什么是“T @ this”意味着在委托声明中？ [英] What does "T @this" mean in a delegate declaration?

问题描述

我刚刚使用达斯汀坎贝尔的WeakEvent课程。尽管一味地使用在互联网上发现的代码和贸易;通常是一个糟糕的主意，它比我以前一起入侵的实现要好得多。它似乎迄今为止工作良好，但为了理解代码，我遇到了以下情况：

  public class WeakEventHandler< T，E> ：IWeakEventHandler< E> 
其中T：class 
其中E：EventArgs 
 {
私人委托void OpenEventHandler（T @ this，object sender，E e）; 
 ... 
  

我习惯使用<$ c来声明委托类型$ c> object sender 和 EventArgs args 参数，那么 T @this 部分实现？显然它声明了一些 WeakEventHandler 的 T 泛型类型，但我从未见过 之前（和谷歌搜索它是可以理解的无望）。

解决方案

> @this 表示您可以使用关键字 this 作为变量。

T 只是第一个开放的泛型类型 WeakEventHandler< T，E> 。

I've just added a weak event implementation to a project using Dustin Campbell's WeakEvent class. Although blindly using Code I Found On The Internet™ is generally a bad idea, it's a far better implementation than what I previously hacked together. It seems to work well so far, but in an effort to understand the code I came across the following:

public class WeakEventHandler<T, E> : IWeakEventHandler<E>
    where T : class
    where E : EventArgs
{
    private delegate void OpenEventHandler(T @this, object sender, E e);
    ...

I'm used to declaring delegates types with just the object sender and EventArgs args arguments, so what does the T @this part achieve? Obviously it is declaring something of WeakEventHandler's T generic type but I've never seen @this before (and googling it is understandably hopeless).

解决方案

The @this means you can use the keyword this as a variable.

The T is simply the first open generic type of WeakEventHandler<T, E>.

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