I was wondering if there's a way to accomplish the following:
In my project, I have defined an interface, let's say IFruit. This interface has a public method GetName(). I also declare an interface IApple which implements IFruit and exposes some other method like GetAppleType() or something. There are more fruits like IBanana, ICherry, whatever.
Now on the outside, I want only to be able to use the actual fruit implementations and not IFruit itself. But I cannot declare the IFruit interface as private or internal, since the inherited interfaces will then say "cannot implement because the base class is less accessible".
I know this is possible with abstract implementations, but that's not an option in this case: I really need to use interfaces. Is there such an option?
Update
I guess my example need some clarification :) I use MEF to load interface implementations. The loaded collections are based upon IApple, IBanana, ICherry, etc. But IFruit itself is useless, I can't use classes based on only that interface. So I was looking for a way to prevent other developers from implementing solely IFruit, thinking that their class will be loaded (which it won't). So basically, it comes down to:
internal interface IFruit
{
public string GetName();
}
public interface IApple : IFruit
{
public decimal GetDiameter();
}
public interface IBanana : IFruit
{
public decimal GetLenght();
}
But that won't compile due to the less accessible base interface.
One way that you can guarantee this doesn't happen unintentionally is to make IFruit internal to your assembly and then use some adaptor to wrap the type appropriately:
public interface IApple { string GetName(); }
public interface IBanana { string GetName(); }
internal interface IFruit { string GetName(); }
class FruitAdaptor: IFruit
{
public FruitAdaptor(string name) { this.name = name; }
private string name;
public string GetName() { return name; }
}
// convenience methods for fruit:
static class IFruitExtensions
{
public static IFruit AsFruit(this IBanana banana)
{
return new FruitAdaptor(banana.GetName());
}
public static IFruit AsFruit(this IApple apple)
{
return new FruitAdaptor(apple.GetName());
}
}
Then:
MethodThatNeedsFruit(banana.AsFruit());
You could also easily extend this to lazily call GetName on the adapted object, if the name could change over time.
Another option could be to have a DEBUG-only check that does load all IFruit implementers, and then throws an exception if one of them doesn't actually implement IBanana/IApple. Since it sounds like these classes are for internal use inside your company, this should stop anyone from accidentally implementing the wrong thing.
It isn't really possible to do what you're trying, but you can put people off using the IFruit interface with an [Obsolete] attribute, with message to say why.
On your IBanana, IApple, ... interfaces, disable the obsolete warning from appearing.
[Obsolete]
public interface IFruit {
...
}
#pragma warning disable 612
public interface IBanana : IFruit {
...
}
#pragma warning restore 612
If you have somewhow in your code (assuming that I correctly understand your state), something like this:
public class WaterMellon : IFruit, IVegetables...
{
}
and you want to be able to consumer of your framework access only to a methods of IFruit, there is no other known method to me then simply cast.
IFruit fruit = new WaterMelon();
fruit. //CAN ACCESS ONLY TO FRUIT IMPLEMNTATION AVAILABLE IN WATERMELON
If this is not what you're asking for, please clarify.
Related
In the following sample class "SomeClass" does not implement "ISomeInterface". Why can't I implement this by passing a more derived interface which does implement the base requirement. Whatever instance would be passed it would still implement the base, am I missing something?
namespace Test
{
public interface IBaseInterface
{
void DoBaseStuff();
}
public interface IChildInterface : IBaseInterface
{
void DoChildStuff();
}
public interface ISomeInterface
{
void DoSomething(IBaseInterface baseInterface);
}
public class SomeClass : ISomeInterface
{
public void DoSomething(IChildInterface baseInterface)
{
}
}
}
This restriction exists because the ISomeInterface expects that any IBaseInterface will satisfy the contract. That is, if you have the following:
public interface IBase {}
public interface IChildA : IBase {}
public interface IChildB : IBase {}
And an interface that expects IBase:
public interface IFoo { void Bar(IBase val); }
Then restricting this in a derived class as you would like:
public class Foo : IFoo { public void Bar(IChildA val) {} }
Would create the following problem:
IChildB something = new ChildB();
IFoo something = new Foo();
something.Bar(something); // This is an invalid call
As such, you're not implementing the contract you said you would.
