- 浏览: 1591245 次
- 性别:
- 来自: 上海
文章分类
- 全部博客 (289)
- java 语法基础 (51)
- spring (8)
- mvc struct /Ant --build.xml (8)
- SOA (0)
- oracle 9i/10g (23)
- sql server 2000-2005 (3)
- 数据库基础知识 (6)
- 设计模式与软件架构 (10)
- Hibernate 持久化 (9)
- J2SE/J2EE/J2ME/AJAX 技术 (8)
- JSF 技术 (3)
- JAVA 图形化 (0)
- JMS (40)
- Eclipse 3.2 IDE 开发技巧 (13)
- 项目处理方法集合 (2)
- html/jsp/javascript (2)
- Unix/Linux (9)
- j2me/ARM/windriver/嵌入式 (4)
- 电信科学 (8)
- jsp (1)
- c/c++ (1)
- LZW压缩算法(java) (2)
- Android (77)
- 版本管理git/svn (2)
最新评论
-
huihai:
有demo吗?
NamingStrategy实现动态表名映射 -
cangbaotu:
推荐给大家一些有用的爬虫源码:https://github.c ...
网络爬虫(源代码参考) -
tuspark:
除了.classpath文件以外,.project文件也应该了 ...
Eclipse .classpath文件浅谈 -
tuspark:
造成eclipse自动关闭的原因有很多,这里有很多介绍:ecl ...
eclipse 自动关闭 解决方案 -
DEMONU:
网上都是这些,这种文章。。。
ActiveMQ中的消息持久性
All programming languages have their share of idioms. Many are useful to know and use, and programmers spend valuable time creating, learning, and implementing them. The problem is that some idioms are later proven not to be all that they were purported, or to simply not work as described. The Java programming language contains several useful programming idioms. It also contains some that further study has shown should not be used. Double-checked locking is one such idiom that should never be used. In this article, Peter Haggar examines the roots of the double-checked locking idiom, why it was developed, and why it doesn't work.
<script type="text/javascript"></script>
Editor's note: This article refers to the Java Memory Model before it was revised for Java 5.0; statements about memory ordering may no longer be correct. However, the double-checked locking idiom is still broken under the new memory model. For more information on the memory model in Java 5.0, see "Java theory and practice: Fixing the Java Memory Model" Part 1 and Part 2.
The Singleton creation pattern is a common programming idiom. When used with multiple threads, you must use some type of synchronization. In an effort to create more efficient code, Java programmers created the double-checked locking idiom to be used with the Singleton creation pattern to limit how much code is synchronized. However, due to some little-known details of the Java memory model, this double-checked locking idiom is not guaranteed to work. Instead of failing consistently, it will fail sporadically. In addition, the reasons for its failure are not obvious and involve intimate details of the Java memory model. These facts make a code failure due to double-checked locking very difficult to track down. In the remainder of this article, we'll examine the double-checked locking idiom in detail to understand just where it breaks down.
To understand where the double-checked locking idiom originated, you must understand the common singleton creation idiom, which is illustrated in Listing 1:
Listing 1. Singleton creation idiom
import java.util.*; class Singleton { private static Singleton instance; private Vector v; private boolean inUse; private Singleton() { v = new Vector(); v.addElement(new Object()); inUse = true; } public static Singleton getInstance() { if (instance == null) //1 instance = new Singleton(); //2 return instance; //3 } } |
The design of this class ensures that only one Singleton
object is ever created. The constructor is declared private
and the getInstance()
method creates only one object. This implementation is fine for a single-threaded program. However, when multiple threads are introduced, you must protect the getInstance()
method through synchronization. If the getInstance()
method is not protected, it is possible to return two different instances of the Singleton
object. Consider two threads calling the getInstance()
method concurrently and the following sequence of events:
- Thread 1 calls the
getInstance()
method and determines thatinstance
isnull
at //1. - Thread 1 enters the
if
block, but is preempted by thread 2 before executing the line at //2. - Thread 2 calls the
getInstance()
method and determines thatinstance
isnull
at //1. - Thread 2 enters the
if
block and creates a newSingleton
object and assigns the variableinstance
to this new object at //2. - Thread 2 returns the
Singleton
object reference at //3. - Thread 2 is preempted by thread 1.
