我正在尝试编写一个简单的代码来按顺序打印数字。场景就像
Thread Number
T1 1
T2 2
T3 3
T1 4
T2 5
T3 6
T1 7
T2 8
T3 9
...and so on.
这是
public class ThreadNumberPrinter {
Object monitor = new Object();
AtomicInteger number = new AtomicInteger(1);
public static void main(String[] args) {
ThreadNumberPrinter tnp = new ThreadNumberPrinter();
Thread t1 = new Thread(tnp.new Printer(1, 3));
Thread t2 = new Thread(tnp.new Printer(2, 3));
Thread t3 = new Thread(tnp.new Printer(3, 3));
t3.start();
t1.start();
t2.start();
}
class Printer implements Runnable {
int threadId;
int numOfThreads;
public Printer(int id, int nubOfThreads) {
threadId = id;
this.numOfThreads = nubOfThreads;
}
public void run() {
print();
}
private void print() {
try {
while (true) {
Thread.sleep(1000l);
synchronized (monitor) {
if (number.get() % numOfThreads != threadId) {
monitor.wait();
} else {
System.out.println("ThreadId [" + threadId
+ "] printing -->"
+ number.getAndIncrement());
monitor.notifyAll();
}
}
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
但是在第二个线程运行并打印数字2之后，所有线程都进入等待阶段并且没有任何内容被打印出来。我不确定我做错了什么。 任何帮助将不胜感激。
答案 0 :(得分：8)
嗯，问题是modulo 3 % 3是0。将您的threadId更改为0..2而不是1..3，并希望它可以正常工作。
答案 1 :(得分：3)
public class TestClass {
private volatile Integer count = 1;
private volatile Integer threadIdToRun = 1;
private Object object = new Object();
public static void main(String[] args) {
TestClass testClass = new TestClass();
Thread t1 = new Thread(testClass.new Printer(1));
Thread t2 = new Thread(testClass.new Printer(2));
Thread t3 = new Thread(testClass.new Printer(3));
t1.start();
t2.start();
t3.start();
}
class Printer implements Runnable {
private int threadId;
public Printer(int threadId) {
super();
this.threadId = threadId;
}
@Override
public void run() {
try {
while (count <= 20) {
synchronized (object) {
if (threadId != threadIdToRun) {
object.wait();
} else {
System.out.println("Thread " + threadId + " printed " + count);
count += 1;
if (threadId == 1)
threadIdToRun = 2;
else if (threadId == 2)
threadIdToRun = 3;
else if (threadId == 3)
threadIdToRun = 1;
object.notifyAll();
}
}
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
以上程序提供输出
Thread 1 printed 1
Thread 2 printed 2
Thread 3 printed 3
Thread 1 printed 4
Thread 2 printed 5
Thread 3 printed 6
Thread 1 printed 7
Thread 2 printed 8
Thread 3 printed 9
Thread 1 printed 10
Thread 2 printed 11
Thread 3 printed 12
Thread 1 printed 13
Thread 2 printed 14
Thread 3 printed 15
Thread 1 printed 16
Thread 2 printed 17
Thread 3 printed 18
Thread 1 printed 19
Thread 2 printed 20
答案 2 :(得分：1)
下面的代码使用通知下一个线程打印数字然后将其递增1然后再次通知下一个线程然后进入等待状态直到某个线程通知它的逻辑。 例如。 T1首先打印该值，然后生成布尔值＆＃34; second＆＃34;如果T2打印下一个数字，则为true。打印数字后T2为boolean＆＃34; third＆＃34;适用于T3。 T3通过制作boolean＆＃34; first＆＃34;来做同样的事情。如果T1打印下一个数字，则为true。
T1 - ＆gt; T2 - ＆gt; T3 - ＆gt; T1 - ＆gt; T2 - ＆gt; T3 - ＆gt; ........等等。
public class Test{
public static volatile int i = 0;
public static void main(String[] args) throws InterruptedException {
Object monitor = new Object();
Notifier notifier = new Notifier(monitor);
Thread thread1 = new Thread(notifier, "T1");
Thread thread2 = new Thread(notifier, "T2");
Thread thread3 = new Thread(notifier, "T3");
thread1.start();
thread2.start();
thread3.start();
}
}
class Notifier implements Runnable {
private Object monitor = null;
private static int i = 1;
private static boolean first = true;
private static boolean second = false;
private static boolean third = false;
public Notifier(Object objcurr) {
