java - 同一类的不同对象的同步方法

Question

这个问题已经有答案了:

Java synchronized method lock on object, or method? (11 个回答)

已关闭 6 年前。

我有课c具有3个同步功能m1 , m2 , m3 。我创建了同一类的 3 个不同实例 c1 , c2 , c3每个都在不同的线程上运行 t1 , t2 , t3分别。如果t1访问 m1可以t2 , t3访问 m1 ??

Answer 1

视情况而定。如果方法是静态的，那么它们在类对象上同步并且交错是不可能的。如果它们是非静态的，那么它们会在 this 对象上同步，并且可以进行交错。以下示例应阐明该行为。

import java.util.ArrayList;
import java.util.List;

class Main {
  public static void main(String... args) {
    List<Thread> threads = new ArrayList<Thread>();

    System.out.println("----- First Test, static method -----");
    for (int i = 0; i < 4; ++i) {
      threads.add(new Thread(() -> {
        Main.m1();
      }));
    }
    for (Thread t : threads) {
      t.start();
    }
    for (Thread t : threads) {
      try {
        t.join();
      } catch (InterruptedException e) {
        e.printStackTrace();
      }
    }

    System.out.println("----- Second Test, non-static method -----");
    threads.clear();
    for (int i = 0; i < 4; ++i) {
      threads.add(new Thread(() -> {
        new Main().m2();
      }));
    }
    for (Thread t : threads) {
      t.start();
    }
    for (Thread t : threads) {
      try {
        t.join();
      } catch (InterruptedException e) {
        e.printStackTrace();
      }
    }

    System.out.println("----- Third Test, non-static method, same object -----");
    threads.clear();
    final Main m = new Main();
    for (int i = 0; i < 4; ++i) {
      threads.add(new Thread(() -> {
        m.m2();
      }));
    }
    for (Thread t : threads) {
      t.start();
    }
    for (Thread t : threads) {
      try {
        t.join();
      } catch (InterruptedException e) {
        e.printStackTrace();
      }
    }
  }

  public static synchronized void m1() {
    System.out.println(Thread.currentThread() + ": starting.");
    try {
      Thread.sleep(1000);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }
    System.out.println(Thread.currentThread() + ": stopping.");
  }

  public synchronized void m2() {
    System.out.println(Thread.currentThread() + ": starting.");
    try {
      Thread.sleep(1000);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }
    System.out.println(Thread.currentThread() + ": stopping.");
  }
}

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