.Net：访问常量时是否会调用静态构造函数？

发布于 2024-11-15 03:26:59 字数 420 浏览 3 评论 0原文

所以这就是我的想法......

public class MyClass
{
    public const string MyConstant = "MyConstantValue";

    private static MyClass DefaultInstance;

    static MyClass()
    {
         DefaultInstance = new MyClass();
    }
}

...

NotificationService.RegisterForNotification(MyClass.MyConstant, Callback);

这行得通还是我需要使用诸如 static readonly ~~property~~ 字段之类的东西来触发静态构造函数？

原文

So here is what I'm thinking...

public class MyClass
{
    public const string MyConstant = "MyConstantValue";

    private static MyClass DefaultInstance;

    static MyClass()
    {
         DefaultInstance = new MyClass();
    }
}

...

NotificationService.RegisterForNotification(MyClass.MyConstant, Callback);

Will this work or do I need to use something like a static readonly ~~property~~ field to trigger the static constructor?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

我一直都在从未离去 2024-11-22 03:26:59

使用常量不一定会导致成员访问，这会导致调用静态构造函数。编译器被允许（甚至鼓励）在编译时替换常量的值。

尽管 readonly 建议使用字段，而不是属性，但您建议的 static readonly 解决方法应该没问题。当属性没有 setter 时，它们是只读的，不涉及 readonly 关键字。

一个简单的例子：

class HasSConstructor
{
    internal const int Answer = 42;
    static HasSConstructor()
    {
        System.Console.WriteLine("static constructor running");
    }
}

public class Program
{
    public static void Main()
    {
        System.Console.WriteLine("The answer is " + HasSConstructor.Answer.ToString());
    }
}

.NET 4.0下的输出：

答案是42

静态构造函数永远不会运行！

Use of a constant doesn't necessarily result in a member access which would cause the static constructor to be called. The compiler is allowed to (encouraged, even) substitute the value of the constant at compile time.

Your suggested workaround of static readonly should be ok, although readonly suggests a field, not a property. Properties are read-only when they have no setter, the readonly keyword isn't involved.

A simple example:

class HasSConstructor
{
    internal const int Answer = 42;
    static HasSConstructor()
    {
        System.Console.WriteLine("static constructor running");
    }
}

public class Program
{
    public static void Main()
    {
        System.Console.WriteLine("The answer is " + HasSConstructor.Answer.ToString());
    }
}

Output under .NET 4.0:

The answer is 42

The static constructor never runs!

回复收藏 0 原文

驱逐舰岛风号 2024-11-22 03:26:59

在创建第一个实例或引用任何静态成员之前，会自动调用静态构造函数。请参阅此处：MSDN：静态构造函数

顺便说一句，常量字段本质上是静态的，但正如所指出的，它们可以（并且可能会）被值本身替换。

回复收藏 0 原文

匿名。 2024-11-22 03:26:59

如果您只是访问公共常量，则不会调用静态构造函数。例如，考虑这个类：

class Testo
{
    public const string MyValue = "Hello, world";
    public static readonly int xyzzy;

    static Testo()
    {
        Console.WriteLine("In static constructor");
        xyzzy = 27;
    }
}

现在，在另一个类中，执行：

Console.WriteLine(Testo.MyValue);
Console.WriteLine(Testo.xyzzy);

您的输出将是：

Hello, world
In static constructor
27

常量在编译时从类中提升。因此，Testo.MyValue 的值在运行时不会引用 Testo 类。在调用需要初始化的内容之前，不会调用静态构造函数。

因此，是的，如果您想确保调用构造函数，则需要访问类似 static readonly 的内容。

The static constructor will not be called if you're just accessing public constants. For example, consider this class:

class Testo
{
    public const string MyValue = "Hello, world";
    public static readonly int xyzzy;

    static Testo()
    {
        Console.WriteLine("In static constructor");
        xyzzy = 27;
    }
}

Now, in another class, execute:

Console.WriteLine(Testo.MyValue);
Console.WriteLine(Testo.xyzzy);

Your output will be:

Hello, world
In static constructor
27

Constants are hoisted from the class at compile time. So the value of Testo.MyValue does not reference the Testo class at runtime. The static constructor is not called until you call something that requires initialization.

So, yes, you need to access something like a static readonly if you want to ensure that the constructor is called.

回复收藏 0 原文