this.monitor = objcurr;
}
@Override
public void run() {
try {
while (true) {
synchronized (monitor) {
String Tname = Thread.currentThread().getName();
if (first && Tname.equalsIgnoreCase("T1")) {
print();
first = false;
second = true;
monitor.notifyAll();
monitor.wait();
} else if (second && Tname.equalsIgnoreCase("T2")) {
print();
second = false;
third = true;
monitor.notifyAll();
monitor.wait();
} else if (third && Tname.equalsIgnoreCase("T3")) {
print();
third = false;
first = true;
monitor.notifyAll();
monitor.wait();
} else {
monitor.wait();
}
}
Thread.sleep(1000);
}
} catch (Exception e) {
e.printStackTrace();
}
}
private void print() {
System.out.println(Thread.currentThread().getName() + " - " + Notifier.i++);
}
答案 3 :(得分：1)
虽然这是使用线程的一种不好的方法，但是如果我们仍然想要它，那么通用的解决方案就是拥有一个工作线程来存储它的id：
class Worker extends Thread {
private final ResourceLock resourceLock;
private final int threadNumber;
private final AtomicInteger counter;
private volatile boolean running = true;
public Worker(ResourceLock resourceLock, int threadNumber, AtomicInteger counter) {
this.resourceLock = resourceLock;
this.threadNumber = threadNumber;
this.counter = counter;
}
@Override
public void run() {
while (running) {
try {
synchronized (resourceLock) {
while (resourceLock.flag != threadNumber) {
resourceLock.wait();
}
System.out.println("Thread:" + threadNumber + " value: " + counter.incrementAndGet());
Thread.sleep(1000);
resourceLock.flag = (threadNumber + 1) % resourceLock.threadsCount;
resourceLock.notifyAll();
}
} catch (Exception e) {
System.out.println("Exception: " + e);
}
}
}
public void shutdown() {
running = false;
}
}
ResourceLock类将存储标志和最大线程数：
class ResourceLock {
public volatile int flag;
public final int threadsCount;
public ResourceLock(int threadsCount) {
this.flag = 0;
this.threadsCount = threadsCount;
}
}
然后主类可以使用它如下：
public static void main(String[] args) throws InterruptedException {
final int threadsCount = 3;
final ResourceLock lock = new ResourceLock(threadsCount);
Worker[] threads = new Worker[threadsCount];
final AtomicInteger counter = new AtomicInteger(0);
for(int i=0; i<threadsCount; i++) {
threads[i] = new Worker(lock, i, counter);
threads[i].start();
}
Thread.sleep(10000);
System.out.println("Will try to shutdown now...");
for(Worker worker: threads) {
worker.shutdown();
}
}
在经过一段时间的延迟之后，我们可能会停止计数，并且在工作人员中关闭方法会提供此规定。
答案 4 :(得分：0)
package com.sourav.mock.Thread;
import java.util.concurrent.atomic.AtomicInteger;
public class ThreeThreadComunication implements Runnable {
AtomicInteger counter;
int[] array;
static final Object mutex = new Object();
public ThreeThreadComunication(int[] array, AtomicInteger counter){
this.counter = counter;
this.array = array;
}
@Override
public void run() {
int i = 0;
while(i < array.length){
synchronized(mutex){
if(Integer.parseInt(Thread.currentThread().getName()) == counter.get()){
System.out.println(array[i]);
if(counter.get() == 3){
counter.getAndSet(1);
}else{
int c = counter.get();
counter.getAndSet(++c);
}
i++;
}
mutex.notifyAll();
try {
mutex.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
package com.sourav.mock.Thread;
import java.util.concurrent.atomic.AtomicInteger;
public class ThreeThreadComunicationTest {
public static void main(String[] args) {
AtomicInteger counter = new AtomicInteger(1);
int[] array1 = new int[]{1, 4, 7};
int[] array2 = new int[]{2, 5, 8};
int[] array3 = new int[]{3, 6, 9};