- Thread 1 starts where it left off and executes line //2 which results in another
Singleton
object being created. - Thread 1 returns this object at //3.
The result is that the getInstance()
method created two Singleton
objects when it was supposed to create only one. This problem is corrected by synchronizing the getInstance()
method to allow only one thread to execute the code at a time, as shown in Listing 2:
Listing 2. Thread-safe getInstance() method
public static synchronized Singleton getInstance() { if (instance == null) //1 instance = new Singleton(); //2 return instance; //3 } |
The code in Listing 2 works fine for multithreaded access to the getInstance()
method. However, when you analyze it you realize that synchronization is required only for the first invocation of the method. Subsequent invocations do not require synchronization because the first invocation is the only invocation that executes the code at //2, which is the only line that requires synchronization. All other invocations determine that instance
is non-null
and return it. Multiple threads can safely execute concurrently on all invocations except the first. However, because the method is synchronized
, you pay the cost of synchronization for every invocation of the method, even though it is only required on the first invocation.
In an effort to make this method more efficient, an idiom called double-checked locking was created. The idea is to avoid the costly synchronization for all invocations of the method except the first. The cost of synchronization differs from JVM to JVM. In the early days, the cost could be quite high. As more advanced JVMs have emerged, the cost of synchronization has decreased, but there is still a performance penalty for entering and leaving a synchronized
method or block. Regardless of the advancements in JVM technology, programmers never want to waste processing time unnecessarily.
Because only line //2 in Listing 2 requires synchronization, we could just wrap it in a synchronized block, as shown in Listing 3:
Listing 3. The getInstance() method
public static Singleton getInstance() { if (instance == null) { synchronized(Singleton.class) { instance = new Singleton(); } } return instance; } |
The code in Listing 3 exhibits the same problem as demonstrated with multiple threads and Listing 1. Two threads can get inside of the if
statement concurrently when instance
is null
. Then, one thread enters the synchronized
block to initialize instance
, while the other is blocked. When the first thread exits the synchronized
block, the waiting thread enters and creates another Singleton
object. Note that when the second thread enters the synchronized
block, it does not check to see if instance
is non-null
.
|
To fix the problem in Listing 3, we need a second check of instance
. Thus, the name "double-checked locking." Applying the double-checked locking idiom to Listing 3 results in Listing 4.
Listing 4. Double-checked locking example
public static Singleton getInstance() { if (instance == null) { synchronized(Singleton.class) { //1 if (instance == null) //2 instance = new Singleton(); //3 } } return instance; } |
The theory behind double-checked locking is that the second check at //2 makes it impossible for two different Singleton
objects to be created as occurred in Listing 3. Consider the following sequence of events:
- Thread 1 enters the
getInstance()
method. - Thread 1 enters the
synchronized
block at //1 becauseinstance
isnull
. - Thread 1 is preempted by thread 2.
- Thread 2 enters the
getInstance()
method. - Thread 2 attempts to acquire the lock at //1 because
instance
is stillnull
. However, because thread 1 holds the lock, thread 2 blocks at //1. - Thread 2 is preempted by thread 1.
- Thread 1 executes and because instance is still
null
at //2, creates aSingleton
object and assigns its reference toinstance
. - Thread 1 exits the
synchronized
block and returns instance from thegetInstance()
method. - Thread 1 is preempted by thread 2.
- Thread 2 acquires the lock at //1 and checks to see if
instance
isnull
. - Because
instance
is non-null
, a secondSingleton
object is not created and the one created by thread 1 is returned.
The theory behind double-checked locking is perfect. Unfortunately, reality is entirely different. The problem with double-checked locking is that there is no guarantee it will work on single or multi-processor machines.
The issue of the failure of double-checked locking is not due to implementation bugs in JVMs but to the current Java platform memory model. The memory model allows what is known as "out-of-order writes" and is a prime reason why this idiom fails.
|
To illustrate the problem, you need to re-examine line //3 from Listing 4 above. This line of code creates a Singleton
object and initializes the variable instance
to refer to this object. The problem with this line of code is that the variable instance
can become non-null
before the body of the Singleton
constructor executes.
