This question already has answers here:
Why an inherited interface can't be converted to its base interface in generic context?
(2 answers)
Closed 3 years ago.
There is an interface which inherits the generic interface.
Here is generic interface
public interface IBaseService<TEntity> where TEntity : BaseEntity
{
}
Interface which inherits generic interface
public interface IChatService :IBaseService<Messages>
{
}
Then i have a class which takes the generic interface in the constructor as a parameter.
public class BaseApi
{
IBaseService<BaseEntity> _baseService;
public BaseApi(IBaseService<BaseEntity> baseService)
{
_baseService = baseService;
}
}
When i pass the interface object which implements the generic interface it doesn't allow me to pass the object of the interface, it asks to pass generic interface.
Here is error:
Any solution?
Thanks in advance, sorry for bad english!
This is quite logical. Let's simplify matters a bit by renaming our types and creating a new example:
public class BaseApi
{
public BaseApi(IBucket<Fruit> baseService) { ... }
}
And our IBucket<Messages> represents a concrete type of fruit (apple):
public interface IChatService :IBucket<Apple>
{
}
A bucket of apples is not the same as a bucket of fruit because you can put an orange in the latter and you cannot do this with the former.
To deal with this you could either change the signature of IChatService:
public interface IChatService :IBaseService<BaseEntity>
{
}
Or pass the expected IBaseService<Messages> type to BaseApi's ctor.
You can fix this with covariance. But only if your argument can be covariant
public interface IBaseService<out TEntity> where TEntity : BaseEntity
{
}
https://learn.microsoft.com/en-us/dotnet/standard/generics/covariance-and-contravariance
Related
This question already has an answer here:
C# using Generics both in the Interface and its implementor class
(1 answer)
Closed 3 years ago.
I have a class that has type constraints:
public class DataManager<TEntity> : IDisposable where TEntity : LogRecord, new()
{
public TEntity GetRecord()
{
...
}
}
It has a derived class which uses ServerLogRecord derived LogRecord as TEntity:
public class ServerDataManager : DataManager<ServerLogRecord>
{
}
I want to abstract DataManager's public functions into an interface, so I can make it unittestable and decouple the implementation with client code, and because the public function returns TEntity, so I have to put the type constraints in interface:
public interface IDataManager<out TEntity> where TEntity : LogRecord, new()
{
TEntity GetRecord();
}
Then I have a question, how should I declare my implantation class to keep it generic? I tried this solution suggested:
public class DataManager<TEntity> : IDataManager<TEntity>, IDisposable
However I got this error:
error CS0314: The type 'TEntity' cannot be used as type parameter 'TEntity' in the generic type or method 'IDataManager<TEntity>'. There is no boxing conversion or type parameter conversion from 'TEntity' to 'LogRecord'.
but reapply constraints works:
public class DataManager<TEntity> : IDataManager<TEntity>, IDisposable where TEntity : LogRecord, new()
So is "rewriting type constraints in implementation class" the only way to keep DataManager generic?
If I understand what was asked, Just have the class implement the interface
public class DataManager<TEntity> : IDataManager<TEntity>, IDisposable { ... }
If the intention is not to have the implementation as a generic as well and you have you entity then
public class DataManager : IDataManager<MyClass>, IDisposable { ... }
Provided MyClass satisfies the constraint.
when implementing interfaces, you are expected to provide the real classes which are going to be used
in the implementation not generic types because most times, you would not be calling the implementation but
the interfaces.
consider this code
//Where K is key
//where T is a generic entity which may be a programming language
public interface ICoder<K, T>
{
Task Program();
Task LearnNewLanguage(T Language);
K GetCoderId();
}
and your implementation
public class CoderImplementation<TDbContext> : ICoder<string, ProgrammingLanguage> where TDbContext : DbContext
{
private readonly TDbContext _edu;
public CoderImplemtation(TDbContext edu)
{
_edu = edu;
}
public Task Program()
{
...start programming by 12 to 6
}
public Task LearnNewLanguage(string Name)
{
_edu.AddLanguage(Name);
}
}
By telling it that TDbContext is of type DbContext you can be able to access methods which a DbContext exposes
even though that DbContext is not known
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Does C# support return type covariance?
Why can’t I implement an Interface this way?
Consider the following:
public interface IAnimal {
}
public class Animal : IAnimal {
}
public interface ICage {
IAnimal SomeAnimal {get;}
}
public class Cage : ICage{
public Animal SomeAnimal { get; set; }
}
I've read a lot of stuff on covariance and contravariance for IEnumerable, but I'm not sure how to get the above code to work. I get the error "Cage does not implement interface member IAnimal". Since it defined Animal, which is more defined than IAnimal, it seems like covariance should take care of me.
What am I missing? Thanks in advance.
That's not currently possible in C#.
It theoretically possible for the language designers to add it, they just haven't [yet]. They may or may not decide to add it to a potential future version of C#.
