标签：etc 创建 解决方法 opera app new 初始化 tps int
一、多线程不安全方式实现
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public sealed class SingleInstance { private static SingleInstance instance; private SingleInstance() { } public static SingleInstance Instance { get { if (null == instance) { instance = new SingleInstance(); } return instance; } } } |
sealed表示SingleInstance不能被继承。其实构造函数私有化已经达到了这个效果，私有的构造函数不能被继承。为了可读性，可以加个sealed。
不安全的单例指的是在多线程环境下可能有多个线程同时进入if语句，创建了多次单例对象。
二、安全的单例模式
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public sealed class SingleInstance { private static volatile SingleInstance instance; private static readonly object obj = new object(); private SingleInstance() { } public static SingleInstance Instance { get { if (null == instance) { lock (obj) { if (null == instance) { instance = new SingleInstance(); } } } return instance; } } } |
加锁保护，在多线程下可以确保实例值被创建一次。缺点是每次获取单例，都要进行判断，涉及到的锁和解锁比较耗资源。
三、只读属性式
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public sealed class SingleInstance { private static readonly SingleInstance instance = new SingleInstance(); private SingleInstance() { } public static SingleInstance Instance { get { return instance; } } } |
借助readonly属性，instance只被初始化一次，同样达到了单例的效果。在Main函数执行第一句话之前，instance其实已经被赋值了，并不是预期的 只有到访问Instance变量时才创建对象。
如下代码：
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class Program { static void Main(string[] args) { Console.WriteLine("Begin"); var temp = SingleInstance.instance; ; } } public sealed class SingleInstance { public static readonly SingleInstance instance = new SingleInstance(); private SingleInstance() { Console.WriteLine("初始化初始化！"); } public static SingleInstance Instance { get { return instance; } } } |
输出：
在执行第一句代码之前，实例已经被初始化。
解决方法是在SingleInstance中加上静态构造函数。
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public sealed class SingleInstance { public static readonly SingleInstance instance = new SingleInstance(); static SingleInstance() { } private SingleInstance() { Console.WriteLine("初始化初始化！"); } public static SingleInstance Instance { get { return instance; } } } |
在运行输出：
四、使用Lazy
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public sealed class SingleInstance { private static readonly Lazy<SingleInstance> instance = new Lazy<SingleInstance>(() => new SingleInstance()); private SingleInstance(){} public static SingleInstance Instance { get { return instance.Value; } } } |
Lazy默认是线程安全的。MSDN描述如下：
Will the lazily initialized object be accessed from more than one thread? If so, the Lazy<T> object might create it on any thread. You can use one of the simple constructors whose default behavior is to create a thread-safe Lazy<T> object, so that only one instance of the lazily instantiated object is created no matter how many threads try to access it. To create a Lazy<T> object that is not thread safe, you must use a constructor that enables you to specify no thread safety.
五、泛型单例
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class Program{ static void Main(string[] args) { Console.WriteLine("Begin"); mySingle.Instance.age = 500; Console.WriteLine(mySingle.Instance.age); }}public abstract class SingleInstance<T>{ private static readonly Lazy<T> _instance = new Lazy<T>(() => { var ctors = typeof(T).GetConstructors(BindingFlags.Instance| BindingFlags.NonPublic| BindingFlags.Public); if (ctors.Count() != 1) throw new InvalidOperationException(String.Format("Type {0} must have exactly one constructor.", typeof(T))); var ctor = ctors.SingleOrDefault(c => c.GetParameters().Count() == 0 && c.IsPrivate); if (ctor == null) throw new InvalidOperationException(String.Format("The constructor for {0} must be private and take no parameters.", typeof(T))); return (T)ctor.Invoke(null); }); public static T Instance { get{ return _instance.Value;} }}public class mySingle : SingleInstance<mySingle>{ private mySingle() { } public int age;} |
标签：etc 创建 解决方法 opera app new 初始化 tps int
原文地址：https://www.cnblogs.com/asdyzh/p/9968858.html