Huh? That statement might be contradictory to everything you thought possible, but it is, in fact, the case. Before explaining how this happens, accept this fact for a moment while examining how this breaks the double-checked locking idiom. Consider the following sequence of events with the code in Listing 4:
- Thread 1 enters the
getInstance()
method. - Thread 1 enters the
synchronized
block at //1 becauseinstance
isnull
. - Thread 1 proceeds to //3 and makes instance non-
null
, but before the constructor executes. - Thread 1 is preempted by thread 2.
- Thread 2 checks to see if instance is
null
. Because it is not, thread 2 returns theinstance
reference to a fully constructed, but partially initialized,Singleton
object. - Thread 2 is preempted by thread 1.
- Thread 1 completes the initialization of the
Singleton
object by running its constructor and returns a reference to it.
This sequence of events results in a period of time where thread 2 returned an object whose constructor had not executed.
To show how this occurs, consider the following pseudo code for the line: instance =new Singleton();
mem = allocate(); //Allocate memory for Singleton object. instance = mem; //Note that instance is now non-null, but //has not been initialized. ctorSingleton(instance); //Invoke constructor for Singleton passing //instance. |
This pseudo code is not only possible, but is exactly what happens on some JIT compilers. The order of execution is perceived to be out of order, but is allowed to happen given the current memory model. The fact that JIT compilers do just this makes the issues of double-checked locking more than simply an academic exercise.
To demonstrate this, consider the code in Listing 5. It contains a stripped-down version of the getInstance()
method. I've removed the "double-checkedness" to ease our review of the assembly code produced (Listing 6). We are interested only in seeing how the line instance=new Singleton();
is compiled by the JIT compiler. In addition, I've provided a simple constructor to make it clear when the constructor is run in the assembly code.
Listing 5. Singleton class to demonstrate out-of-order writes
class Singleton { private static Singleton instance; private boolean inUse; private int val; private Singleton() { inUse = true; val = 5; } public static Singleton getInstance() { if (instance == null) instance = new Singleton(); return instance; } } |
Listing 6 contains the assembly code produced by the Sun JDK 1.2.1 JIT compiler for the body of the getInstance()
method from Listing 5.
Listing 6. Assembly code produced from code in Listing 5
;asm code generated for getInstance 054D20B0 mov eax,[049388C8] ;load instance ref 054D20B5 test eax,eax ;test for null 054D20B7 jne 054D20D7 054D20B9 mov eax,14C0988h 054D20BE call 503EF8F0 ;allocate memory 054D20C3 mov [049388C8],eax ;store pointer in ;instance ref. instance ;non-null and ctor ;has not run 054D20C8 mov ecx,dword ptr [eax] 054D20CA mov dword ptr [ecx],1 ;inline ctor - inUse=true; 054D20D0 mov dword ptr [ecx+4],5 ;inline ctor - val=5; 054D20D7 mov ebx,dword ptr ds:[49388C8h] 054D20DD jmp 054D20B0 |
Note: To reference the lines of assembly code in the following explanation, I refer to the last two values of the instruction address because they all begin with 054D20
. For example, B5
refers to test eax,eax
.
The assembly code is produced by running a test program that calls the getInstance()
method in an infinite loop. While the program runs, run the Microsoft Visual C++ debugger and attach it to the Java process representing the test program. Then, break the execution and find the assembly code representing the infinite loop.
The first two lines of assembly code at B0
and B5
load the instance
reference from memory location 049388C8
into eax
and test for null
. This corresponds to the first line of the getInstance()
method in Listing 5. The first time this method is called, instance
is null
and the code proceeds to B9
. The code at BE
allocates the memory from the heap for the Singleton
object and stores a pointer to that memory in eax
. The next line, C3
, takes the pointer in eax
and stores it back into the instance reference at memory location 049388C8
. As a result, instance
is now non-null
and refers to a valid Singleton
object. However, the constructor for this object has not run yet, which is precisely the situation that breaks double-checked locking. Then at line C8
, the instance
pointer is dereferenced and stored in ecx
. Lines CA
and D0
represent the inline constructor storing the values true
and 5
into the Singleton
object. If this code is interrupted by another thread after executing line C3
but before completing the constructor, double-checked locking fails.
