静态变量初始化

问题描述

今天，我与同事进行了讨论，并总结了以下几点.如果一切正确或需要修改，请稍加说明.

Today I had a discussion with my colleague and concluded following points. Kindly throw some light if all are correct or some modification is required.

1. 当未在类中定义静态构造函数时，将在使用静态字段之前对其进行初始化.
2. 在类中定义静态构造函数时，将在使用静态字段之前或在实例创建之前(在实例创建之前)初始化静态字段.
3. 如果在静态方法内未访问任何静态字段，则调用该静态方法.仅当在该类中定义了静态构造函数时，才会初始化静态字段.
4. 如果可能的话，应该避免在类中使用静态构造函数.

推荐答案

1.-3.您无法确切知道它何时发生，因此不能依赖它.静态构造函数可以让您稍微控制一下在调用它时会发生什么.

1.-3.You cannot exactly know when it happens and so you cannot depend on it. A static constructor will give you a little control what happens when it get called.

public class UtilityClass
{
  //
  // Resources
  //

  // r1 will be initialized by the static constructor
  static Resource1 r1 = null;

  // r2 will be initialized first, as static constructors are
  // invoked after the static variables are initialized
  static Resource2 r2 = new Resource2();

  static UtilityClass()
  {
    r1 = new Resource1();
  }

  static void f1(){}
  static void f2(){}
}

4.静态构造函数很慢

4.Static constructors are slow

静态构造函数执行的确切时间取决于实现，但是要遵循以下规则:

The exact timing of static constructor execution is implementation-dependent, but is subject to the following rules:

• 类的静态构造函数在的任何实例之前执行类已创建.
• 类的静态构造函数在任何静态之前执行该班的成员是
  引用.
• 类的静态构造函数在该类的静态字段初始化器(如果有)之后执行.
• 类的静态构造函数在执行期间最多执行一次单个程序实例.
• 两个之间的执行顺序两个的静态构造函数
  没有指定不同的类.

• The static constructor for a classexecutes before any instance of theclass is created.
• The static constructor for a classexecutes before any of the staticmembers for the class are
  referenced.
• The static constructor for a classexecutes after the static field initializers (if any) for the class.
• The static constructor for a classexecutes, at most, one time duringa single program instantiation.
• The order of execution between twostatic constructors of two
  different classes is not specified.

这篇关于静态变量初始化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！