Displaying a form from a dynamically loaded DLL

Question

This is an extension of a question I previously asked here.

Long story short, I dynamically load a DLL and make a type out of it with the following code:

Answer 1

Yes, you aren't actually specifying any code to run outside the class initializer. For instance, with forms you have to actually show them.

You could modify your code to the following...

Assembly assembly = Assembly.LoadFile("C:\\test.dll");
Type type = assembly.GetType("test.dllTest");
Form form = Activator.CreateInstance(type) as Form;
form.ShowDialog();

Answer 2

I would go with:

Assembly assembly = Assembly.LoadFile("C:\\test.dll");
Type type = assembly.GetType("test.dllTest");
object obj = Activator.CreateInstance(type);
Form form = obj as Form;
if (form != null)
    form.Show(); //or ShowDilaog() whichever is needed

Other error checking/handling should be added; however at the very least I would ensure the conversion works.

Answer 3

If a class belongs to Form then the Assembly.GetType() returns NULL. If a class belongs to User Control then I can see that the type is returned.

Also the syntax should be as:

Type type = assembly.GetType("Assemblytest.clsTest");

where

• clsTest will be the name of class (of a user control)
• Assemblytest is the name of assembly without the .dll extention.

Answer 4

You have to do something with the form you've just created:

Assembly assembly = Assembly.LoadFile("C:\\test.dll");
Type type = assembly.GetType("test.dllTest");
Form form = (Form)Activator.CreateInstance(type);
form.ShowDialog(); // Or Application.Run(form)