The best workaround would probably be:
public class Cage : ICage
{
public Animal SomeAnimal { get; set; }
IAnimal ICage.SomeAnimal
{
get { return SomeAnimal }
}
}
Quoted from Eric Lippert, in response to me asking the same question here:
Why can't I implement an Interface this way?
C# does not support return type covariance for the purposes of
interface implementation or virtual method overrding. See this
question for details:
Does C# support return type covariance?
C# does support generic covariance and contravariance of interfaces
and delegate types that are constructed wtih reference types for the
type arguments as of C# 4.
And C# does support return type covariance when converting a method
that returns a reference type to a delegate type whose return type is
a compatible reference type. (And similarly it supports parameter type
contravariance.)
If this subject interests you, I have written a great many articles
discussing various versions of variance that C# does and does not
support. See
http://blogs.msdn.com/b/ericlippert/archive/tags/covariance+and+contravariance/
for details.
To get covariance you should have something like this:
public interface IAnimal {
}
public class Lion : IAnimal {}
public class Sheep : IAnimal {}
// note the "out" on the T type parameter
public interface ICage<out T> where T:IAnimal {
T SomeAnimal {get;}
}
public class Cage<T> : ICage<T> where T:IAnimal {
public T SomeAnimal { get; set; }
}
you can now do this:
// without covariance on ICage you can't assign a 'Cage<Sheep>' to 'ICage<IAnimal>'
ICage<IAnimal> sheeps = new Cage<Sheep>() {SomeAnimal=new Sheep()};
ICage<IAnimal> lions = new Cage<Lion>() {SomeAnimal=new Lion()};
or this (creating a heterogeneous list of Cage<IAnimals> with both Cages of Sheeps and Cages of Lions), which is an equivalent but probably more useful example:
// without covariance on ICage it errors: cannot convert from 'Cage<Sheep>' to 'ICage<IAnimal>'
var zoo = new List<ICage<IAnimal>>{
new Cage<Sheep> {SomeAnimal=new Sheep()},
new Cage<Lion> {SomeAnimal=new Lion()},
};
if you try removing the out from the declaration of ICage you will see the difference.
public interface IAnimal {
}
public class Animal : IAnimal {
}
public interface ICage {
IAnimal SomeAnimal {get;}
}
public class Cage : ICage{
public Animal SomeAnimal { get; set; }
}
public class AnotherAnimal : IAnimal {
}
Cage c = new Cage();
ICage ic = (ICage)c;
ic.Animal = new AnotherAnimal();
this would be invalid because AnotherAnimal implements IAnimal, but it is not an Animal.
EDIT
the above would only be relevant if there was a setter defined in the interface. Because there is not, the proper answer here is that the desired functionality cannot be achieved in C#; it is a feature not included in the language.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Why isn't there generic variance for classes in C# 4.0?
Example:
interface foo<out T> where T : BaseThing { }
compiles
class foo<out T> where T : BaseThing { }
does not.
Is this just unsupported, or is there some reason why it can never work or doesn't make logical sense?
Edit: Here's what I Wanted to do in case someone was wondering...
class BaseThing { }
class DerivedThing : BaseThing { }
class foo<out T> where T : BaseThing { }
class bar : foo<DerivedThing> { }
private void test()
{
foo<BaseThing> fooInstance = new bar();
}
The out modifier tells the compiler that the type parameter can be covariant. This means that use of types of T could be more-derived. Since you're using a specific class (e.g. a constructed generic type), there is only one instance of the concrete type so there is not a more-derived type in play. Multiple types can implement an interface like foo, which means you're dealing with potentially different types of T, in which case one of those T types may be more-derived.
I have an interface that has a generic method with two type parameters. I want to partially explicitly implement that generic method in a class. Is this possible? Some example code below:
public interface ISomeInterface
{
TResultType Results<TResultsType,TSearchCriteriaType>(TSearchCriteriaType searchCriteria);
}
public class SomeConcrete : ISomeInterface
{
public TResultsType Results<TResultsType, ConcreteSearchCriteria>(ConcreteSearchCriteria searchCriteria)
{
return (TResultsType)Results;
}
}
Do I have to explicitly implement both type parameters to make this work?
Do I have to explicitly implement both type parameters to make this work?
In order to implement this interface, your class must allow ANY types to be used for that method. (Any types which fit the constraints defined in the interface, which, in this case, since there are no constraints, means any type.)
You can't restrict the interface within a specific class implementing it, since this is a generic method (not a generic type), and there is no constraints which cause this to work properly.
In order to do what you wish, I think, you'd need to make the interface generic:
public interface ISomeInterface<TSearchCriteriaType>
{
TResultType Results<TResultsType>(TSearchCriteriaType searchCriteria);
}
You can then implement it as:
public class SomeConcrete : ISomeInterface<ConcreteSearchCriteria>
{
public TResultsType Results<TResultsType>(ConcreteSearchCriteria searchCriteria)
{
var results = GenerateResults();
return (TResultsType)results;
}
}
By making the interface generic on the search criteria, you allow your class to implement it based on a specific type for the search criteria.
