Lambda匿名函数作为参数 [英] LambdaAnonymous Function as a parameter

问题描述

我是 C# 的新手.只是玩弄它.不是为了真正的目的.

I'm a very new to C#. Just playing around with it. Not for a real purpose.

void makeOutput( int _param)
{
    Console.WriteLine( _param.ToString());
}

//... 
// Somewhere in a code
{
    makeOutput(     /* some not c# code for an example for what do I want */ function : int () { return 0; }     );
}

是否可以使用真正的匿名函数(意味着返回结果)?

Is it possible to use a REAL anonymous functions (means returning result)?

我不想使用诸如

// Somewhere in a code
{
    Func<int> x = () => { return 0; };

    makeOutput( x())
}

我也不想更改方法参数类型，例如

Also I DO NOT want to change method parameter type such as

void makeOutput( Func<int> _param)
{
}

这是很常见的决定.

一切都好.我只是明白我想要不可能的事情.我想声明匿名函数并在同一个地方执行它.注意:直接声明和直接调用没有通用包装器.

Everything is alright. I just understood that I wanted impossible things. I wanted to declare anonymous function and execute it in the same place. Note: DIRECT declaring and DIRECT call without generic wrapper.

// flash-like (as3) code    /// DOES NOT COMPILE
makeOutput(    (function : int(){ return 0; })()   );

推荐答案

是的.
它被称为委托.

Yes.
It's called a delegate.

委托是(或多或少)普通类型；您可以像任何其他类型一样将它们传递给函数.

Delegates are (more-or-less) normal types; you can pass them to functions just like any other type.

void makeOutput(Func<int> param) {
    Console.WriteLine(param());
}

makeOutput(delegate { return 4; });
makeOutput(() => { return 4; });
makeOutput(() => 4);

<小时>

您的编辑问题没有意义.

C# 是类型安全的.
如果方法不希望函数作为参数，则不能将方法作为参数.

C# is type-safe.
If the method doesn't want a function as a parameter, you cannot give it a method as a parameter.

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