In this situation, you have two simple options:
Adjust IFoo to be generic, and accept a T that is a derivation of IBase:
public interface IFoo<T> where T : IBase { void Bar(T val); }
public class Foo : IFoo<IChildA> { public void Bar(IChildA val) {} }
Of course, this means that Foo can no longer accept any IBase (including IChildB).
Adjust Foo to implement IFoo, with an additional utility method for void Bar(IChildA val):
public class Foo : IFoo
{
public void Bar(IBase val) {}
public void Bar(IChildA val) {}
}
This has an interesting side-effect: whenever you call ((IFoo)foo).Bar it will expect IBase, and when you call foo.Bar it will expect IChildA or IBase. This means it satisfies the contract, while also having your derived-interface-specific method. If you want to "hide" the Bar(IBase) method more, you could implement IFoo explicitly:
void IFoo.Bar(IBase val) { }
This creates even more inconsistent behavior in your code, as now ((IFoo)foo).Bar is completely different from foo.Bar, but I leave the decision up to you.
This means, with the second version in this section, that foo.Bar(new ChildB()); is now invalid, as IChildB is not an IChildA.
Why can't I implement this by passing a more derived interface which does implement the base requirement. Whatever instance would be passed it would still implement the base, am I missing something?
This is not allowed because of the reasoning I mentioned above, IFoo.Bar expects any IBase, whereas you want to further constrain the type to IChildA, which is not a super-interface of IBase, and even if it were it would not be allowed because it violates the interface implementation, though you could more easily define a second method at that point that does what you want.
Keep in mind that when you implement an interface, you subscribe to a contract, and C# will not let you violate that contract.
This violates the Liskov substitution principle.
ISomeInterface guarantees that the method can be called with any IBaseInterface instance. Your implementation cannot limit that to only accept IChildInterface interfaces.
From MSDN:
When a class or struct implements an interface, the class or struct must provide an implementation for all of the members that the interface defines
This method in the derived
void DoSomething(IChildInterface baseInterface)
Does not have the same signature as the one in the interface:
void DoSomething(IBaseInterface baseInterface)
IChildInterface and IBaseInterface are not the same types. Therefore your derived class does not implement all methods of the interface and you get the compilation error.
For a possible the logic behind having this as a restriction instead of the compiler understanding the inheritance see Liskov's substitution principle as in SLakes answer
You should change some interface to use some type which implements IBaseInterface,
then change the method signatures to use whichever child your SomeClass wants.
public interface ISomeInterface<TSomeChild> where TSomeChild : IBaseInterface
{
void DoSomething(TSomeChild baseInterface);
}
public class SomeClass : ISomeInterface<IChildInterface>
{
public void DoSomething(IChildInterface baseInterface)
{
}
}
If you could do that, then you could do this:
IAnimal cat = new Cat();
IAnimalTrainer dogTrainer = new DogTrainer();
dogTrainer.Train(cat);
An IAnimalTrainer can train any IAnimal. But a DogTrainer can only train Dogs. Thus it's illegal for DogTrainer to implement the IAnimalTrainer interface.
public interface IFoo
public class Foo : IFoo
public interface ILang<T> where T:IFoo
public class Lang : ILang<Foo>
But now, I want something like this with wildcard
public class CarroMontadora<T> where T:ILang<?>
Beacause this does not work with my interface:
public class MYCLASS<T> where T:ILang<IFoo>
And this is not too smart to do, because is not so generics:
public class MYCLASS<T,U> where T:ILang<U> where U: IFoo
You will have to add another generic parameter to your new class:
public class CarroMontadora<T1, T2>
where T2 : IFoo
where T1 : ILang<T2>
{
}
Good options already listed but let me give you another one.
