new Thread(new RunThreadsInOrder());
Thread t2 = new Thread(new RunThreadsInOrder());
Thread t3 = new Thread(new RunThreadsInOrder());
Thread t4 = new Thread(new RunThreadsInOrder());
Thread t5 = new Thread(new RunThreadsInOrder());
Thread t6 = new Thread(new RunThreadsInOrder());
Thread t7 = new Thread(new RunThreadsInOrder());
t7.start();
t6.start();
t5.start();
t4.start();
t3.start();
t2.start();
t1.start();
}
}
答案 2 :(得分:1)
我被要求在一次访谈中写一个类似的程序,其中添加的条件是它应该是可扩展的,我们可以提供自己的线程数,并且他们应该打印字符,第一个线程打印'A'然后后续线程打印B,C,D等。我就是这样做的。
public class AlternateCharPrinter {
public static char ch = 65;
private static void createAndStartThreads(int count) {
Object lock = new Object();
for (int i = 0; i < count; i++) {
new Thread(new AlternateCharRunner((char) (65 + i), lock)).start();
}
}
public static void main(String[] args) {
createAndStartThreads(4);
}
}
class AlternateCharRunner implements Runnable {
private char ch;
private Object lock;
private static int runnerCount;
public AlternateCharRunner(char ch, Object lock) {
this.ch = ch;
this.lock = lock;
runnerCount++;
}
@Override
public void run() {
while (true) {
synchronized (lock) {
while (ch != AlternateCharPrinter.ch) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(AlternateCharPrinter.ch++);
if (AlternateCharPrinter.ch == (65 + runnerCount)) {
AlternateCharPrinter.ch = 65;
}
lock.notifyAll();
}
}
}
}
答案 3 :(得分:0)
您需要替换
if (notifyAllExample.status!=1)
与
while (notifyAllExample.status!=1)
和其他2个类一样。如果没有,那么一旦等待退出线程继续而不知道它是否轮到它。
答案 4 :(得分:0)
替换:
if(notifyAllExample.status!=1){
notifyAllExample.wait();
}
使用:
while(notifyAllExample.status!=1){
notifyAllExample.wait();
}
所有班级都相应。
答案 5 :(得分:0)
public class Main {
public static void main(String[] args) throws IOException{
Thread t1 = new Thread(new A(), "1");
Thread t2 = new Thread(new A(), "2");
Thread t3 = new Thread(new A(), "3");
t1.start();
try{
t1.join();
}catch (Exception e){
}
t2.start();
try{
t2.join();
}catch (Exception e){
}
t3.start();
try{
t3.join();
}catch (Exception e){
}
}
}
class A implements Runnable{
public void run(){
System.out.println(Thread.currentThread().getName());
}
}
或者您可以使用Executor Framework
public class Sequence {
int valve = 1;
public static void main(String[] args){
Sequence s = new Sequence();
ExecutorService es = Executors.newFixedThreadPool(3);
List<Runnable> rList = new ArrayList<>();
rList.add(new A(s));
rList.add(new B(s));
rList.add(new C(s));
for(int i = 0; i < rList.size(); i++){
es.submit(rList.get(i));
}
es.shutdown();
}
}
class A implements Runnable{
Sequence s;
A(Sequence s){
this.s = s;
}
public void run(){
synchronized (s) {
for (int i = 0; i < 10; i++) {
while (s.valve != 1) {
try {
s.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("A");
s.valve = 2;
s.notifyAll();
}
}
}
}
class B implements Runnable{
Sequence s;
B(Sequence s){
this.s = s;
}
public void run() {
synchronized (s) {
for (int i = 0; i < 10; i++) {
while (s.valve != 2) {
try {
s.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("B");
s.valve = 3;
s.notifyAll();
}
}
}
}
class C implements Runnable{
Sequence s;
C(Sequence s){
this.s = s;
}
public void run() {
synchronized (s) {
for(int i = 0; i < 10; i++) {
while (s.valve != 3) {
try {
s.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("C");
s.valve = 1;
s.notifyAll();
}
}
}
}
在第一种情况下,每个线程的连接会导致线程彼此等待。在第二种情况下,列表存储线程,执行程序一个接一个地执行它们,创建3个线程
另一种方法是只存在一个可运行的类,并且线程之间的通信是通过主类中的静态变量和runnable类中的变量完成的。
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.Executor;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
public class Seq {
int i = 1;
public static void main(String[] args){
Seq s = new Seq();
Common c1 = new Common(s, 1);
Common c2 = new Common(s, 2);
Common c3 = new Common(s, 3);
List<Runnable> l = new ArrayList<>();
l.add(c1);
l.add(c2);
l.add(c3);
ExecutorService es = Executors.newFixedThreadPool(3);
for(int i = 0; i < 3; i++){
es.submit(l.get(i));
}
es.shutdown();
}
}
class Common implements Runnable{
Seq s;
int o;
Common(Seq s, int o){
this.s = s;
this.o = o;
}
public void run(){
synchronized (s) {
for (int z = 0; z < 100; z++) {
if(s.i > 3)
s.i = 1;
while (s.i != o) {
try {
s.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(o);
s.i++;
s.notifyAll();
}
}
}
}
答案 6 :(得分:0)
解决这个问题的最简单的解决方案可以是以下方式:
public class PrintInOrder implements Runnable {
private int valueToPrint;
private int id;
private static int turn = 1;
private static int RESET_TURN_THRESHOLD = 3;
public PrintInOrder() {
this.valueToPrint = -1;
}
public PrintInOrder(int id, int val) {
this.id = id;
this.valueToPrint = val;
}
@Override
public void run() {
while(true) {
if (turn == this.id) {
System.out.println(Thread.currentThread().getName() + "::::" + valueToPrint);
turn++;
}
if (turn > RESET_TURN_THRESHOLD) {
turn = 1;
}
}
}
public static void main(String []args) {
Thread t1 = new Thread(new PrintInOrder(1, 1));
t1.setName("THREAD-1");
t1.start();
Thread t2 = new Thread(new PrintInOrder(2, 2));
t2.setName("THREAD-2");
t2.start();
Thread t3 = new Thread(new PrintInOrder(3, 3));
t3.setName("THREAD-3");
t3.start();
}
}
/*
OUTPUT::::
THREAD-1::::1
THREAD-2::::2
THREAD-3::::3
THREAD-1::::1
THREAD-2::::2
THREAD-3::::3
THREAD-1::::1
THREAD-2::::2
THREAD-3::::3
THREAD-1::::1
THREAD-2::::2
THREAD-3::::3
THREAD-1::::1
THREAD-2::::2
THREAD-3::::3
THREAD-1::::1
THREAD-2::::2
THREAD-3::::3
...