ThreeThreadComunication obj1 = new ThreeThreadComunication(array1, counter);
ThreeThreadComunication obj2 = new ThreeThreadComunication(array2, counter);
ThreeThreadComunication obj3 = new ThreeThreadComunication(array3, counter);
Thread t1 = new Thread(obj1, "1");
Thread t2 = new Thread(obj2, "2");
Thread t3 = new Thread(obj3, "3");
t1.start();
t2.start();
t3.start();
}
}
答案 5 :(得分：0)
public class EvenOdd1 {
//public static String str ="str1";
public static void main(String[] args) {
// TODO Auto-generated method stub
EvenOdd1 edd1 = new EvenOdd1();
AbThread tr2 = new AbThread(0,edd1);
AbThread tr3 = new AbThread(1,edd1);
AbThread tr4 = new AbThread(2,edd1);
tr2.start();
tr3.start();
tr4.start();
}
}
class AbThread extends Thread {
int mod;
int mod_count=1;
EvenOdd1 edd1;
public static int count=1;
int num_thread=3;
public AbThread(int mod,EvenOdd1 edd1){
this.mod = mod;
this.edd1 = edd1;
}
public void run()
{
synchronized(edd1)
{
try{
while(true){
while(count%num_thread!=mod)
edd1.wait();
if(count==30)
break;
print();
edd1.wait();
}
}
catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
public void print()
{
int val = mod==1?2*mod_count:(mod==2?3*mod_count:4*mod_count);
System.out.println(Thread.currentThread().getName() + " : " + val);
edd1.notifyAll();
count=count+1;
this.mod_count++ ;
}
}
答案 6 :(得分：0)
public class PrintThreadsSequentially {
static int number = 1;
static final int PRINT_NUMBERS_UPTO = 20;
static Object lock = new Object();
static class SequentialThread extends Thread {
int remainder = 0;
int noOfThreads = 0;
public SequentialThread(String name, int remainder, int noOfThreads) {
super(name);
this.remainder = remainder;
this.noOfThreads = noOfThreads;
}
@Override
public void run() {
while (number < PRINT_NUMBERS_UPTO) {
synchronized (lock) {
while (number % noOfThreads != remainder) { // wait for numbers other than remainder
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(getName() + " value " + number);
number++;
lock.notifyAll();
}
}
}
}
public static void main(String[] args) {
SequentialThread first = new SequentialThread("First Thread", 0, 4);
SequentialThread second = new SequentialThread("Second Thread", 1, 4);
SequentialThread third = new SequentialThread("Third Thread", 2, 4);
SequentialThread fourth = new SequentialThread("Fourth Thread", 3, 4);
first.start(); second.start(); third.start(); fourth.start();
}
}
答案 7 :(得分：0)
ThreadSynchronization类可用于在＆＃39; n＆＃39;之间打印数字。没有。顺序的线程。 逻辑是在每个连续线程之间创建一个公共对象，并使用“等待”，“通知”＃39;按顺序打印数字。 注意：最后一个线程将与第一个线程共享一个对象。
您可以更改＆＃39; maxThreads＆＃39;在运行之前增加或减少程序中线程数的值。
price_tiers_new答案 8 :(得分：0)
这个怎么样？
public class PrintNumbers implements Runnable {
public static final int NO_OF_THREADS = 3;
public static final int MAX_DIGIT = 20;
public static final String THREAD_NAME_PREFIX = "t";
volatile int current = 1;
private Object lock = new Object();
public static void main(String[] args) {
PrintNumbers printNumbers = new PrintNumbers();
for (int i = 1; i <= NO_OF_THREADS; i++) {
new Thread(printNumbers, THREAD_NAME_PREFIX + i).start();
}
}
@Override
public void run() {
String printThread;
while (current < MAX_DIGIT) {
synchronized (lock) {
if (current % NO_OF_THREADS != 0) {
printThread = THREAD_NAME_PREFIX + current % NO_OF_THREADS;
} else {
printThread = THREAD_NAME_PREFIX + NO_OF_THREADS;
}
if (!printThread.equals(Thread.currentThread().getName())) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
if (printThread.equals(Thread.currentThread().getName())) {
