生产者消费者问题
生产者消费者问题
单例模式、排序算法、生产者和消费者、死锁
Synchronized版本 wait notify
public class A {
public static void main(String[] args) {
Data data = new Data();
new Thread(()->{
for (int i = 0;i< 10;i++) {
try {
data.increment();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
},"A").start();
new Thread(()->{
for (int i = 0;i < 10;i++) {
try {
data.decrement();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
},"B").start();
}
}
//判断等待,业务,通知
class Data{
//资源类
private int number = 0;
//+1
public synchronized void increment() throws InterruptedException {
if (number !=0) { //0才往下执行
//等待
this.wait();
}
number ++;
System.out.println(Thread.currentThread().getName()+"=>"+number);
//通知其他线程,我+1完毕了
this.notifyAll();
}
//-1
public synchronized void decrement() throws InterruptedException {
if (number == 0) { //1才会自动往下执行
//等待
this.wait();
}
number --;
System.out.println(Thread.currentThread().getName()+"=>"+number);
//通知其他线程,我-1完毕了
this.notifyAll();
}
}
问题存在,A B C D 4个线程,虚假唤醒
if 改为 while 判断
JUC 版的生产者消费者问题
通过Lock找到Condition
- 代码实现
/**
* 线程时间的通信问题:生产者和消费者问题 等待唤醒,通知唤醒
* 线程交替执行 A B 操作同一个变量 num = 0
* A:num+1
* B:num-1
*/
public class B {
public static void main(String[] args) {
Data2 data = new Data2();
new Thread(()->{
for (int i = 0;i< 10;i++) {
try {
data.increment();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
},"A").start();
new Thread(()->{
for (int i = 0;i < 10;i++) {
try {
data.decrement();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
},"B").start();
new Thread(()->{
for (int i = 0;i < 10;i++) {
try {
data.increment();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
},"c").start();
new Thread(()->{
for (int i = 0;i < 10;i++) {
try {
data.decrement();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
},"D").start();
}
}
//判断等待,业务,通知
class Data2{
//资源类
private int number = 0;
Lock lock = new ReentrantLock();
Condition condition = lock.newCondition();
// condition.await(); //等待
// condition.signalAll(); //唤醒全部
//+1
public void increment() throws InterruptedException {
lock.lock();
try {
while (number !=0) { //0才往下执行
//等待
condition.await();
}
number ++;
System.out.println(Thread.currentThread().getName()+"=>"+number);
//通知其他线程,我+1完毕了
condition.signalAll();
} catch (Exception e) {
e.printStackTrace();
}finally {
lock.unlock();
}
}
//-1
public void decrement() throws InterruptedException {
lock.lock();
try {
while (number == 0) { //1才会自动往下执行
//等待
condition.await();
}
number --;
System.out.println(Thread.currentThread().getName()+"=>"+number);
//通知其他线程,我-1完毕了
condition.signalAll();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
}
任何一个新的技术,绝对不仅仅是覆盖了原来的技术,优势和补充
Condition 精准的通知和唤醒线程
- 代码测试
/**
* A 执行完调用B,B执行完调用C,C执行完调用A
*/
public class C {
public static void main(String[] args) {
Data3 data3 = new Data3();
new Thread(()->{
for (int i = 0; i < 10; i++) {
data3.printA();
}
},"A").start();
new Thread(()->{
for (int i = 0; i < 10; i++) {
data3.printB();
}
},"B").start();
new Thread(()->{
for (int i = 0; i < 10; i++) {
data3.printC();
}
},"C").start();
}
}
class Data3 {
//资源类
private Lock lock = new ReentrantLock();
Condition condition1 = lock.newCondition();
Condition condition2 = lock.newCondition();
Condition condition3 = lock.newCondition();
private int number = 1; //1A 2B 3C
public void printA() {
lock.lock();
try {
//业务代码,判断->执行->通知
while (number != 1) {
//等待
condition1.await();
}
System.out.println(Thread.currentThread().getName()+"=>AAAAAA");
//唤醒指定的人,B
number = 2;
condition2.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
public void printB() {
lock.lock();
try {
//业务代码,判断->执行->通知
while (number !=2) {
//等待
condition2.await();
}
System.out.println(Thread.currentThread().getName()+"=>BBBBBB");
//唤醒C
number = 3;
condition3.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
public void printC() {
lock.lock();
try {
//业务代码,判断->执行->通知
while (number != 3) {
condition3.await();
}
System.out.println(Thread.currentThread().getName()+"=>CCCCCC");
number = 1;
condition1.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
}
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