C# 如何将类作为方法的参数传递？
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How to pass a Class as parameter for a method?
提问by Amin AmiriDarban
I have two classs:
我有两个班级：
Class Gold;
Class Functions;
There is a method ClassGetin class Functions, which has 2 parameters.
I want to send the class Goldas parameter for one of my methods in class Functions.
How is it possible?
ClassGetclass 中有一个方法Functions，它有 2 个参数。我想将该类Gold作为我在 class 中的方法之一的参数发送Functions。这怎么可能？
For example:
例如：
public void ClassGet(class MyClassName, string blabla)
{
MyClassName NewInstance = new MyClassName();
}
Attention:I want to send MyClassNameas string parameter to my method.
注意：我想MyClassName作为字符串参数发送到我的方法。
采纳答案by Jeroen van Langen
The function you're trying to implement already exists (a bit different)
您尝试实现的功能已经存在（有点不同）
Look at the Activator class: http://msdn.microsoft.com/en-us/library/system.activator.aspx
查看 Activator 类：http: //msdn.microsoft.com/en-us/library/system.activator.aspx
example:
例子：
private static object CreateByTypeName(string typeName)
{
// scan for the class type
var type = (from assembly in AppDomain.CurrentDomain.GetAssemblies()
from t in assembly.GetTypes()
where t.Name == typeName // you could use the t.FullName aswel
select t).FirstOrDefault();
if (type == null)
throw new InvalidOperationException("Type not found");
return Activator.CreateInstance(type);
}
Usage:
用法：
var myClassInstance = CreateByTypeName("MyClass");
回答by Guffa
You could send it as a parameter of the type Type, but then you would need to use reflection to create an instance of it. You can use a generic parameter instead:
您可以将它作为 type 的参数发送Type，但随后您需要使用反射来创建它的实例。您可以改用泛型参数：
public void ClassGet<MyClassName>(string blabla) where MyClassName : new() {
MyClassName NewInstance = new MyClassName();
}
回答by Khan
Are you looking for type parameters?
您在寻找类型参数吗？
Example:
例子：
public void ClassGet<T>(string blabla) where T : new()
{
var myClass = new T();
//Do something with blablah
}
回答by Amin AmiriDarban
public void ClassGet(string Class, List<string> Methodlist)
{
Type ClassType;
switch (Class)
{
case "Gold":
ClassType = typeof(Gold); break;//Declare the type by Class name string
case "Coin":
ClassType = typeof(Coin); break;
default:
ClassType = null;
break;
}
if (ClassType != null)
{
object Instance = Activator.CreateInstance(ClassType); //Create instance from the type
}
}