System.out.println(String.format("Thread %s : %s", Thread.currentThread().getName(), current));
current = current + 1;
}
lock.notifyAll();
}
}
}
}
答案 9 :(得分：0)
我已经尝试过使用三个线程按顺序打印的简单方法，并且效果很好。
public class AppPrint123 {
static int count = 1;
static int counter = 1;
static Object lock = new Object();
public static void main(String[] args) {
Thread t1 = new Thread(new Runnable() {
public void run() {
while (true) {
synchronized (lock) {
try {
Thread.sleep(100);
while (count != 1) {
lock.wait();
}
System.out.println(Thread.currentThread().getName() + ": " + counter);
count++;
counter++;
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.notifyAll();
}
}
}
}, "T1");
Thread t2 = new Thread(new Runnable() {
public void run() {
while (true) {
synchronized (lock) {
try {
Thread.sleep(100);
while (count != 2) {
lock.wait();
}
System.out.println(Thread.currentThread().getName() + ": " + counter);
counter++;
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.notifyAll();
}
}
}
}, "T2");
Thread t3 = new Thread(new Runnable() {
public void run() {
while (true) {
synchronized (lock) {
try {
Thread.sleep(100);
while (count != 3) {
lock.wait();
}
System.out.println(Thread.currentThread().getName() + ": " + counter);
count = count - 2;
counter++;
} catch (InterruptedException e) {
e.printStackTrace();
}
lock.notifyAll();
}
}
}
}, "T3");
t1.start();
t2.start();
t3.start();
}
}
如果要使用三个线程依次生成123123123之类的输出，则可以打印 count 变量。
答案 10 :(得分：0)
要做的不好的方法是使用多个线程来实现：
def my_constrain(x):
x = tf.abs(x)
return tf.clip_by_value(x, 0, 0.5)
a = tf.get_variable(name='a', initializer=0., constraint=lambda x: my_constrain(x))
以上，这些值为private static AtomicInteger currentThreadNo = new AtomicInteger(0);
private static int currentNo = 1;
private static final Object lock = new Object();
，因此对于所有辅助对象它们都保持相同。
static输出：
import java.util.concurrent.atomic.AtomicInteger;
public class PrintNumbersUsingNThreads implements Runnable {
private final int threadNo;
private final int totalThreads;
private static AtomicInteger currentThreadNo = new AtomicInteger(0);
private static int currentNo = 1;
private static final Object lock = new Object();
public PrintNumbersUsingNThreads(int threadNo, int totalThreads) {
this.threadNo = threadNo;
this.totalThreads = totalThreads;
}
@Override
public void run() {
while (true) {
while (currentThreadNo.get() % totalThreads != threadNo) {
try {
synchronized (lock) {
lock.wait();
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(Thread.currentThread().getName() + " printing " + currentNo);
currentNo++;
int curr = currentThreadNo.get();
if (curr == totalThreads) {
currentThreadNo.set(1);
} else {
currentThreadNo.incrementAndGet();
}
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
synchronized (lock) {
lock.notifyAll();
}
}
}
public static void main(String[] args) {
int totalThreads = 3;
for(int i = 0; i < totalThreads; i++){
new Thread(new PrintNumbersUsingNThreads(i,totalThreads),"thread"+i).start();
}
}
}
答案 11 :(得分：0)
下面是非常通用的代码。我同意在这种情况下使用多个线程不是一个好习惯
MultipleThreads类实现Runnable {
AtomicInteger integer;
int max_val = 100;
int remainder;
int numofThreads;
public MultipleThreads(AtomicInteger integer, int remainder, int numofThreads) {
this.integer = integer;
this.remainder = remainder;
this.numofThreads = numofThreads;
}
@Override
public void run() {
while (integer.intValue() <= max_val) {
synchronized (integer) {
while (integer.intValue() % numofThreads != remainder) {
