To master the common problems and solutions of Java thread status, specific code examples are required
In Java multi-thread programming, thread status is an important concept. Understanding and mastering thread status can not only help us better understand how multi-threading works, but also help us solve some common threading problems. This article will introduce several common thread status problems and their solutions, and provide corresponding code examples.
public class DeadlockExample { private static Object resource1 = new Object(); private static Object resource2 = new Object(); public static void main(String[] args) { // 线程1 Thread thread1 = new Thread(() -> { synchronized (resource1) { System.out.println("Thread 1: Holding resource 1"); try { Thread.sleep(100); } catch (InterruptedException e) { e.printStackTrace(); } synchronized (resource2) { System.out.println("Thread 1: Holding resource 1 and resource 2"); } } }); // 线程2 Thread thread2 = new Thread(() -> { synchronized (resource2) { System.out.println("Thread 2: Holding resource 2"); try { Thread.sleep(100); } catch (InterruptedException e) { e.printStackTrace(); } synchronized (resource1) { System.out.println("Thread 2: Holding resource 1 and resource 2"); } } }); // 启动线程 thread1.start(); thread2.start(); } }
In the above code, two threads hold resource1 and resource2 resources respectively, and try to obtain the resources held by each other at the same time. When running this program, a deadlock occurs, preventing the program from ending normally.
The way to solve the deadlock problem is to avoid waiting for resources in a loop. For example, you can obtain resources in a fixed order, or set a timeout mechanism to give up competing for resources.
public class ThreadUnsafeExample { private static int count = 0; public static void main(String[] args) { Thread thread1 = new Thread(() -> { for (int i = 0; i < 1000; i++) { count++; } }); Thread thread2 = new Thread(() -> { for (int i = 0; i < 1000; i++) { count++; } }); thread1.start(); thread2.start(); try { thread1.join(); thread2.join(); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println("Count: " + count); } }
In the above code, the two threads perform cumulative operations on the global variable count. Due to the uncertainty of the thread execution order, the final calculated count value is uncertain, which may lead to incorrect results.
The way to solve the problem of thread insecurity is to use synchronization mechanism, such as using the synchronized keyword or Lock lock to protect shared resources.
public class WaitNotifyExample { private static Object lock = new Object(); public static void main(String[] args) { Thread thread1 = new Thread(() -> { synchronized (lock) { try { System.out.println("Thread 1: Waiting for lock"); lock.wait(); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println("Thread 1: Resumed"); } }); Thread thread2 = new Thread(() -> { synchronized (lock) { System.out.println("Thread 2: Acquired lock"); try { Thread.sleep(2000); } catch (InterruptedException e) { e.printStackTrace(); } lock.notify(); System.out.println("Thread 2: Notified"); } }); thread1.start(); thread2.start(); } }
In the above code, thread 1 acquires the lock first and enters the waiting state until thread 2 calls the notify() method to wake up thread 1. When thread 1 is awakened, it will continue to execute the following code.
The way to solve the problem of thread waiting and waking up is to use the wait() and notify() methods in the Object class to cooperate between threads.
To sum up, mastering the common problems and solutions of Java thread status is crucial to understanding multi-threaded programming. Through the above code examples, we can better understand and apply thread status and avoid common threading problems in actual multi-threaded development.
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