Singleton.Instance = Lazy.Value?

Can we say making a class Lazy is like making it Singleton ?

In both case we create the instance via a property and access to the same instance (if i开发者_运维问答t's created) in further usages.

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No, they are not the same.

Lazy initialisation of a variable only affects that variable, it doesn't make the class a singleton, or even reuse instances between variables. If you for example have two variables of the type Lazy<MyClass>, they would still create separate instances of the class.

A singleton usually uses lazy intialisation internally, but it doesn't have to. It could also be implemented using early initalisation, and just return the already created instance.

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Your question is not clear, do you want a class in the singleton pattern with lazy initialization?

public sealed class Singleton
{
   private Singleton() { }
   static Lazy<Singleton> instance = new Lazy<Singleton>(() => new Singleton());

   public static Singleton Instance
      {
          get
          {
              return instance.Value;
          }
      }
}

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No, you can have an object that is lazy but there can be multiple instances of the same object. Singleton would exist once and only once.

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// This is a lazy class - but you can have more than one
class Lazy
{
    private int? _value;
    public int Value {
        get
        {
           return _value ?? (_value = new Random().Next()).Value;
        }
    }
}

you can of course make Value static so that all Lazys have the same Value