如何初始化静态属性是从基类派生的。 [英] How to initial static property is derived from base class.

问题描述

如何初始化静态属性是从基类派生的，如下面的代码。

How to initial static property is derived from base class as below code.

class A
      {
            protected static string keyName { get; set; }

            static A()
            {
                  Initial();
            }
            public static void Initial()
            {
               
                 string a = keyName;
                
            }
      }

class B:A
      {
         
            static B()
            {
                  keyName = "NewId";
            }

        
      }

我的尝试：



我想在初始函数中获取变量keyName的值。

What I have tried:

I want to get value of variable keyName in Initial function.

推荐答案

问题是你不能这样做的：在第一次使用B的实例之前，不会调用B的 static 构造函数。

所以如果是构造A的实例：

The problem is that you can't do that: the static constructor for B is not called until an instance of B is first used.
So if an instance of A is constructed:

A a = new A();

然后 keyName 属性的值永远不会被初始化（因此 null ）。

解决这个问题的一种方法是将 A 声明为 abstract ，这会阻止正在构造A的实例，但即使这样， A 的构造函数也会在 B 的构造函数之前调用（就像你一样）我希望， A 必须在 B 可以使用之前完成）这将是stil l表示 keyName 未在初始被调用之前初始化。



你不能做你想做的事：它是不可能的，因为类 A 构造函数必须在任何派生类构造函数代码可以执行 - 这同样适用于实例和静态构造函数。



你想要实现的是什么？需要这个？

Then the value of the keyName property is never initialized (and thus null).
One way to get round this - sort of - would be to declare the class A as abstract, which would prevent instances of A being constructed, but even then the constructor for A will be called before the constructor for B (as you would expect, A has to be "complete" before B can use it) which would still mean that keyName is not initialised before Initial is called.

You can't do what you want: it just isn't possible because the class A constructor has to be complete before any derived class constructor code can be executed - and that applies equally to instance and static constructors.

What are you trying to achieve that you think you need this?

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