如何在.NET中使用GetMethod区分通用签名和非通用签名？

Question

假设存在如下所述的类X,如何获取非泛型方法的方法信息？下面的代码将引发异常.

using System;

class Program {
    static void Main(string[] args) {
        var mi = Type.GetType("X").GetMethod("Y"); // Ambiguous match found.
        Console.WriteLine(mi.ToString());
    }
}

class X {
    public void Y() {
        Console.WriteLine("I want this one");
    }
    public void Y<T>() {
        Console.WriteLine("Not this one");
    }
}

Answer 1

不要使用GetMethod,使用GetMethods,然后检查IsGenericMethod.

using System;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var mi = Type.GetType("X").GetMethods().Where(method => method.Name == "Y"); 
        Console.WriteLine(mi.First().Name + " generic? " + mi.First().IsGenericMethod);
        Console.WriteLine(mi.Last().Name + " generic? " + mi.Last().IsGenericMethod);
    }
}

class X
{
    public void Y()
    {
        Console.WriteLine("I want this one");
    }
    public void Y<T>()
    {
        Console.WriteLine("Not this one");
    }
}

作为奖励 - 扩展方法:

public static class TypeExtensions
{
    public static MethodInfo GetMethod(this Type type, string name, bool generic)
    {
        if (type == null)
        {
            throw new ArgumentNullException("type");
        }
        if (String.IsNullOrEmpty(name))
        {
            throw new ArgumentNullException("name");
        }
        return type.GetMethods()
            .FirstOrDefault(method => method.Name == name & method.IsGenericMethod == generic);
    }
}

然后就是:

static void Main(string[] args)
{
    MethodInfo generic = Type.GetType("X").GetMethod("Y", true);
    MethodInfo nonGeneric = Type.GetType("X").GetMethod("Y", false);
}