动作<> VS事件行动 [英] Action<> Vs event Action

问题描述

是

public event Action delt = () => { Console.WriteLine("Information"); };

an overloaded version of

Action<int, int> delg = (a, b) => { Console.WriteLine( a + b); }; ?

我的意思是动作<？>委托是事件行动的重载版本

I mean Action<> delegate is an overloaded version of "event Action"?

推荐答案

这不是所谓的过剩。

基本上，有一组类型，声明如下：

Basically, there is a set of types, declared like this:

namespace System {
    delegate void Action();
    delegate void Action<T>(T a);
    delegate void Action<T1, T2>(T1 a1, T2 a2);
    ...
}

每个人是不同类型的，独立的所有其他。编译器知道你说的是哪类型当您尝试通过<的存在与否来引用它;> 类型名称后，和类型参数中的数字<>

Each of them is a different type, independent of all other. The compiler knows which type you mean when you try to reference it by the presence or absence of <> after the type name, and the number of type parameters within <>.

事件是一个完全不同的事情。 ，且不在此发挥任何部分。如果你按事件和委托之间的区别感到困惑，看到这两个问题：的 1 2

event is a different thing altogether, and doesn't play any part in this. If you're confused by the difference between event and delegate, see these two questions: 1 2

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