如何将泛型类型（T）作为参数？

Question

我有这个错误：

Severity    Code    Description Project File    Line
Error   CS0246  The type or namespace name 'T' could not be found (are you missing a using directive or an assembly reference?)

关于此方法的方法签名：

 public static void SendMessage(string queuName, T objeto)
        {
            QueueClient Client =QueueClient.CreateFromConnectionString(connectionString, "Empresa");
            BrokeredMessage message = new BrokeredMessage(objeto);
            message.ContentType = objeto.GetType().Name;
            Client.Send(new BrokeredMessage(message));
        }

Answer 1

public static void SendMessage<T>(string queuName, T objeto)
{
    QueueClient Client =QueueClient.CreateFromConnectionString(connectionString, "Empresa");
    BrokeredMessage message = new BrokeredMessage(objeto);
    message.ContentType = objeto.GetType().Name;
    Client.Send(new BrokeredMessage(message));
}

Answer 2

您忘记指定类型参数。您可以通过两种方式做到这一点：

要么在方法定义中定义它们（在你的情况下这是可行的方法，因为你的方法是静态的）：

public static void SendMessage<T>(string queuName, T objeto)

或者您可以在 class 定义中指定它们（例如方法）：

class MyClass<T>{
    public void SendMessage(string queuName, T objeto){}
}

Answer 3

您示例的正确语法是：

public static void SendMessage<T>(string queuName, T objeto)
{
// Type of T is
Type t = typeof(T);
// Obtain Name
string name = t.Name
// Create another instance of T
object to = Activator.CreateInstance<T>();
// etc.
}

总的来说：

T method<T>(T param) where T: restrictions //new() for example
{ return (T)Activator.CreateInstance<T>(); }

如何将泛型类型（T）作为参数？

How to put generic Type (T) as parameter?

c#

servicebus

azure