Have both ILang<T> AND a simple marker interface ILang - use ILang to enforce that condition and make ILang<T> : ILang
It's not flawlessly type-safe but that API is pretty clear enough that it is unlikely that someone will screw it up.
You cannot have nested constraints on generics on C# without having both type parameters as part of the class (your third example), however you can enforce this in a cleaner fashion by good old method signatures.
For example
public class CarroMontadora<T> where T: IFoo {
public CarroMontadora(ILang<T> param) {
// do stuff
}
}
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.
I have an interface, ICheese, that I would like to be able to extend as I need. It is contained by class that is implementing the IFruit interface:
public interface IFruit<T> where T : ICheese
{
T : cheese { get; }
}
This was done to dodge the fact that C# doesn't have return type covariance; I want to be able to have any implementation of IFruit and know that it has some version of ICheese inside of it.
IFruit talks to a number of ICrackers, who have a reference to IFruit:
public interface ICracker<T> where T : ICheese
{
void Initialize(IFruit<T> fruit);
}
public abstract class Cracker<T> where T : ICheese
{
IFruit<T> fruit
public void Initialize(IFruit<T> fruit)
{
this.fruit = fruit;
}
}
This works so long as the implementation of ICracker specifically defines which kind of ICheese it is going to use:
public class Apple : IFruit<Cheddar> {/* Implementation of Class */}
public class Saltine : Cracker<Cheddar> {/* Implementation of Class */}
In that instance, calling Initialize(new Apple()) on an instance of Saltine works.
However, if we have an ICracker like this:
public class Ritz : Cracker<ICheese> { /* Implementation of Class */ }
I cannot pass an Apple into Rizt's Initialize() function, even though Cheddar inherits from ICheese. Is there a way where I can get this to work or a pattern that would accomplish the same goals? The basic idea is that I want ICrackers to be able to say "I need at least this version of ICheese" and be able to accept any kind of IFruit that provides at least that version of ICheese.
You can make IFruit covariant in T:
public interface IFruit<out T> where T : ICheese
{
T Cheese { get; }
}
I have an interface:
interface IFoo {
int foo(int bar);
}
Can I now extend an existing class to conform to the interface? Say class String. I know I can define the foo() method on strings. But is it possible to go further and tell the compiler that strings can be cast to IFoo?
You can do it with other classes, but not with System.String, because it is sealed.
If you wanted to do this to a non-sealed class, you could simply derive from it, add appropriate constructors, and put the interface as something that your new class implements.
interface IFoo {
int Size {get;}
}
// This class already does what you need, but does not implement
// your interface of interest
class OldClass {
int Size {get;private set;}
public OldClass(int size) { Size = size; }
}
// Derive from the existing class, and implement the interface
class NewClass : OldClass, IFoo {
public NewCLass(int size) : base(size) {}
}
When the class is sealed, your only solution of presenting it as some interface through composition: write a wrapper class implementing your interface, give it an instance of the sealed class, and write method implementations that "forward" calls to the wrapped instance of the target class.
I think the question could be restated as, 'can I use extension methods to make a sealed class implement an interface that it did not before?' As others point out, the String class is sealed. However, I think you have to name what interfaces a class implements in its declaration:
public someClass : IFoo
{
// code goes here
}
So you can't do this directly to String, not just because it is sealed, but because you do not have the source code for it.
The best you can do is make your own class that has-a String and use it like a string. Anything that a String does that you need to do you will either have to do on its String member (thus making it public), or you will have to wrap/reimplement the method you need:
public class betterString : IFoo
{
public String str {get; set;}
public foo(int i)
{
// implement foo
}
}
then later, when using it:
public void someMethod(betterString better)
{
better.foo(77);
System.Console.WriteLine(better.str);
}