I am unable to cast a generic type to another generic type, besides the cast should be valid
What I want to archive is in short (for MyModel implementing IModel, and MyImplementation implementing IImplementation):
IImplementation<IModel> implementation = new MyImplementation<MyModel>();
Assert.IsNull(implementation as IImplementation<IModel>);
This is a bit confusing, as the type should be valid.
Complete conceptual model:
interface IModel {}
class MyModel : IModel {}
interface IImplementation<TModel> where TModel : IModel { }
class MyImplementation<TModel> : IImplementation<TModel>
where TModel : IModel { }
public void CallRegister()
{
var implementation = new MyImplementation<MyModel>();
var instance = CastModel(implementation);
Assert.IsNotNull(instance); //this assert fails!
}
private object CastModel<TModel>(IImplementation<TModel> implementation) where TModel : IModel
{
return implementation as IImplementation<IModel>;
}
I need this cast to enable me to save multiple IImplementations to the same Dictionary<Type, IImplementation<IModel>>, where the key is obtained by doing typeof(TModel).
To do this type safe I don't want to use a Dictionary<Type, object>.
Why does the cast fail? Are there additional resources to this? Its a similar question to Invalid Cast of Type Constrained C# Generic, but it is not explained why there just that it does not work.
What is the best way to archive a functionality similar to the dictionary as explained above if this kind of cast is not possible?
Though Olivier's answer gets the idea across about why this usually goes wrong, there is a way to make this work in your program.
The feature you want is called generic interface covariance. Covariance is the property that if a Cat is an Animal, then an IFoo<Cat> is an IFoo<Animal>.
Covariance in C# only works in the following situations:
The "outer" type is an interface, delegate or array. No classes or structs.
If an interface or delegate, the type must be marked at compile time as supporting covariance. Arrays get (unsafe!) covariance for free.
The "inner" types -- the types that are varying -- are both reference types. You can't say that an IFoo<int> is an IFoo<object> even though an int is an object, because they are not both reference types.
To mark an interface as covariant, you put out before the declaration of the type parameter which you wish to allow to vary:
interface IImplementation<out TModel> where TModel : IModel { }
If you do that, your program will start to work.
HOWEVER, out is a reminder to you that covariance is only safe if T is used in output positions. This is legal:
interface I<out T> {
T M();
}
This is not:
interface I<out T> {
void M(T t);
}
In the first, T is only passed out of things. In the second, it is passed in.
In the first scenario, we cannot use covariance to introduce a type hole. We have an I<Cat> and we cast it to I<Animal>, and now M returns an Animal, but that's OK, because we already know that it will return a Cat, and a Cat is an Animal.
But in the second scenario, we have the opposite situation. If we allowed an I<Cat> to be converted to I<Animal> then we have an M that can take a Turtle, but the real implementation can only handle Cats. That's why C# will make this illegal.
So go forth and use covariance, but remember that you have to certify to the compiler that you want it, and that it is safe under all circumstances. If you don't want it, or it is not safe, then you don't get to have covariance, and you'll have to find a different solution to your problem.
This kind of conversion is not allowed for good reasons. Let's take an example where the problem is more obvious. We have the classes Animal, Cat : Animal and Dog : Animal. Now let's do this:
List<Animal> list = new List<Cat>(); // Seems to be possible at first glance.
// An now comes the problem:
list.Add(new Dog()); // Seems to be possible as well.
But wait! The list is in reality a list of cats! And we are trying to add a dog. Even adding new Animal() to list, which is statically typed as List<Animal>, would not work.
Therefore two types T<A> and T<B> are not assignment compatible in C#, even if A and B are!
You need another approach.
What you can do is to wrap your dictionary in a class with a generic method having a generic type constraint.
public class MyImplementationDict
{
private readonly Dictionary<Type, object> _internalDict = new Dictionary<Type, object>();
public void Add<T>(IImplementation<T> item)
where T : IModel
{
_internalDict.Add(typeof(T), item);
}
...
}
Related
I'm writing a method that returns a different instance of a class based on the type passed in, for a load process.
The method I'm trying to write will be used like:
Loader<MyType> loader = context.GetLoader<MyType>();
loader.Load(myTypeArray);
Internally, I'm planning on keeping a Dictionary<Type, T> with the loader as the value. However, I'm not sure what type T should be.
