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How to create a new object instance from a Type


By : Matt
Date : March 29 2020, 07:55 AM
help you fix your problem The Activator class within the root System namespace is pretty powerful.
There are a lot of overloads for passing parameters to the constructor and such. Check out the documentation at:
code :
ObjectType instance = (ObjectType)Activator.CreateInstance(objectType);

ObjectType instance = (ObjectType)Activator.CreateInstance("MyAssembly","MyNamespace.ObjectType");


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Create an instance of an unknown type of object and TryUpdateModel


By : Alex Lombry
Date : March 29 2020, 07:55 AM
This might help you I'm using MVC and i have a controller action which handles several different view models, each view model has validation and i'd like the controller to check the validation. , You will need to write a custom model binder for this to work:
code :
public class MyModelBinder : DefaultModelBinder
{
    protected override object CreateModel(ControllerContext controllerContext, ModelBindingContext bindingContext, Type modelType)
    {
        var typeValue = bindingContext.ValueProvider.GetValue("viewType");
        var type = Type.GetType(
            (string)typeValue.ConvertTo(typeof(string)), 
            true
        );
        var model = Activator.CreateInstance(type);
        bindingContext.ModelMetadata = ModelMetadataProviders.Current.GetMetadataForType(() => model, type);
        return model;
    }
}
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult WizardCaseResult([ModelBinder(typeof(MyModelBinder))]object model)
{
    ...
}
ViewType vm = new ViewType();
TryUpdateModel<ViewType>(vm);

Create instance with type object during runtime


By : Obi Osonduagwueki
Date : March 29 2020, 07:55 AM
like below fixes the issue First step is to retrieve the actual type you want to create. Most probably, the name of the type is stored in the database as a String. So you call
code :
var typeName = "System.Int32";
var type = System.Type.GetType(typeName);
var obj = System.Activator.CreateInstance(type);
var i = (IFormattable)obj;
Console.WriteLine(i.ToString("0", System.Globalization.CultureInfo.InvariantCulture));

How to create a new instance of a object type


By : user2295992
Date : March 29 2020, 07:55 AM
will help you I have a base class where I inherit my forms. , You don't need to cast it as anything; object is fine:
code :
object instance = Activator.CreateInstance(type);

Using the class type of an object to create a new instance


By : Robert Martinez Mart
Date : March 29 2020, 07:55 AM
like below fixes the issue To realise the line Clazz user = (Clazz) o;, in the save method, where Clazz is the class of the object passed to the save method, so in this case User, you can make use of generics in the signature of save as follows:
code :
public <T> void save(T user) throws IllegalAccessException, InstantiationException, InvocationTargetException {
    int i = 0;
    List<Object> args = new ArrayList<>();
    args.add(0,"Example");
    T newUser = (T)user.getClass().getConstructors()[i].newInstance(args);
    // Exciting things...
}

Determine a object's type based on a string and create a new instance?


By : Kanishka Dilshan
Date : March 29 2020, 07:55 AM
To fix this issue You can scan current assembly (or whatever assembly that contains your shapes) for types that inherited from Shape, and use Activator.CreateInstance to create the instance. It's better if you cache the name and type in a dictionary, so you can quickly lookup the type that you want.
code :
var type_str = "Circle";

// TypeChache should be static field
if (TypeChache == null)
{
    var assem = Assembly.GetExecutingAssembly();
    TypeChache = assem.GetTypes()
                 .Where(x => typeof(Shape).IsAssignableFrom(x))
                 .ToDictionary(x=>x.Name, x=>x);
}

var type = TypeChache[type_str];    
var inst = (Shape) Activator.CreateInstance(type);
DoSomething(inst);
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