I am trying to design an interface, so that it has a generic type of id and a generic method that returns the type of the class that implements this interface. For example:
public interface IEntity <IDType, MethodReturnType>
{
IDType ID {get; set;}
MethodReturnType Get();
}
public class Model : IEntity<int, Model>
{
int ID {get; set; }
Model Get() { // do something }
}
My question is, it seems silly to put in Model as the second type parameter of IEntity, because I am already in a Model's class, it should be some intelligence way to figure out what type it is (although using generic type requires it to be determined before compile time).
Is it any other solution that can help me to get rid of the Model type while retain the Get method definition in the interface?
In this context, there are two typical ways to go about designing your classes and interfaces. I'll stray slightly from your exact example to try to make the answer more general.
Which option to choose really depends on how you want your classes and interfaces to be used.
Option 1
Make your interface generic, so that the interface members have knowledge of the exact type.
public interface IEntity<TDescription>
{
TDescription Get();
}
public class MyModel : IEntity<MyDescription>
{
MyDescription Get() { ... }
}
public class MyDescription { ... }
This means that when you use your interface IEntity<TDescription> you need to know TDescription at the time of use. The benefit is that you get more compile-time type checking.
Option 2
Do not make your interface generic and instead have your interface members use interfaces as well.
public interface IEntity
{
IDescription Get();
}
public interface IDescription { ... }
public class MyModel : IEntity
{
MyDescription Get() { ... }
IDescription IEntity.Get() { return this.Get(); }
}
public class MyDescription : IDescription { ... }
This is more flexible, but it also means less compile-time type checking.
You can determine the inherited class type with this.GetType() but that will not allow you to create generic functions/parameters/etc as you are doing.
So to your answer, no, you can't unless you will not use that type in any way (you can still get the base type of the class and use it but cannot set it as return type/param type/etc).
Related
I am getting an object of type Company<IDesignation>. Now I want to cast it to ICompany<Manager>. Run time I know that IDesignation is nothing but "Manager" type.
Is this what you are looking for?
Company comp = new Company();
Manager mgner = new Manager(comp.getManager());
IDesignation manager = mgner;
ICompany company = (ICompany)manager;
Assuming Company is:
public class Company: ICompany, IDesignation //or something?
Using either Generic Type Casting (what u are trying to do) or simply Casting an Interface or object depends on whether you will perform this task explicitly or implicitly (maybe class has pre-defined function to cast it) and as your comment has pointed out... maybe by user or runtime matters or whether or not and/or how you need to instantiate your object so I would really need to see the class implementation in order to be able to provide something that uses type casting in the way you want to perform it.
What you need is Contravariance, i.e. your IEntityDelta generic type parameter needs to be made contravariant.
The only way to do that and for this to work is to have:
public interface IEntityDelta<in T> : IEntityDelta where T : IEntity
Note the in T in the definition.
Check out the in (Generic Modifier) (C# Reference)
Or this Understanding Covariant and Contravariant interfaces in C#
If you're not the creator of that interface and if the IEntityDelta<> is defined w/o the in modifier you're out of luck.
And just to mention that adding in/out modifiers is easier said than done. For that to compile your methods, properties etc. need to satisfy
the conditions of contravariance (or covariance in case of 'out') on that generic type (T) parameter.
And this what your classes, interfaces look like based on your info (which was terrible btw. next time you need to dedicate a bit more time
in providing minimal but complete code that makes sense):
public interface IEntityDelta<in T> : IEntityDelta
where T : IEntity
{
void MakeDelta(T entity); // this is allowed
//T Entity { get; set; } // this won't work
}
public class EntityDelta<T> : IEntityDelta<T>
where T : class, IEntity
{
public T Entity { get; set; }
public EntityDelta(T entity) => Entity = entity;
public void MakeDelta(T entity) { }
}
public interface IEntityDelta { }
public abstract class Entity : IEntity { }
public class Order : Entity { }
public interface IEntity { }
...and usage:
var order = new Order();
EntityDelta<IEntity> orderDelta = new EntityDelta<IEntity>(order);
IEntityDelta<IEntity> idelta = orderDelta;
IEntityDelta<Order> iOrderDelta = orderDelta;
I'm writing an SDK which has an OOP structure for implementing data types;
first an interface
then an abstract implementation
finally an abstract generic implementation
People can choose to implement either the interface, or derive from either of the classes.
public interface IGoo
{
IGoo Duplicate();
...