I could make T object, but I'd prefer to keep it as some sort of Loader<>. I know I'll need to cast it either way, but using Loader<> would give a little more type safety.
Is there any good way to implement this?
You have two choices. Both require you to add an interface.
Add a plain ILoader (with no type arguments) interface, and implement it in your loader as class Loader<T> : ILoader. The common interface will allow you to store all loaders in a List<ILoader>.
interface ILoader {}
class Loader<T> : ILoader {}
void Example()
{
var loaders = new List<ILoader>();
loaders.Add( new Loader<Foo>() );
loaders.Add( new Loader<Bar>() );
}
Define a new covariant interface ILoader<out T> and implement it in your Loader as class Loader<T> : ILoader<T>. The covariance will allow you to store all loaders in a List<ILoader<object>>. This option only works if T is a reference type, and is only viable if none of your interface methods accept T as an argument.
interface ILoader<out T>
{
//void SetLoader(T input); //Does not compile due to covariance
T GetLoader(); //Read is allowed
}
class Loader<T> : ILoader<T>
{
public void SetLoader(T input) {} //Works when not in interface
public T GetLoader() { return default(T); }
}
void Example()
{
var loaders = new List<ILoader<object>>();
loaders.Add( new Loader<Foo>() );
loaders.Add( new Loader<Bar>() );
}
Code sample on DotNetFiddle
I could make T object, but I'd prefer to keep it as some sort of Loader<>. I know I'll need to cast it either way, but using Loader<> would give a little more type safety.
No it will not; this sentence contradicts itself! You say both "I know I'll need to cast" so there is already zero compile-time type safety, and that you can get more type safety, but zero is not more than zero!
Remember the function of a representation-preserving cast is precisely "I know a fact that will be true at runtime that the compiler does not know at compile time". That's the purpose of a cast.
There is no way to represent in the C# type system "this is a thing of type C<?>", so don't try to find one. You've already embraced run-time type checking, so go for it. You have an invariant that your dict[typeof(T)] produces something of type Loader<T>; the only person that can maintain that invariant is you, so don't even try to rely on the compiler to do it for you.
I've got an interface structure that looks like this:
At the most basic level is an IDataProducer with this definition:
public interface IDataProducer<out T>
{
IEnumerable<T> GetRecords();
}
and an IDataConsumer that looks like this:
public interface IDataConsumer<out T>
{
IDataProducer<T> Producer { set; }
}
Finally, I've got an IWriter that derives off of IDataConsumer like so:
public interface IWriter<out T> : IDataConsumer<T>
{
String FileToWriteTo { set; }
void Start();
}
I wanted to make IWriter's generic type T covariant so that I could implement a Factory method to create Writers that could handle different objects without having to know what type would be returned ahead of time. This was implemented by marking the generic type "out". The problem is, I'm having a compile error on IDataConsumer because of this:
Invalid variance: The type parameter 'T' must be contravariantly valid on 'IDataConsumer<T>.Producer'. 'T' is covariant.
I'm not really sure how this can be. It looks to me like the generic type is marked as covariant through the whole chain of interfaces, but it is very possible I don't totally understand how covariance works. Can someone explain to me what I am doing wrong?
The problem is that your Producer property is write-only. That is, you are actually using T in a contravariant way, by passing the value that is generic on type T into the implementer of the interface, rather than the implementer passing it out.
One of the things I like best about the way the C# language design team handled the variance feature in generic interfaces is that the keywords used to denote covariant and contravariant type parameters are mnemonic with the way the parameters are used. I always have a hard time remembering what the words "covariant" and "contravariant" mean, but I never have any trouble remembering what out T vs. in T means. The former means that you promise to only return T values from the interface (e.g. method return values or property getters), while the latter means that you promise to only accept T values into the interface (e.g. method parameters or property setters).
You broke that promise by providing a setter for the Producer property.
Depending on how these interfaces are implemented, it's possible what you want is interface IDataConsumer<in T> instead. That would at least compile. :) And as long as the IDataConsumer<T> implementation really is only consuming the T values, that would probably work. Hard to say without a more complete example.
