c＃ - 通用接口类型

Question

i've made a mini project to get with reflection all the interfaces from the dll i've imported, that inherits from my "IBase" interface like this

 Type[] services = typeof(DataAccess.IXXX).Assembly.GetTypes();
 foreach (Type SC in services)
        {
            if (SC.IsInterface)
            {
                if (SC.GetInterfaces().Contains(typeof(DataAccess.IBase)))
                {
                    file.WriteLine(SC.Name);
                }
             }
         }

The problem is that a lot of my interfaces contains generics

public interface IExample<TKey, Tvalue, TCount> : IBase

But my SC.Name write that like this

IExample'3

Can you help me ?

Answer 1

IExample'3 is the internal name of the of an interface with 3 generic type arguments (as you have probably already guessed). To get the generic type arguments of a class or interface use the Type.GetGenericArguments

You can use something like this to print the correct name

var type = typeof(IExample<int, double>);
var arguments = Type.GetGenericArguments(type);
if(arguments.Any())
{
    var name = argument.Name.Replace("'" + arguments.Length, "");
    Console.Write(name + "<");
    Console.Write(string.Join(", ", arguments.Select(x => x.Name));     
    Console.WriteLine(">")
}
else
{
    Console.WriteLine(type.Name);
}

Answer 2

我认为Type.GetGenericArguments方法就是你所需要的

Answer 3

As you can see, the .NET name property doesn't show you the generic parameter types as part of the name. You have to get the parameter types from GetGenericArguments.

Here is a method returns the name of a generic type in the C# style. It is recursive, so it can handle generics that have generic types as parameters ie IEnumerable<IDictionary<string, int>>

using System.Collections.Generic;
using System.Linq;

static string FancyTypeName(Type type)
{
    var typeName = type.Name.Split('`')[0];
    if (type.IsGenericType)
    {
        typeName += string.Format("<{0}>", string.Join(",", type.GetGenericArguments().Select(v => FancyTypeName(v)).ToArray()));
    }
    return typeName;
}