This question already has an answer here: What does 'synchronized' mean? 15 answers If T1 gets a lock on method do() ie method is under synchronized block. and other portion of program say method display() is not synchronized then other threads can access this method. So your or is correct. Straight from the Java documentation: It is not possible for two invocations of synchron
这个问题在这里已经有了答案： “同步”是什么意思？ 15个答案 如果T1在方法do()上获得锁定，即方法在同步块之下。 和程序的其他部分say方法display()不同步，则其他线程可以访问此方法。 所以你的或者是正确的。 直接从Java文档： 同一对象上的同步方法的两次调用不可能交错。 当一个线程正在执行一个对象的同步方法时，所有其他线程调用同一对象的同步方法块（挂起执行），直到第一个线程完成对象。 所以你后面的
Possible Duplicate: What does 'synchronized' mean? What is the purpose of Java synchronization, and how should I use it? Java Tutorial: Synchronization. A generic answer to your generic question. Don't like it? Please elaborate a little. Cheers. If you mean how to implement synchronization (as in Java), it is language dependent. Generally it can implemented using the OS l
可能重复： “同步”是什么意思？ Java同步的目的是什么？我应该如何使用它？ Java教程：同步。 对通用问题的一般答案。 不喜欢它？ 请详细说明一下。 干杯。 如果你的意思是如何实现同步（如在Java中），它是语言相关的。 通常它可以使用操作系统级锁来实现（如信号量，互斥锁等）
This question already has an answer here: What does 'synchronized' mean? 15 answers Learning Java, use of synchronized keyword 2 answers
这个问题在这里已经有了答案： “同步”是什么意思？ 15个答案 学习Java，使用synchronized关键字2的答案
This question already has an answer here: What does 'synchronized' mean? 15 answers There is no synchronized keyword in C++. There is one in Java, though, where for methods it means the following two things: It is not possible for two invocations of synchronized methods on the same object to interleave. When one thread is executing a synchronized method for an object, all other
这个问题在这里已经有了答案： “同步”是什么意思？ 15个答案 C ++中没有synchronized关键字。 然而，Java中有一个方法意味着以下两件事： 同一对象上的同步方法的两次调用不可能交错。 当一个线程正在执行一个对象的同步方法时，所有其他线程调用同一对象的同步方法块（挂起执行），直到第一个线程完成对象。 当一个同步方法退出时，它会自动建立一个与之后调用同一个对象的同步方法的after-before关系。 这保证了
I'm learning about deadlocks in Java, and there's this sample code from Sun's official tutorial: Alphonse and Gaston are friends, and great believers in courtesy. A strict rule of courtesy is that when you bow to a friend, you must remain bowed until your friend has a chance to return the bow. Unfortunately, this rule does not account for the possibility that two friends might bow
我正在学习Java中的死锁问题，并且有Sun官方教程中的示例代码： 阿尔方斯和加斯顿是朋友，也是很好的礼节。 严格的礼貌规则是，当你向朋友鞠躬时，你必须保持鞠躬，直到你的朋友有机会归还弓。 不幸的是，这条规则没有考虑到两个朋友可能同时向对方低头的可能性。 public class Deadlock { static class Friend { private final String name; public Friend(String name) { this.name = nam
I have to take few actions depending upon the if condition. Say I have an enum "VoucherType" Now I have a code, that is executed depending upon the condition:- private boolean verifyGiveAwayAccounting(GiveAwayMoneyVerificationEvent event) { if(event.getVoucherType().equals(VoucherType.GIVE_AWAY_MONEY_ON_SIGNUP)){ someAction(); } return verifySystemAccountTransac
根据条件，我必须采取一些行动。 假设我有一个枚举“VoucherType” 现在我有一个代码，根据条件执行： - private boolean verifyGiveAwayAccounting(GiveAwayMoneyVerificationEvent event) { if(event.getVoucherType().equals(VoucherType.GIVE_AWAY_MONEY_ON_SIGNUP)){ someAction(); } return verifySystemAccountTransaction(event); } 如果事件类型为“GIVE_AWAY_MONEY_ON_SIGNUP”，我必须执行som
I have an enum with some states in it: enum State { A, B, C, D } and an object that has a corresponding state: class MyObject { State state; } I need to write an algorithm that takes two MyObject instances and does something depending on the particular states of those instances: void doWork(MyObject o1, MyObject o2) { if (o1.state == A && o2.state == A)
我有一个枚举与它的一些国家： enum State { A, B, C, D } 以及具有相应状态的对象： class MyObject { State state; } 我需要编写一个带有两个MyObject实例的算法，并根据这些实例的特定状态执行一些操作： void doWork(MyObject o1, MyObject o2) { if (o1.state == A && o2.state == A) { // do something } else if (o1.state == A && o2.state == B
How can I check equality on a value of an enum? I tried the following, but it fails. WHy? class enum Test { LETTER("1A"); private String value; public Test(String value) { this.value = value; } @Override public void toString() { return value; } } String teststring = "1A". Sysout(teststring.equals(Test.LETTER)); //false. Why??? The comparison is
我如何检查枚举值的平等？ 我尝试了以下，但它失败了。 为什么？ class enum Test { LETTER("1A"); private String value; public Test(String value) { this.value = value; } @Override public void toString() { return value; } } String teststring = "1A". Sysout(teststring.equals(Test.LETTER)); //false. Why??? 由于您正在比较String实例和Enum实例，所以比较返
I just got to read the following code somewhere : public class SingletonObjectDemo { private static SingletonObjectDemo singletonObject; // Note that the constructor is private private SingletonObjectDemo() { // Optional Code } public static SingletonObjectDemo getSingletonObject() { if (singletonObject == null) { singletonObject = new SingletonOb
我只是读了下面的代码： public class SingletonObjectDemo { private static SingletonObjectDemo singletonObject; // Note that the constructor is private private SingletonObjectDemo() { // Optional Code } public static SingletonObjectDemo getSingletonObject() { if (singletonObject == null) { singletonObject = new SingletonObjectDemo(); }
I've noticed that the following snippet... @Override public boolean equals(Object otherObject) { ... } ...is not allowed for an Enum, since the method equals(Object x) is defined as final in Enum . Why is this so? I cannot think of any use case which would require overriding equals(Object) for Enum. I'm just curious to know the reasoning behind this behavior. Anything but retu
我注意到下面的代码片段... @Override public boolean equals(Object otherObject) { ... } ...不允许Enum，因为方法equals(Object x)在Enum定义为final 。 这是为什么？ 我想不出任何需要重写Enum的equals(Object)用例。 我只是想知道这种行为背后的原因。 除了return this == other任何东西都是违反直觉的，并且违反最不惊讶的原则。 当且仅当它们是相同的对象并且覆盖这种行为的能力容易出错时，两个枚举常量预