Peter's answer is correct. To add to it: it helps to try out some examples and see what goes wrong. Suppose the code you had originally was allowed by the compiler. We could then say:
class TigerConsumer : IDataConsumer<Tiger>
{
public IDataProducer<Tiger> p;
public IDataProducer<Tiger> Producer { set { p = value; } }
... and so on ...
}
class GiraffeProducer : IDataProducer<Giraffe>
{
public IEnumerable<Giraffe> GetRecords() {
yield return new Giraffe();
}
TigerConsumer t = new TigerConsumer();
IDataConsumer<Mammal> m = t; // compatible with IDataConsumer<Mammal>
m.Producer = new GiraffeProducer(); // compatible with IDataProducer<Mammal>
foreach(Tiger tiger in t.p.GetRecords())
// And we just cast a giraffe to tiger
Every step on the way here is perfectly typesafe but the program is plainly wrong. Either one of those conversions has to be illegal, or one of the interfaces is not safe for covariance. We wish all those conversions to be legal, and therefore we must detect the lack of type safety in your interface declarations.
I have two legacy C# user controls I need to work together.
I have an existing dialog onto which I need to add an existing generic usercontrol.
I have tried to sample the hierarchy below
interface Foo<T> {}
interface Bar<T>
{
T DataObject { get; set; }
}
public class ClassA<T> where T : Foo<T>
{
public ClassA(T dataObject)
{
//Do stuff if T implements Bar<T> - Pseudocode ahead
if(var T is Bar<T>)
{
var x = new ClassB<T>();
//x is typesafe, and I can set DataObject
x.DataObject = dataObject;
}
}
}
public class ClassB<T> where T : Bar<T>
{
T DataObject { get; set; }
}
The existing dialog, ClassA currently don't have any generic contraints, but could easily be changed to require T implementing Foo<T>.
The userControl, ClassB is based on another interface, Bar<T>. In practice, Bar<T> objects always implements Foo<T> - in theory of course not.
Are there any construction I can use to declare an object of type ClassB, and get compile time validation?
The construction above will give me a compile error stating:
The type 'T' cannot be used as type parameter 'T' in the generic type
of method ClassB<T>. There is no implicit reference conversion from
'T' for Bar<T>
I can make the ClassB object with Reflection, setting the properties using Reflection as well - but I would prefer a compile time solution.
But in my current situation with two existing dialogs - i'm not sure I am able to.
Any help is appreciated - also just if it is stating what I expect, that it can't be done.
--EDIT
Trying to elaborate a bit.
The problem rises when I have a ClassC that implements both Foo<T> and Bar<T>
public class ClassC<T> : Foo<T>, Bar<T>
{
T DataProperty
}
If I make an instance of ClassA<ClassC>, that is T in the specific instance is ClassC - then is there a way in code I can use T in creating an instance of ClassB- in this case T in ClassA does live up to the ClassB constraints, since T is ClassC.
I havent figured out how or if possible - tend to believe I can't.
As I wrote above, I have a solution based on reflection, i'm just not fan of using reflection and getting run-time validation only. But in this case with two legacy objects that need to work together I might be running out of options.
First of all, your types are a bit weird. They are kind-of recursive, with ClassB<T> requiring a T that implements Bar<T> which has the same structure as ClassB<T>. Maybe you meant to ClassB<T> to implement Bar<T> instead of requiring it as a type parameter?
Anyway you cannot do this. In order to be able to write ClassB<T>, the compiler needs to ensure that T is a valid type parameter for ClassB<> at runtime. This can only be the case when the type parameters on ClassA<T> for T are at least as restrictive as the ones from ClassB<T>.
Unfortunately, even a hard type check which would ensure that this is the case will not allow you to write ClassB<T>.
So without being able to write ClassB<T>, you will not be able to get static type safety at compile-time. So even when you create an instance of ClassB<T> (which you can), you won’t be able to access DataProperty on it since you will not be able to cast it as a ClassB<T>.