}
public abstract class Goo : IGoo
{
IGoo IGoo.Duplicate() {
return Duplicate();
}
abstract public Goo Duplicate();
...
}
public abstract class Goo<T> : Goo
{
abstract public Goo<T> Duplicate(); ??????
...
}
I'd like to re-implement the Duplicate method so that it always returns the most specific type possible. I.e. when you call Duplicate on an IGoo instance, you get another IGoo. If you call it on Goo, you get Goo, if you call it on -say- Goo<int>, you get Goo<int>. And all Duplicate() methods always call the most specific implementation.
Is this possible? Is it only possible when you can implement an interface explicitly? In which case, should I not make Goo<int> derive from Goo, but have it implement IGoo instead and type all the low-level functionality twice?
What about the following?
public interface IObj
{
IObj Duplicate();
}
public abstract class Obj : IObj
{
public Obj()
{
}
public virtual IObj Duplicate()
{
return this;
}
}
public abstract class ObjT<T> : Obj
{
public ObjT()
{
}
public override IObj Duplicate()
{
return this;
}
}
public class ObjImpl : Obj
{
}
public class ObjTImpl : ObjT<int>
{
}
I understand that you want it to return the most specific type possible in any inheriting class but it actually is. It's boxing the inheriting type into the interface (or a raw object if you where to return objects instead of interface types. If you run the following test in a console app you will see the proper type is represented:
namespace TestConsole
{
class Program
{
static void Main(string[] args)
{
ObjImpl a = new ObjImpl();
ObjTImpl b = new ObjTImpl();
Console.WriteLine(a.Duplicate().GetType());
Console.WriteLine(b.Duplicate().GetType());
Console.ReadLine();
}
}
}
// outputs:
// ObjImpl
// ObjTImpl
The idea of redefining abstracts of abstracts goes against the purpose of abstract polymorphism. If the derived types do not intend to implement the inherited abstract member, they should not be inheriting it.
Although the example I gave above would require casting to access any child class-specific members, it would be the proper way to do it in this approach. The runtime needs to know what types it should expect to deal with.
There is always dynamics you could play around with but to be honest I haven't played around with dynamics with generics and inheritance as I suspect I would make my compiler cry, and when it cries, I cry, a little bit deep down inside... lol
It is only possible when you implement the interface explicitly. That's because the return type of a method is not part of its signature - which the compiler checks when overloading. Therefore, otherwise identical methods which only differ in their return type are syntactically not possible.
This is an extension to my previous question: Type safety through inheritance
I created a new question instead of updating my old as this one dives deeper into this topic.
My original purpose was to declare a method that returns an object of the implementing classes type. One such method might look like GetSimpleClone() in the following:
I have a base interface IEntity that declares the method GetSimpleClone() and is implemented by several other interfaces, for example IPerson.
public interface IEntity<T> : where T : IEntity<T>
{
T GetSimpleClone();
}
public interface IPerson : IEntity<IPerson>
{
}
The interface IAddress also implements IEntity. However another interface IVenue inherits from IAddress. As I described in the beginning, I want the GetSimpleClone() method in IAddress to return an object of type IAddress while the same method in IVenue should return an object of type IVenue. Therefore the declaration of IAddress differs from IPerson as it has to declare a generic type itself:
public interface IAddress<T> : IEntity<T> where T : IAddress<T>
{
}
public interface IVenue : IAddress<IVenue>
{
}
Now the problem is that IPersonhas a reference to IAddress and understandably the compiler forces me to define the generic type of IAddress.
public interface IPerson : IEntity<IPerson>
{
IAddress<"Compiler: Define Type!!"> Address { get; set; }
}
I really can't see a solution to this problem and would appreciate any help from you, even if it is just to say that there is simply no solution. :)
Thanks in advance...