Not all JIT compilers generate the code as above. Some generate code such that instance
becomes non-null
only after the constructor executes. Both the IBM SDK for Java technology, version 1.3 and the Sun JDK 1.3 produce code such as this. However, this does not mean you should use double-checked locking in these instances. There are other reasons it could fail. In addition, you do not always know which JVMs your code will run on, and the JIT compiler could always change to generate code that breaks this idiom.
|
Double-checked locking: Take two
Given that the current double-checked locking code does not work, I've put together another version of the code, shown in Listing 7, to try to prevent the out-of-order write problem you just saw.
Listing 7. Attempting to solve the out-of-order write problem
public static Singleton getInstance() { if (instance == null) { synchronized(Singleton.class) { //1 Singleton inst = instance; //2 if (inst == null) { synchronized(Singleton.class) { //3 inst = new Singleton(); //4 } instance = inst; //5 } } } return instance; } |
Looking at the code in Listing 7 you should realize that things are getting a little ridiculous. Remember, double-checked locking was created as a way to avoid synchronizing the simple three-line getInstance()
method. The code in Listing 7 has gotten out of hand. In addition, the code does not fix the problem. Careful examination reveals why.
This code is trying to avoid the out-of-order write problem. It tries to do this by introducing the local variable inst
and a second synchronized
block. The theory works as follows:
- Thread 1 enters the
getInstance()
method. - Because
instance
isnull
, thread 1 enters the firstsynchronized
block at //1. - The local variable
inst
gets the value ofinstance
, which isnull
at //2. - Because
inst
isnull
, thread 1 enters the secondsynchronized
block at //3. - Thread 1 then begins to execute the code at //4, making
inst
non-null
but before the constructor forSingleton
executes. (This is the out-of-order write problem we just saw.) - Thread 1 is preempted by Thread 2.
- Thread 2 enters the
getInstance()
method. - Because
instance
isnull
, thread 2 attempts to enter the firstsynchronized
block at //1. Because thread 1 currently holds this lock, thread 2 blocks. - Thread 1 then completes its execution of //4.
- Thread 1 then assigns a fully constructed
Singleton
object to the variableinstance
at //5 and exits bothsynchronized
blocks. - Thread 1 returns
instance
. - Thread 2 then executes and assigns
instance
toinst
at //2. - Thread 2 sees that
instance
is non-null
, and returns it.
The key line here is //5. This line is supposed to ensure that instance
will only ever be null
or refer to a fully constructed Singleton
object. The problem occurs where theory and reality run orthogonal to one another.
The code in Listing 7 doesn't work because of the current definition of the memory model. The Java Language Specification (JLS) demands that code within a synchronized
block not be moved out of a synchronized
block. However, it does not say that code not in a synchronized
block cannot be moved into a synchronized
block.
A JIT compiler would see an optimization opportunity here. This optimization would remove the code at //4 and the code at //5, combine it and generate the code shown in Listing 8:
Listing 8. Optimized code from Listing 7
public static Singleton getInstance() { if (instance == null) { synchronized(Singleton.class) { //1 Singleton inst = instance; //2 if (inst == null) { synchronized(Singleton.class) { //3 //inst = new Singleton(); //4 instance = new Singleton(); } //instance = inst; //5 } } } return instance; } |
If this optimization takes place, you have the same out-of-order write problem we discussed earlier.
|
Another idea is to use the keyword volatile
for the variables inst
and instance
. According to the JLS (see Resources), variables declared volatile
are supposed to be sequentially consistent, and therefore, not reordered. But two problems occur with trying to use volatile
to fix the problem with double-checked locking:
- The problem here is not with sequential consistency. Code is being moved, not reordered.
- Many JVMs do not implement
volatile
correctly regarding sequential consistency anyway.
