C# 我该如何制作DoSomething？？？方法一种通用方法？

是否可以将

DoSomething

方法作为单一通用方法

我的想法是：

//I was thinking
//DoSomethingAny(c1.Method1, "an_id");

这是密码

using System;

public class Program
{
    static MyClass c1;
    public static void Main()
    {
        c1 = new MyClass();
        Console.WriteLine("Hello World");
        DoSomething1("Test1");      
        DoSomething2("Test2");
        DoSomething3("Test3");

        //I was thinking
        //DoSomethingAny(c1.Method1, "an_id");
    }

    static int DoSomething1(string id)
    {
        //much more code above, but identical in all methods
        var x = c1.Method1(id);
        //much more code below, but identical in all methods
        return x;
    }

    static int DoSomething2(string id)
    {
        var x = c1.Method2(id);
        return x;
    }

    static int DoSomething3(string id)
    {
        var x = c1.Method3(id);
        return x;
    }
}

public class MyClass
{
    public int Method1(string id)
    {
        Console.WriteLine("Method 1 Do Work");
        return 1;
    }

    public int Method2(string id)
    {
        Console.WriteLine("Method 2 Do Work");
        return 1;
    }

    public int Method3(string id)
    {
        Console.WriteLine("Method 3 Do Work");
        return 1;
    }
}

---

实际上，您可以使用委托作为参数进行重构，因此您只能编写一次

DoSomething

方法：

public static void Main()
{
  c1 = new MyClass();
  Console.WriteLine("Hello World");
  DoSomething("Test1", c1.Method1);
  DoSomething("Test2", c1.Method2);
  DoSomething("Test3", c1.Method3);
}

static int DoSomething(string id, Func<string, int> action)
{
  //much more code above, but identical in all methods
  int x = action(id);
  //much more code below, but identical in all methods
  return x;
}

---

为什么要浪费时间和混乱的堆积如山的回答，当问题已经是如此明显的许多现有的问题？提供帮助和个性化。。。为了个性化，我的朋友。在这里，我不知道是否有一个相关的和改编的副本，为这个特定的问题。我没有搜索过，也不知道搜索什么…评论不是为了进行广泛的讨论；这段对话已经结束。

static T DoSomething<T>(string id, Func<string, T> action)
{
  //much more code above, but identical in all methods
  var x = action(id);
  //much more code below, but identical in all methods
  return x;
}