一、多线程不安全方式实现
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
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
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:
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。
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
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
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; }