The second point is worth expanding upon. Consider the code in Listing 9:
Listing 9. Sequential consistency with volatile
class test { private volatile boolean stop = false; private volatile int num = 0; public void foo() { num = 100; //This can happen second stop = true; //This can happen first //... } public void bar() { if (stop) num += num; //num can == 0! } //... } |
According to the JLS, because stop
and num
are declared volatile
, they should be sequentially consistent. This means that if stop
is ever true
, num
must have been set to 100
. However, because many JVMs do not implement the sequential consistency feature of volatile
, you cannot count on this behavior. Therefore, if thread 1 called foo
and thread 2 called bar
concurrently, thread 1 might set stop
to true
before num
is set to 100
. This could lead thread 2 to see stop
as true
, but num
still set to 0
. There are additional problems with volatile
and the atomicity of 64-bit variables, but this is beyond the scope of this article. See Resources for more information on this topic.
|
The bottom line is that double-checked locking, in whatever form, should not be used because you cannot guarantee that it will work on any JVM implementation. JSR-133 is addressing issues regarding the memory model, however, double-checked locking will not be supported by the new memory model. Therefore, you have two options:
- Accept the synchronization of a
getInstance()
method as shown in Listing 2. - Forgo synchronization and use a
static
field.
Option 2 is shown in Listing 10:
Listing 10. Singleton implementation with static field
class Singleton { private Vector v; private boolean inUse; private static Singleton instance = new Singleton(); private Singleton() { v = new Vector(); inUse = true; //... } public static Singleton getInstance() { return instance; } } |
The code in Listing 10 does not use synchronization and ensures that the Singleton
object is not created until a call is made to the static getInstance()
method. This is a good alternative if your objective is to eliminate synchronization.
|
You might wonder about the String
class given the issue of out-of-order writes and a reference becoming non-null
prior to the constructor executing. Consider the following code:
private String str; //... str = new String("hello"); |
The String
class is supposed to be immutable. However, given the out-of-order write problem we discussed previously, could that cause a problem here? The answer is it could. Consider two threads with access to the String str
. One thread could see the str
reference refer to a String
object in which the constructor has not run. In fact, Listing 11 contains code that shows this occurring. Note that this code breaks only with older JVMs that I tested. Both the IBM 1.3 and Sun 1.3 JVMs produce immutable String
s as expected.
Listing 11. Example of a Mutable String
class StringCreator extends Thread { MutableString ms; public StringCreator(MutableString muts) { ms = muts; } public void run() { while(true) ms.str = new String("hello"); //1 } } class StringReader extends Thread { MutableString ms; public StringReader(MutableString muts) { ms = muts; } public void run() { while(true) { if (!(ms.str.equals("hello"))) //2 { System.out.println("String is not immutable!"); break; } } } } class MutableString { public String str; //3 public static void main(String args[]) { MutableString ms = new MutableString(); //4 new StringCreator(ms).start(); //5 new StringReader(ms).start(); //6 } } |
This code creates a MutableString
class at //4 that contains a String
reference shared by two threads at //3. Two objects are created, StringCreator
and StringReader
, on two separate threads at lines //5 and //6, passing a reference to the MutableString
object. The StringCreator
class enters an infinite loop and creates String
objects with the value "hello" at //1. The StringReader
also enters an infinite loop and checks to see if the current String
object has the value "hello" at //2. If it doesn't, the StringReader
thread prints out a message and stops. If the String
class is immutable, you should never see any output from this program. The only way for StringReader
to see the str
reference to be anything other than a String
object with "hello" as its value is if the problem of out-of-order writes occurs.
Running this code on old JVMs like Sun JDK 1.2.1 results in the out-of-order write problem, and thus, a non-immutable String
.
|
In an effort to avoid costly synchronization in singletons, programmers, quite ingeniously, invented the double-checked locking idiom. Unfortunately, it was not until this idiom was in fairly wide use that it became apparent that it is not a safe programming construct due to the current memory model. Work is underway to redefine areas of the memory model that are weak. However, even under the newly proposed memory model, double-checked locking will not work. The best solution to this problem is to accept synchronization or use a static field
.