So in order to solve this, you would either have to access DataProperty using reflection only, or call a method inside ClassA<T> that does have the type constraint. I’ll show you both solutions:
public class ClassA<T>
where T : Foo<T>
{
public ClassA(T dataObject)
{
if (typeof(Bar<T>).IsAssignableFrom(typeof(T)))
{
// method 1, calling a generic function
MethodInfo mi = typeof(ClassA<T>).GetMethod("SetBDataObject").MakeGenericMethod(typeof(Bar<T>));
mi.Invoke(this, new object[] { dataObject });
// method 2, doing it all with reflection
Type type = typeof(ClassB<>).MakeGenericType(typeof(T));
object x = Activator.CreateInstance(type);
type.GetProperty("DataObject").SetValue(x, dataObject);
}
}
public object SetBDataObject<TB> (TB obj)
where TB : Bar<TB>
{
var x = new ClassB<TB>();
x.DataObject = obj;
return x;
}
}
The first thing that probably is confusing in your code, is that you have used the same letter T as the Type parameter in both classes ClassA<T> and ClassB<T>.
I'll start by stating the obvious:
when you call var x = new ClassB<T>(); the constraint for T here is in the context of ClassA<T> (i.e. T : Foo<T>), while new ClassB<T>() expects T to match the constraint of T : Bar<T>.
It seems to me that underlying reason for the issue you are having is a design problem. It looks like you a little mix up between types and classes.
Lets walk it through:
from the Gang of Four Design Patterns book:
An objects's class defines how the object is implemented .The class
defines object's internal state and the implementation of its
operations.
In contrast, an objects's type only refers to its interface -the set
of requests to which it can respond.
An object can have many type, and object of different classes can have
the same type.
The usage of interfaces in your code implies coding against types (that's good!).
Checking for if (dataObject is Bar<T>) and upon the result constructing a ClassB<U> where !typeof(U).Equals(typeof(T) implies heavily relying on implementation (e.g. class).
If you ask me, I think you should try one of the following:
Use the factory pattern for constructing ClassB. In the dedicated factory you can add some more logics and verifications in order to decide how to construct it (from your code, it is not clear since the types do not match...).
If possible, resolve the relation between Foo<T> and Foo<T> and declare the constraints in the interfaces. In that case, both interfaces should have same constraints fto T
ive run into a problem - these is my class structure
public interface IModel{}
public interface IGenericMapper<T> where T : IModel {...}
public class ActualModel:IModel {...}
public class ActualMapper: IGenericMapper<ActualModel> {...}
My actual code to initialse the mapper is:
IGenericMapper<IModel> mapper;
mapper= new ActualMapper();
It does not compile. I get the error
Cannot implicitly convert type 'ActualMapper' to 'IGenericMapper'.
An explicit conversion exists (are you missing a cast?)
When I do cast it using
mapper= new ActualMapper() as IGenericMapper<IModel>;
the mapper does not get initialized properly (it comes back as NULL)
What am I missing - since ActualMapper() implements IGeneric Mapper and its type impliments `IModel' why can it not initialize mapper.
Is there another way to structure this so achieve what I need?
Thank you so much
Note the solution people have proposed gives me other compilation errors as the mapping Interface has the following members
T GetModel(busO bBusinessObject);
busO SetBusObject(T source, busO target);
apparently you cant have the generic type as an input parameter when its declared at "out"
Pretty sure you're moving into the realm of covarience here with the Generic change;
try this:
public interface IModel{}
public interface IGenericMapper< out T> where T : IModel{}
public class ActualModel : IModel{}
public class ActualMapper : IGenericMapper<ActualModel> {}
and then:
IGenericMapper<IModel> blah = new ActualMapper();
with out the 'out T' the best you can do is:
IGenericMapper<ActualModel> blah = new ActualMapper();
This is a rabbit hole, so be careful especially if you ever try to mix the two :)
http://msdn.microsoft.com/en-us/library/ee207183.aspx
[Edit]
If you want to be able to downcast the generic T, then it has to be out and you cannot use it as an input. You can, however, move some of it to real time in your implementation; i.e. to a check to see if you can cast it to a model type.
interface IGenericMapper<out TModel, in TKeyOrIdent>
TModel GetModel(TKeyOrIdent bBusinessObject);
void SetModel(object model, TKeyOrIdent target);
you have to define
IGenericMapper<out T>
to support your scenario, but this applies other limitations.