try {
integer.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
if (integer.intValue() > max_val)
break;
System.out.println("printing :" + Thread.currentThread().getName() + " : " + integer.getAndIncrement());
integer.notifyAll();
}
}
}
}
公共类ThreadSynchronization {
public static void main(String[] args) {
AtomicInteger at = new AtomicInteger(1);
MultipleThreads th1 = new MultipleThreads(at, 1, 5);
MultipleThreads th2 = new MultipleThreads(at, 2, 5);
MultipleThreads th3 = new MultipleThreads(at, 3, 5);
MultipleThreads th4 = new MultipleThreads(at, 4, 5);
MultipleThreads th5 = new MultipleThreads(at, 0, 5);
new Thread(th1).start();
new Thread(th2).start();
new Thread(th3).start();
new Thread(th4).start();
new Thread(th5).start();
}
}
答案 12 :(得分：0)
另外，请确保使用 pthread_cond_broadcast 而不是 phread_cond_signal
#include "stdio.h"
#include "stdlib.h"
#include "pthread.h"
pthread_mutex_t count_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t condition_var = PTHREAD_COND_INITIALIZER;
void *functionCount1();
void *functionCount2();
void *functionCount3();
int count = 0;
#define COUNT_DONE 10
void main()
{
pthread_t thread1, thread2, thread3;
pthread_create( &thread1, NULL, &functionCount1, NULL);
pthread_create( &thread2, NULL, &functionCount2, NULL);
pthread_create( &thread3, NULL, &functionCount3, NULL);
pthread_join( thread1, NULL);
pthread_join( thread2, NULL);
pthread_join( thread3, NULL);
exit(0);
}
// Let me write what I think
// we can;t do he %2 %3 since multiple threads may satisfy the conditions.
// count = 0; count % 3 = 0
// count = 1; count % 3 = 1
// count = 2; count % 3 = 2
// count = 3; cooun % 3 = 0
// Print odd numbers
void *functionCount1()
{
for(;;) {
// Lock mutex and then wait for signal to relase mutex
pthread_mutex_lock( &count_mutex );
if ( count % 3 == 0 ) {
printf("Counter value functionCount1: %d\n",count);
count++;
pthread_cond_broadcast( &condition_var );
} else {
pthread_cond_wait( &condition_var, &count_mutex );
}
if ( count >= COUNT_DONE ) {
pthread_mutex_unlock( &count_mutex );
return(NULL);
}
pthread_mutex_unlock( &count_mutex );
}
}
// print even numbers
void *functionCount2()
{
for(;;) {
// Lock mutex and then wait for signal to relase mutex
pthread_mutex_lock( &count_mutex );
if ( count % 3 == 1 ) {
printf("Counter value functionCount2: %d\n",count);
count++;
pthread_cond_broadcast( &condition_var );
} else {
pthread_cond_wait( &condition_var, &count_mutex );
}
if( count >= COUNT_DONE ) {
pthread_mutex_unlock( &count_mutex );
return(NULL);
}
pthread_mutex_unlock( &count_mutex );
}
}
// print even numbers
void *functionCount3()
{
for(;;) {
// Lock mutex and then wait for signal to relase mutex
pthread_mutex_lock( &count_mutex );
if ( count % 3 == 2 ) {
printf("Counter value functionCount3: %d\n",count);
count++;
pthread_cond_broadcast( &condition_var );
} else {
pthread_cond_wait( &condition_var, &count_mutex );
}
if( count >= COUNT_DONE ) {
pthread_mutex_unlock( &count_mutex );
return(NULL);
}
pthread_mutex_unlock( &count_mutex );
}
}
答案 13 :(得分：0)
任意数量线程的通用解决方案-
对于 3 个线程逻辑 =MIN % 3 != 提醒
对于 4 个线程= MIN % 4 != 提醒
/< strong>代码 >***/
公共类 PrintSequenceRunnable1 实现 Runnable {
def save(self, *args, **kwargs):
self.run_validators()
super().save(*args, **kwargs)
def run_validators(self) -> None:
for field_name, field_value in model_to_dict(self).items():
model_field = getattr(UserSearchHistory, field_name)
field = getattr(model_field, 'field', object())