- In his book Practical Java Programming Language Guide (Addison-Wesley, 2000), Peter Haggar covers a multitude of Java programming topics including an entire chapter on multithreading issues and programming techniques.
- The Java Language Specification, Second Edition by Bill Joy, et. al. (Addison-Wesley, 2000) is the definitive technical reference for the Java programming language.
- The Java Virtual Machine Specification, Second Edition by Tim Lindholm and Frank Yellin (Addison-Wesley, 1999) is the definitive document on Java compilers and runtime environments.
- Visit Bill Pugh's Java Memory Model Web site for a wealth of information on this important topic.
- For more information on
volatile
and 64-bit variables, see Peter Haggar's article "Does Java Guarantee Thread Safety?" in the June 2002 issue of Dr. Dobb's Journal . - JSR-133 deals with the revision to Java platform's memory model and thread specification.
- Java software consultant Brian Goetz examines when to use synchronization in "Threading lightly: Synchronization is not the enemy" (developerWorks, July 2001).
- In "Threading lightly: Sometimes it's best not to share" (developerWorks, October 2001), Brian Goetz examines
ThreadLocal
and offers tips for exploiting its power. - In "Writing multithreaded Java applications" (developerWorks, February 2001), Alex Roetter introduces the Java Thread API, outlines issues involved in multithreading, and offers solutions to common problems.
- Allen Holub proposes significant changes and additions to the Java language in "If I were king: A proposal for fixing the Java programming language's threading problems" (developerWorks, October 2000).
- Find other Java technology resources on the developerWorks Java technology zone.
|
Peter Haggar is a Senior Software Engineer with IBM in Research Triangle Park, North Carolina, and the author of the book Practical Java Programming Language Guide published by Addison-Wesley. In addition, he has published numerous articles on Java programming. He has a broad range of programming experience, having worked on development tools, class libraries, and operating systems. At IBM, Peter works on emerging Internet technology and is currently focused on high-performance Web services. Peter is a frequent technical speaker on Java technology at numerous industry conferences. He has worked for IBM for more than 14 years and received a B.S. in Computer Science from Clarkson University. Contact Peter at haggar@us.ibm.com. |
发表评论
-
Observer 观察者
2010-08-31 22:54 2025在Java中通过Observable类 ... -
设计模式之Command
2008-11-17 15:20 1038Command模式是最让我疑惑 ... -
设计模式之Flyweight(享元)
2008-11-11 13:37 1067模式定义: 避免大量拥有相同内容的小类的开销(如耗费内存),使 ... -
设计模式之Bridge(桥模式)
2008-11-11 10:41 2772Bridge模式定义 :将抽象 ... -
设计模式之Adapter(适配器)
2008-11-07 16:03 1220适配器模式定义:将两个不兼容的类纠合在一起使用,属于结构型模式 ... -
设计模式之Proxy(代理)
2008-11-06 17:25 1356理解并使用设计模式,能 ... -
设计模式之Factory
2008-11-06 17:18 1124厂模式定义:提供创建对象的接口. 为何使用?工厂模式是我们最常 ... -
简化DAO设计的思想
2008-09-08 14:22 2122当应用中需要使用到上十张表时,DAO的维护变得日益困难,主要表 ... -
从GOF的设计模式说起
2008-06-30 21:30 1562设计模式对于做软件者来说是必修的内功之一,尤其现在面向对象已经 ...
相关推荐
设计模式之 Singleton(单态/单件) 阎宏博士讲解:单例(Singleton)模式 保证一个类只有一个实例,并提供一个访问它的全局访问点 设计模式之 Factory(工厂方法和抽象工厂) 使用工厂模式就象使用 new 一样频繁. ...
3. 双重检查锁定(Double-Checked Locking,DCL): 这种方式结合了懒汉式和线程安全的优点,既延迟了实例的创建,又保证了线程安全。 ```java public class Singleton { private volatile static Singleton ...
考虑到线程安全,现代Java中通常采用双重检查锁定(Double-Checked Locking)或静态内部类等方式来实现线程安全的单例。 在"adv-java-factory-singleton"项目中,我们可以期待看到如何在Java中实现和应用这些模式的...
