Detailed explanation of Java multi-threading interview questions

Question 1

A thread is executing a synchronized method in an object. Can thread B execute a non-synchronized method in the same object at the same time?

Yes, the two threads require different resources to run and do not need to preempt.

Case 1,

package duoxiancheng2;

/**
 * @author yeqv
 * @program A2
 * @Classname Ms1
 * @Date 2022/2/7 19:08
 * @Email w16638771062@163.com
 */
public class Ms1 {
    //A线程正在执行一个对象中的同步方法，B线程是否可以同时执行同一个对象中的非同步方法？
    Object a = new Object();

    public static void main(String[] args) {
        var t = new Ms1();
        new Thread(() -> t.a1()).start();//A线程
        new Thread(() -> t.a2()).start();//B线程
    }

    void a1() {
        synchronized (a) {
            System.out.println("同步方法");
        }
    }

    void a2() {
        System.out.println("非同步方法");
    }
}

Run result:

Question 2

Same as above, can thread B be executed at the same time? Another synchronized method in the same object?

No, two threads need a common resource to execute. The common resource is synchronized and can only be occupied by one thread at the same time.

Case 2,

package duoxiancheng2;

import java.util.concurrent.TimeUnit;

/**
 * @author yeqv
 * @program A2
 * @Classname Ms2
 * @Date 2022/2/7 19:25
 * @Email w16638771062@163.com
 */
public class Ms2 {
    //同上，B线程是否可以同时执行同一个对象中的另一个同步方法？
    Object a = new Object();
    public static void main(String[] args) {
        var t = new Ms2();
        new Thread(() -> t.a1()).start();//A线程
        new Thread(() -> t.a2()).start();//B线程
    }
    void a1() {
        synchronized (a) {
            System.out.println("进入同步方法1");
            try {
                TimeUnit.SECONDS.sleep(10);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("同步方法1结束");
        }
    }
    void a2() {
        synchronized (a) {
            System.out.println("进入同步方法2");
            try {
                TimeUnit.SECONDS.sleep(10);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("同步方法2结束");

        }
    }
}

Running result:

Thread A runs first and occupies resources.

After thread A finishes running and releases resources, thread B can enter execution

Thread B completes execution

Question 3

Will the lock be released if the thread throws an exception?

Yes, the resource will be released immediately after the thread throws an exception.

Case 3,

package duoxiancheng2;

import java.util.concurrent.TimeUnit;

/**
 * @author yeqv
 * @program A2
 * @Classname Ms3
 * @Date 2022/2/7 19:41
 * @Email w16638771062@163.com
 */
public class Ms3 {
    //线程抛出异常会释放锁吗？
    Object a = new Object();

    public static void main(String[] args) {
        var t = new Ms3();
        new Thread(() -> t.a1()).start();//A线程
        new Thread(() -> t.a2()).start();//B线程
    }

    void a1() {
        int c = 3;
        int b;
        synchronized (a) {
            System.out.println("进入同步方法1");
            try {
                b = c / 0;
                System.out.println(b);
                TimeUnit.SECONDS.sleep(10);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("同步方法1结束");
        }
    }

    void a2() {
        synchronized (a) {
            System.out.println("进入同步方法2");
            try {
                TimeUnit.SECONDS.sleep(10);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("同步方法2结束");

        }
    }
}

Result: As soon as an exception occurs in the method, the resources will be released immediately. Thread two starts executing

Question four

Write a program to prove that the AtomicInteger class is more efficient than synchronized

synchronized is more efficient

Case 1

package duoxiancheng2;

import java.util.concurrent.atomic.AtomicInteger;

/**
 * @author yeqv
 * @program A2
 * @Classname Ms4
 * @Date 2022/2/7 20:04
 * @Email w16638771062@163.com
 */
public class Ms4 {

    AtomicInteger n = new AtomicInteger(10000);
    int num = 10000;

    public static void main(String[] args) {

        var t = new Ms4();
        new Thread(t::minus, "T1").start();
        new Thread(t::minus, "T2").start();
        new Thread(t::minus, "T3").start();
        new Thread(t::minus, "T4").start();
        new Thread(t::minus, "T5").start();
        new Thread(t::minus, "T6").start();
        new Thread(t::minus, "T7").start();
        new Thread(t::minus, "T8").start();

    }

    void minus() {
        var a = System.currentTimeMillis();
        while (true) {
           /* if (n.get() > 0) {
                n.decrementAndGet();
                System.out.printf("%s 售出一张票，剩余%d张票。 %n", Thread.currentThread().getName(), n.get());
            } else {
                break;
            }*/
            synchronized (this) {
                if (num > 0) {
                    num--;
                    System.out.printf("%s 售出一张票，剩余%d张票。 %n", Thread.currentThread().getName(), num);
                } else {
                    break;
                }


            }


        }
        var b = System.currentTimeMillis();
        System.out.println(b - a);
    }
}

synchronized result:

AtomicInteger result:

Question 5

Write a program to prove that multiple methods of the AtomXXX class do not constitute atomicity

package demo16;

import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.atomic.AtomicInteger;

/**
 * 写一个程序证明AtomXXX类的多个方法并不构成原子性
 */
public class T {
    AtomicInteger count = new AtomicInteger(0);

    void m() {
        for (int i = 0; i < 10000; i++) {
            if (count.get() < 100 && count.get() >= 0) { //如果未加锁,之间还会有其他线程插进来
                count.incrementAndGet();
            }
        }
    }

    public static void main(String[] args) {
        T t = new T();
        List<Thread> threads = new ArrayList<>();
        for (int i = 0; i < 10; i++) {
            threads.add(new Thread(t::m, "thread" + i));
        }
        threads.forEach(Thread::start);
        threads.forEach((o) -> {
            try {
                //join()方法阻塞调用此方法的线程,直到线程t完成，此线程再继续。通常用于在main()主线程内，等待其它线程完成再结束main()主线程。
                o.join(); //相当于在main线程中同步o线程，o执行完了，main线程才有执行的机会
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        });
        System.out.println(t.count);
    }
}

Question 6

Write a program to start 100 threads in the main thread, 100 After the thread completes, the main thread prints "Done"

package cn.thread;

import java.util.concurrent.CountDownLatch;

/**
 * 写一个程序，在main线程中启动100个线程，100个线程完成后，主线程打印“完成”
 *
 * @author webrx [webrx@126.com]
 * @version 1.0
 * @since 16
 */
public class T12 {
    public static void main(String[] args) {
        CountDownLatch latch = new CountDownLatch(100);
        for (int i = 0; i < 100; i++) {
            new Thread(() -> {
                String tn = Thread.currentThread().getName();
                System.out.printf("%s : 开始执行...%n", tn);
                System.out.printf("%s : 执行完成，程序结束。%n", tn);
                latch.countDown();
            }, "T" + i).start();
        }

        try {
            latch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("---------------------------------------");
        System.out.println("100个线程执行完了。");
        String tn = Thread.currentThread().getName();
        System.out.printf("%s : 执行完成，程序结束。%n", tn);
    }
}

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