I think this is what you're looking for:
public class SomeConcrete : ISomeInterface
{
TResultsType ISomeInterface.Results<TResultsType, TSearchCriteriaType>(TSearchCriteriaType searchCriteria)
{
return Results<TResultsType>((ConcreteSearchCriteria)(object)searchCriteria);
}
public TResultsType Results<TResultsType>(ConcreteSearchCriteria searchCriteria)
{
// return something
}
}
Note that this will fail at run time, though the compiler can see no problems:
ISomeInterface someInterface = new SomeConcrete();
var result = someInterface.Results<object, SomeOtherSearchCriteria>(null);
More generally, you're no longer implementing the interface as might be expected. You might want to make it ISomeInterface<TResultsType, TSearchCriteriaType>, then make SomeConcrete<TResultsType> : ISomeInterface<TResultsType, ConcreteSearchCriteria>. That way you are implementing the interface as expected, declaring it all strongly.
You can't specialize a generic method like that. If it's generic in the interface, it has to be generic in the implementing classes. If you move the type parameters to the interface itself, then you can do it (and you can also have variance annotations, making it covariant in the result type and contravariant in the search criteria type):
public interface ISomeInterface<out TResultsType, in TSearchCriteriaType> {
TResultsType Results(TSearchCriteriaType searchCriteria);
}
public class SomeConcrete<TResultsType> :
ISomeInterface<TResultsType, ConcreteSearchCriteria> {
public TResultsType Results(ConcreteSearchCriteria searchCriteria) {
...
}
}
Now your class is generic on the result type.
You can also have a class specialized in a certain result type, and generic in the criteria type:
public class IntSearcher<TSearchCriteriaType> :
ISomeInterface<int, TSearchCriteriaType> {
public int Results(TSearchCriteriaType searchCriteria) {
...
}
}
I think this is a more consistent design, since both generic parameters are at the same level (and it looks like they should go together). But only you can say whether this is better for your code or not.
Yes, I think you do, otherwise you are not directly implementing the interface.
Your Implementation is not as generic as the interface contract.
Given these base classes and interfaces
public abstract class Statistic : Entity, IStatistic
{
protected abstract IStatisticsRepository<IStatistic> Repository {get;}
...
public class AverageCheckTime : Statistic
...
public interface IStatisticsRepository<T> : IRepository<T> where T : IStatistic
...
public interface IAverageCheckTimeRepository : IStatisticsRepository<AverageCheckTime>
...
public class AverageCheckTimeRepository : StatisticRepository<AverageCheckTime>, IAverageCheckTimeRepository
...
public class RepositoryFactory
{
public static IAverageQueueTimeRepository AverageQueueTimeRepository
{
get { return CurrentServiceLocator.GetInstance<IAverageQueueTimeRepository>(); }
}
Why does AverageCheckTime's implementation throw an invalid cast exception:
protected override IStatisticsRepository<IStatistic> Repository
{
get { return (IStatisticsRepository<IStatistic>)RepositoryFactory.AverageCheckTimeRepository; }
}
How do I cast an instance of IAverageCheckTimeRepository as an IStatisticsRepository<IStatistic> which I assumed it already was?
OK, I've made these changes...which makes me wonder if I've gone over the top with the generics in the first place
public interface IStatisticsHelper
{
void GenerateStatistics();
List<IStatistic> BuildReport();
}
...
public interface IStatisticsRepository<T> : IRepository<T>, IStatisticsHelper where T : IStatistic
{
}
...
public abstract class Statistic : Entity, IStatistic
{
protected abstract IStatisticsHelper Repository { get; }
...
public class AverageCheckTime : Statistic
{
protected override IStatisticsHelper Repository
{
get { return RepositoryFactory.AverageCheckTimeRepository; }
}
No, C# 3 does not support generic variance. C# 4 does, but you would have to declare that IStatisticsRepository is covariant in T:
public interface IStatististicsRepository<out T> : IRepository<T>
where T : IStastistic
Variance isn't safe in general - it depends on how the generic type parameter is used. C# 4 supports both covariance and contravariance for type arguments which are reference types, but only when the generic type involved is an interface or a delegate, and only when the type parameter is used in the appropriate way within the interface/delegate.
Without seeing the declaration for IRepository<T>, we can't tell whether or not it's safe. For example, if IRepository<T> contains a method like this:
void Save(string id, T value);
then it wouldn't be safe, because you'd be able to write:
IStatisticsRepository<IStatistic> repo = RepositoryFactory.AverageCheckTimeRepository;
IStatistic foo = new SomeOtherStastisticType();
repo.Save("Foo", foo);
That would be trying to save a SomeOtherStatisticType value in an AverageCheckTimeRepository, which violates type safety. It's only safe to make the interface covariant in T if values of type T only come "out" of the interface. (There are some wrinkles around exactly what that means, mind you...)
For a lot more information on this, see Eric Lippert's blog series on the topic.