validators = getattr(field, 'validators', list())
for validator_func in validators:
if field_value is not None:
validator_func(field_value)
}
/**************************************************** ******************/
答案 14 :(得分：-1)
public class ThreadTask implements Runnable {
private int counter;
private int threadID;
private final Object lock;
private int prev;
public ThreadTask(Object obj, int threadid, int counter){
this.lock = obj; // monitor
this.threadID = threadid; //id of thread
this.counter = counter;
this.prev =threadid + 1;
}
public void run(){
while(counter<100){
synchronized(lock){
if(counter == this.prev && this.threadID % 3 == this.threadID){
System.out.println("T" + this.threadID + " = " + this.prev);
this.prev = this.prev + 3;
}
counter++;
lock.notifyAll();
try{
lock.wait();
}catch(Exception e){
e.printStackTrace();
}
}
}
}
}
public class ThreadMain {
static volatile int counter = 1;
public static void main(String args[]) throws InterruptedException{
final Object lock = new Object();
ThreadTask first = new ThreadTask(lock, 0, counter);
ThreadTask second = new ThreadTask(lock, 1, counter);
ThreadTask third = new ThreadTask(lock, 2, counter);
Thread t1 = new Thread(first, "first");
Thread t2 = new Thread(second, "second");
Thread t3 = new Thread(third, "third");
t1.start();
t2.start();
t3.start();
t1.join();
t2.join();
t3.join();
}
}
答案 15 :(得分：-1)
package ThreadCoreConcepts;
import java.util.ArrayList;
import java.util.List;
/**
* 3 Thread T1,T2,T3 will print output {1,2,3 4,5,6 7,8,9} Where T1 will print
* {1,4,7} , T2 will print { 2,5,8} and T3 will print {3,6,9}
*
* @author harsmahe
*
*/
public class ThreeThreadSequenceGen {
private volatile static int value = 1;
public static void main(String args[]) throws InterruptedException {
ThreeThreadSequenceGen gen = new ThreeThreadSequenceGen();
Object mutex = new Object();
Thread t1 = new Thread(gen.new RunThread(1, mutex));
t1.setName("1");
Thread t2 = new Thread(gen.new RunThread(2, mutex));
t2.setName("2");
Thread t3 = new Thread(gen.new RunThread(3, mutex));
t3.setName("3");
t1.start();
t2.start();
t3.start();
}
class RunThread implements Runnable {
private int start = 0;
private Object mutex;
private List<Integer> list = new ArrayList<Integer>();
public RunThread(final int start, Object mutex) {
// TODO Auto-generated constructor stub
this.start = start;
this.mutex = mutex;
}
@Override
public void run() {
try {
while (value <= 9) {
// while (true) {
// TODO Auto-generated method stub
int name = Integer.valueOf(Thread.currentThread().getName());
// System.out.println("[" + Thread.currentThread().getName()
// + "]");
// notifyAll();
synchronized (mutex) {
if (name == 1 && value == start) {
list.add(value);
System.out.println("[" + Thread.currentThread().getName() + "]" + value);
start = start + 3;
value++;
mutex.notifyAll();
mutex.wait();
} else if (name == 2 && value == start) {
System.out.println("[" + Thread.currentThread().getName() + "]" + value);
list.add(value);
start = start + 3;
value++;
mutex.notifyAll();
mutex.wait();
} else if (name == 3 && value == start) {
System.out.println("[" + Thread.currentThread().getName() + "]" + value);
list.add(value);
start = start + 3;
value++;
mutex.notifyAll();
if (value < 9) {
mutex.wait();
}
} else {
mutex.notifyAll();
// mutex.wait();
}
}
}
} catch (Exception e) {
e.printStackTrace();
} finally {
// System.out.println(list);
}
}
}
}
enter code here