为了解决这一问题,可以采用双重检查锁定(Double-Checked Locking)模式来确保线程安全性: ```cpp class Singleton { public: static Singleton* Instance() { if (_instance == nullptr) { lock_guard<mutex>...
- 可以通过双重检查锁定(Double-Checked Locking)优化性能。 **线程安全问题及解决方案** - **问题**:在多线程环境中,若不采取措施,则可能会导致单例模式失效,即生成多个实例。 - **解决方案**: - 使用`...
3. **双检锁/双重校验锁定(Double-checked Locking)**: 这种方式在多线程环境下性能较好,只有在首次创建实例时才进行同步。 ```cpp class Singleton { private: Singleton() {} volatile static bool ...
懒汉式的一个改进是**双重检查锁定(Double-Checked Locking, DCL)**,它尝试减少不必要的同步开销: ```java public class Singleton { private volatile static Singleton instance = null; private Singleton...
3. 双重检查锁定(Double-Checked Locking) 这是一种在多线程环境下保证线程安全的懒汉式实现,它减少了同步的开销。 ```csharp public sealed class Singleton { private static volatile Singleton instance; ...
单例模式需要注意线程安全问题,如双重检查锁定(Double-Checked Locking)等实现方式。 - **抽象工厂(Abstract Factory)**:提供一个创建一系列相关或相互依赖对象的接口,而无需指定它们的具体类。 - **建造者...
在这个例子中,`Instance`属性使用了双重检查锁定(Double-Checked Locking)以提高性能。第一次检查是为了避免不必要的锁定,第二次检查是在多线程环境中确保线程安全地创建单例。 **Mutex** 除了单例模式,C#还...
双重检查锁定(Double-Checked Locking) 这是一种比较复杂的实现方式,旨在解决懒汉式的效率问题和线程安全问题。 ```java public class Singleton { private static volatile Singleton instance = null; ...
在C++中实现单例,需要注意线程安全问题,可能会使用到静态成员变量、双重检查锁定(Double-Checked Locking)以及C++11的std::call_once和std::once_flag等工具来确保正确性。 第5章可能会讲解“工厂模式”...
3. **双重检查锁定(Double-Checked Locking)**:在懒汉式的基础上优化,减少同步开销。 ```java public class Singleton { private volatile static Singleton instance; private Singleton() {} public static...
在Java中,可以使用双重检查锁定(Double-Checked Locking)、静态内部类或枚举等方式实现单例。 创建型模式还包括工厂方法、抽象工厂、建造者、原型和工厂模式。 结构型模式包括适配器、桥接、组合、装饰器、...
在Java中,通过私有构造器和静态方法来实现,线程安全的单例可以使用枚举或双重检查锁定(double-checked locking)来创建。例如: ```java public enum Singleton { INSTANCE; } // 或者使用双重检查锁定 public...
3. 双重检查锁定(Double-Checked Locking):既实现了延迟初始化,又保证了线程安全,是一种优化的懒汉式。 ```cpp class Singleton { public: static Singleton& getInstance() { if (nullptr == instance) {...
常见的单例实现方式有懒汉式(lazy initialization)、饿汉式(eager initialization)和双重检查锁定(double-checked locking)。这些实现方式各有优缺点,如懒汉式线程安全但可能导致多次初始化,饿汉式在类加载...
在Java中,通常使用双重检查锁定(double-checked locking)或静态内部类实现。 理解并应用这些设计原则,不仅可以提高代码质量,还能帮助你更好地理解和使用设计模式。例如,工厂模式和单例模式是Java中常见的设计...
在Java中,可以通过静态内部类、枚举或双检锁(Double-Checked Locking)等方式实现单例。 其次,工厂模式(Factory Pattern)是创建型设计模式,它提供了一种创建对象的最佳方式,通过一个公共接口来创建特定类型...
更高级的实现,如双重检查锁定(Double-Checked Locking, DCL),试图在提高性能的同时保证线程安全。然而,DCL在早期的Java版本中由于并发问题并不总是安全的,但在Java 5之后引入的内存模型改进已经解决了这个问题...