Simply IGenericMapper<IModel> != IGenericMapper<ActualModel> even if ActualModel : IModel
In most scenarios it makes sense to have a base interface that isn't a generic. See IList<T> for example, it implements IList.
You can then choose on implementation to explicitly implement interface members. See List<T>.GetEnumerator() : IEnumerable
Do not use generics under the assumption that you can eliminate casting everywhere. I've tried it C# simply doesn't have the required features for this.
I would suggest an IGenericMapper interface as base interface for IGenericMapper<T> then do your generic code on IGenericMapper and finally (and at that point you already have the type) cast it back to the specific type.
I'm looking for a way to define "MethodA" below, such that it returns a class definition (System.Type), of which an instance of said type implements "InterfaceB"
interface IMyInterface
{
TType MethodA<TType, TInterface>()
: where TType : System.Type
: where [instanceOf(TType)] : TInterface
}
(Note: instanceOf is not real, of course...)
I suspect it's not possible to get this kind of verification at compile time. I'm hoping someone out there will prove me wrong.
Thanks in advance for any guidance.
EDIT: I've updated this in hopes of being more specific that what is returned is a System.Type, of which this later code can execute:
var classType = myInterface.MethodA<(something, ISomeInterface)>();
ISomeInterface = new classType(); //--Assuming default constructor
I haven't really focused on this part yet, just more curious about the theoretical construction of my primary quesiton.
There are two interpretations of your question; one would be trivial, one impossible, so I'll go forward and cover both.
You want to return an instance of a type which implements both System.Type and TInterface.
This is trivial: Just use where TType : Type and where TType : TInterface.
You want to return an instance of System.Type that represents a type which inherits from TInterface.
This is impossible to specify in the .NET (and C#) type-system.
The type system can only resolve information from the type hierarchy themselves, but not enforce "contracts" like restricted run-time property values. There are a few hacks regarding default-constructors etc., but as far as I know not even testing for existing methods is possible (unlike C++ templates, for instance, not to speak of Qi et al.).
Update
Please check the comment from Michael Graczyk.
Also: I just found out that there are code-contract checkers (static and run-time) for .NET: Microsoft DevLabs Code-Contracts for .NET. I have never used them, not even new about them, but that looks interesting!
However, even without looking, I'm quite sure that overload-resolution etc. will not be able to use such information.
In this situation:
// We have an interface...
interface InterfaceB {}
// And this class implements the interface.
class ImplementsB : InterfaceB {}
// But this class does not.
class DoesNotImplementB {}
You could define MethodA as:
static Type MethodA<TClass, TInterface>()
where TClass : TInterface
{
return typeof(TClass);
}
Then the following will work:
Type t = MethodA<ImplementsB, InterfaceB>();
But this gives a compile-time error:
Type t = MethodA<DoesNotImplementB, InterfaceB>();
The type 'DoesNotImplementB' cannot be used as type parameter 'TClass' in the generic type or method 'MethodA<TClass,TInterface>()'. There is no implicit reference conversion from 'DoesNotImplementB' to 'InterfaceB'.
So, this way you are sure that the result of MethodA is a Type of a class that implements TInterface. Given that Type object, you can instantate it later like this:
public object Instantiate(Type type)
{
// Call the default constructor.
// You can change this to call any constructor you want.
var constructor = type.GetConstructor(Type.EmptyTypes);
var instance = constructor.Invoke(new object[0]);
return instance;
}
If you know that your Type is compatible with some interface TInterface, then you can avoid a cast with an additional method like this:
public TInterface Instantiate<TInterface>(Type type)
{
return (TInterface)Instantiate(type);
}
However, if type is a Type that somehow does not implement TInterface, you'll get an InvalidCastException at run-time. There is no way to constrain Type to be a type that implements a particular interface at compile-time. However, at run-time you can check it to avoid the InvalidCastException exception:
public TInterface Instantiate<TInterface>(Type type)
{
if (!typeof(TInterface).IsAssignableFrom(type))
throw new Exception("Wrong type!");
return (TInterface)Instantiate(type);
}
Note that typeof(TType) is an expression that results in a Type object, so everywhere you see typeof() you could replace it with any Type variable and vice versa.
Is this what you wanted to know?