e-h*_*der 5 c# arguments assemblies createinstance
我现在拥有的(成功加载插件)是这样的:
Assembly myDLL = Assembly.LoadFrom("my.dll");
IMyClass myPluginObject = myDLL.CreateInstance("MyCorp.IMyClass") as IMyClass;
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这仅适用于具有不带参数的构造函数的类.如何将参数传递给构造函数?
Mag*_*oll 12
你不能.而是使用Activator.CreateInstance,如下面的示例所示(请注意,Client命名空间位于一个DLL中,而Host位于另一个DLL中.两者必须位于同一目录中才能使代码生效.)
但是,如果你想创建一个真正可插入的接口,我建议你使用一个Initialize方法,它接受你的接口中的给定参数,而不是依赖于构造函数.这样你就可以要求插件类实现你的接口,而不是"希望"它接受构造函数中接受的参数.
using System;
using Host;
namespace Client
{
public class MyClass : IMyInterface
{
public int _id;
public string _name;
public MyClass(int id,
string name)
{
_id = id;
_name = name;
}
public string GetOutput()
{
return String.Format("{0} - {1}", _id, _name);
}
}
}
namespace Host
{
public interface IMyInterface
{
string GetOutput();
}
}
using System;
using System.Reflection;
namespace Host
{
internal class Program
{
private static void Main()
{
//These two would be read in some configuration
const string dllName = "Client.dll";
const string className = "Client.MyClass";
try
{
Assembly pluginAssembly = Assembly.LoadFrom(dllName);
Type classType = pluginAssembly.GetType(className);
var plugin = (IMyInterface) Activator.CreateInstance(classType,
42, "Adams");
if (plugin == null)
throw new ApplicationException("Plugin not correctly configured");
Console.WriteLine(plugin.GetOutput());
}
catch (Exception e)
{
Console.Error.WriteLine(e.ToString());
}
}
}
}
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