To accomplish that, you will need to introduce yet another generic:
public interface IPerson<T> : IEntity<IPerson<T>>
where T : IAddress<T>
{
IAddress<T> Address { get; set; }
}
The better solution I could think of is just to put aside the safety thing and create a good base IAddress interface with a common methods, like getAddressString, and use it for all types of addresses you may want to have.
I want to create a re-usable library. I was going to use extension methods however I run into some issues in some cases for the client to have to specify in the calling method the types.
QUESTION - If I use an abstract base class as the basis, can I specify an attribute/property in the class to be generic (e.g. the key property might be an 'int' in one case, or a 'string' in another)?
Yes.
public abstract class MyBase<T>
{
public abstract T GetSomething();
}
public class MyConcreteType : MyBase<int>
{
public override int GetSomething()
{
return 3;
}
}
Or, what exactly do you mean ?
(Trying it out would have given you the answer faster then posting it on SO, I think ... )
Yes, you can do the following:
public abstract class Class<T>
{
T value;
T Prop { get; set;}
}
I was wondering if anyone could tell me if this kind of behaviour is possible in C# 4.0
I have an object hierarchy I'd like to keep strongly typed. Something like this
class ItemBase {}
class ItemType<T> where T : ItemBase
{
T Base { get; set; }
}
class EquipmentBase : ItemBase {}
class EquipmentType : ItemType<EquipmentBase> {}
What I want to be able to do to have something like this
ItemType item = new EquipmentType();
And I want item.Base to return type ItemBase. Basically I want to know if it's smart enough to strongly typed generic to a base class without the strong typing. Benefit of this being I can simply cast an ItemType back to an EquipmentType and get all the strongly typedness again.
I may be thinking about this all wrong...
You're talking about covariance which would allow you to do:
ItemType<object> item = new EquipmentType();
You couldn't do this in C# 4 because of the following reasons:
Generic covariance only works on interfaces, arrays, and delegate types, not base classes
Your ItemType class uses T as an in/out type parameter meaning it receives a T and also returns a T.
Number 2 is the main issue because if it were allowed, then the following code would have to be compilable, yet fail at runtime.
// this will not work
ItemType<object> item = new EquipmentType();
item.Base = new Object(); // this seems ok but clearly isn't allowed
Covariance and Contravariance FAQ
No, because ItemType as far as the compiler is concerned is a separate type from ItemType<EquipmentBase> or ItemType<Foo>. All three are treated as unique types and they cannot represent one another.
In your class declarations, you declared it as ItemType<T> and so ItemType would be an undefined type which would not compile.
At best, you could use an ItemType<EquipmentBase> object to represent EquipmentType or any other class derived from ItemType<EquipmentBase> but not ItemType<PersonType>.
I don't think that the new features of C# 4.0 will help you out there. However, there is a way around this which already works since generics were introduced: you create an abstract base class with the same name as the generic class and put all members which you want and which don't need to accept or return an argument of the generic type, like so:
class ItemBase {
}
abstract class ItemType {
public ItemBase Base {
get { return GetItemBase(); }
set { SetItemBase(value); }
}
protected abstract void SetItemBase(ItemBase value);
protected abstract ItemBase GetItemBase();
}
class ItemType<T> : ItemType where T : ItemBase {
protected override sealed void SetItemBase(ItemBase value) {
Base = (T) value;
}
protected override sealed ItemBase GetItemBase() {
return Base;
}
public new T Base { get; set; }
}