*/
答案 7 :(得分:0)
这是我的解决方法-
我创建了三个线程,每个线程都知道它需要打印什么以及打印之后的内容。
我还创建了一个NLock类,其中包含需要打印的下一个单词。
只要线程能够获取NLock锁,它就会检查 如果轮到他了,那么它会打印单词并设置要在NLock中打印的下一个值,否则会等到轮到他为止
public class SynchronizeThreeThreads {
public static void main(String args[]) throws InterruptedException {
NLock lock=new NLock("A");
Thread a =new Thread(new PrintInOrder("A","B",lock));
Thread b =new Thread(new PrintInOrder("B","C",lock));
Thread c =new Thread(new PrintInOrder("C","A",lock));
a.start();
b.start();
c.start();
c.join(); // Once all is done main thread will exit
System.out.println("Done");
}
}
class NLock{
private String value;
public NLock(String value) {
this.value=value;
}
public String getValue() {
return value;
}
public void setValue(String next) {
this.value=next;
}
}
class PrintInOrder implements Runnable{
private String word;
private String next;
private NLock lock;
public PrintInOrder(String word, String next,NLock lock){
this.word=word;
this.next=next;
this.lock=lock;
}
@Override
public void run() {
int i=0;
while(i<3) {
synchronized (lock) {
try {
//Check if it's my turn
if(lock.getValue().equals(word)) {
System.out.println(this.word);
//Set what next needs to be printed
//So that when that thread wakes up it knows that it's his turn
lock.setValue(next);
i++;
lock.notifyAll();
Thread.sleep(100);
}
else //Nope not my turn wait
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
下面是输出
A 乙 C 一种 乙 C 一种 乙 C 完成
答案 8 :(得分:-1)
我认为使用join实现这一点更简单。 例如:
public static void main(String[] args) {
final Thread t1 = new Thread("t1") {
@Override
public void run() {
System.out.println("i am thread: " + Thread.currentThread().getName());
}
};
final Thread t2 = new Thread(t1, "t2") {
@Override
public void run() {
t1.start();
try {
t1.join();
} catch ( InterruptedException e ) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("i am thread: " + Thread.currentThread().getName());
}
};
Thread t3 = new Thread(t2, "t3") {
@Override
public void run() {
t2.start();
try {
t2.join();
} catch ( InterruptedException e ) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("i am thread: " + Thread.currentThread().getName());
}
};
t3.start();
}
答案 9 :(得分:-1)
这是我的解决方案,请尝试让我知道
package thread;
class SyncPrinter {
public static void main(String[] args) {
SyncPrinterAction printAction1 = new SyncPrinterAction(new int[]{1,5,9,13}, true);
SyncPrinterAction printAction2 = new SyncPrinterAction(new int[]{2,6,10,14}, true);
SyncPrinterAction printAction3 = new SyncPrinterAction(new int[]{3,7,11,15}, true);
SyncPrinterAction printAction4 = new SyncPrinterAction(new int[]{4,8,12,16}, false);
printAction1.setDependentAction(printAction4);
printAction2.setDependentAction(printAction1);
printAction3.setDependentAction(printAction2);
printAction4.setDependentAction(printAction3);
new Thread(printAction1, "T1").start();;
new Thread(printAction2, "T2").start();
new Thread(printAction3, "T3").start();
new Thread(printAction4, "T4").start();
}
}
class SyncPrinterAction implements Runnable {
private volatile boolean dependent;
private SyncPrinterAction dependentAction;
int[] data;
public void setDependentAction(SyncPrinterAction dependentAction){
this.dependentAction = dependentAction;
}
public SyncPrinterAction( int[] data, boolean dependent) {
this.data = data;
this.dependent = dependent;
}
public SyncPrinterAction( int[] data, SyncPrinterAction dependentAction, boolean dependent) {
this.dependentAction = dependentAction;
this.data = data;
this.dependent = dependent;
}
@Override
public void run() {
synchronized (this) {
for (int value : data) {
try {
while(dependentAction.isDependent())
//System.out.println("\t\t"+Thread.currentThread().getName() + " :: Waithing for dependent action to complete");
wait(100);
} catch (InterruptedException e) {
e.printStackTrace();
}
dependentAction.setDependent(true);
System.out.println(Thread.currentThread().getName() + " :: " +value);
dependent = false;
}
}
}
private void setDependent(boolean dependent) {
this.dependent = dependent;
}
private boolean isDependent() {
return dependent;
}
}