多線程交替打印ABC以及兩個線程交叉打印100以內奇偶數

一、兩個線程交叉打印100以內奇偶數，打印結束後線程正常結束。

實現方式一：直接使用Lock就可以。初始化線程時在線程中設置一個標識，0代表打印偶數，1代表打印奇數。
定義一個全局變量num代表要打印的數字，在線程中判斷當前要打印的數字是不是和本線程一直，即奇數=奇數線程，偶數對應偶數線程，如果一直則打印，num++，同時喚醒另外的線程；否則進行等待。

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
 
public class Test14 {
    public static void main(String[] args) throws Exception {
        Test14 t = new Test14();
        t.run();
 
    }
    Lock lock = new ReentrantLock();
    Condition condition = lock.newCondition();//獲取condition對象
    int num = 1;
 
    void run() throws Exception {
        new Thread(new PrintThread(0)).start();
        new Thread(new PrintThread(1)).start();
    }
 
    class PrintThread implements Runnable {
 
        int id;
        public PrintThread(int id) {
            this.id = id;
        }
        public void run() {
            while (num <= 100) {
                lock.lock();//獲取鎖，上鎖
                try {
                    if (num % 2 == id) {
                        System.out.println("id " + id + " - " + num);
                        num ++;
                        condition.signalAll();//喚醒所有等待線程。能夠從等待方法返回的線程必須獲得與Condition相關的鎖。
                    } else {
                        condition.await();//造成當前線程在接到信號或被中斷之前一直處於等待狀態。
                    }
                } catch (Exception e) {
                    e.printStackTrace();
                } finally {
                    lock.unlock();//解鎖
                }
            }
        }
    }
}

 

實現方式二：

 

二、3個線程交替打印ABC

1.通過ReentrantLock我們可以很方便的進行顯式的鎖操作，即獲取鎖和釋放鎖，對於同一個對象鎖而言，統一時刻只可能有一個線程拿到了這個鎖，此時其他線程通過lock.lock()來獲取對象鎖時都會被阻塞，直到這個線程通過lock.unlock()操作釋放這個鎖後，其他線程才能拿到這個鎖。
 

import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class ABC_Lock {
    private static Lock lock = new ReentrantLock();// 通過JDK5中的Lock鎖來保證線程的訪問的互斥
    private static int state = 0;//通過state的值來確定是否打印

    static class ThreadA extends Thread {
        @Override
        public void run() {
            for (int i = 0; i < 10;) {
                try {
                    lock.lock();
                    while (state % 3 == 0) {// 多線程併發，不能用if，必須用循環測試等待條件，避免虛假喚醒
                        System.out.print("A");
                        state++;
                        i++;
                    }
                } finally {
                    lock.unlock();// unlock()操作必須放在finally塊中
                }
            }
        }
    }

    static class ThreadB extends Thread {
        @Override
        public void run() {
            for (int i = 0; i < 10;) {
                try {
                    lock.lock();
                    while (state % 3 == 1) {// 多線程併發，不能用if，必須用循環測試等待條件，避免虛假喚醒
                        System.out.print("B");
                        state++;
                        i++;
                    }
                } finally {
                    lock.unlock();// unlock()操作必須放在finally塊中
                }
            }
        }
    }

    static class ThreadC extends Thread {
        @Override
        public void run() {
            for (int i = 0; i < 10;) {
                try {
                    lock.lock();
                    while (state % 3 == 2) {// 多線程併發，不能用if，必須用循環測試等待條件，避免虛假喚醒
                        System.out.print("C");
                        state++;
                        i++;
                    }
                } finally {
                    lock.unlock();// unlock()操作必須放在finally塊中
                }
            }
        }
    }

    public static void main(String[] args) {
        new ThreadA().start();
        new ThreadB().start();
        new ThreadC().start();
    }
}

2.ReentrantLock搭配的通行方式是Condition，如下：

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class ABC_Condition {
    private static Lock lock = new ReentrantLock();
    private static Condition A = lock.newCondition();
    private static Condition B = lock.newCondition();
    private static Condition C = lock.newCondition();

    private static int count = 0;

    static class ThreadA extends Thread {
        @Override
        public void run() {
            try {
                lock.lock();
                for (int i = 0; i < 10; i++) {
                    while (count % 3 != 0)//注意這裏是不等於0，也就是說在count % 3爲0之前，當前線程一直阻塞狀態
                        A.await(); // A釋放lock鎖
                    System.out.print("A");
                    count++;
                    B.signal(); // A執行完喚醒B線程
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                lock.unlock();
            }
        }
    }

    static class ThreadB extends Thread {
        @Override
        public void run() {
            try {
                lock.lock();
                for (int i = 0; i < 10; i++) {
                    while (count % 3 != 1)
                        B.await();// B釋放lock鎖，當前面A線程執行後會通過B.signal()喚醒該線程
                    System.out.print("B");
                    count++;
                    C.signal();// B執行完喚醒C線程
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                lock.unlock();
            }
        }
    }

    static class ThreadC extends Thread {
        @Override
        public void run() {
            try {
                lock.lock();
                for (int i = 0; i < 10; i++) {
                    while (count % 3 != 2)
                        C.await();// C釋放lock鎖
                    System.out.print("C");
                    count++;
                    A.signal();// C執行完喚醒A線程
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                lock.unlock();
            }
        }
    }

    public static void main(String[] args) throws InterruptedException {
        new ThreadA().start();
        new ThreadB().start();
        new ThreadC().start();
    }
}

 

參考：https://blog.csdn.net/qq_40664693/article/details/104916479

           https://blog.csdn.net/xiaokang123456kao/article/details/77331878