A colleague of mine posted a question on an internal forum which got me thinking about whether this was possible through C#. Basically, he's got an interface as follows:
public interface IProvider<T>
{
T GetT();
}
Is it possible to use something that implements that interface as a type parameter to another generic class and have access to the type T without re-specifying it? For example:
public class Foo<P> where P : IProvider<T>
{
P p;
T GetInnerT() { return p.GetT(); }
}
This does not compile, because the type T is not defined and hence can't be used as a parameter for IProvider. Is something like this even possible? Just curious!
No, he'd need Foo to be generic in both P and T:
public class Foo<P, T> where P : IProvider<T>
otherwise there's no T for the IProvider<T> constraint to be aware of - the fact that it's part of the declaration of IProvider<T> is coincidental. The T in the above declaration is entirely separate (from the compiler's point of view) from the T in IProvider<T>. For example, this would be an equivalent declaration:
public class Foo<TProvider, TProvided> where TProvider : IProvider<TProvided>
Another thing to bear in mind is that an implementation of IProvider<T> could implement it multiple times, for different type arguments:
public class BigProvider : IProvider<string>, IProvider<int>, IProvider<char>
Now what would Foo<BigProvider> mean? It would be ambiguous... whereas with my declaration above, you'd do:
var x = new Foo<BigProvider, int>();
to mean the int-providing aspect of BigProvider.
No, it's not possible, because your definition doesn't allow for the provision of type T when declaring Foo. Consider:
var x = new Foo<string>();
What is T in this case? There is no way to know at compile-time nor at runtime.
As you yourself said: the type T is not defined, and for the class to be valid you have to provide a means of definition. For example:
public class Foo<P, T> where P : IProvider<T>
Related
I'm new to C# and programming in general and wondering about the notation <T>
The question has been asked before, for example here: What does "T" mean in C#?
I just wanted to get some clarification to extend on that.
Can <T> be anything? I understand it is a naming convention by MS and could be named whatever you want, and it is for generic types, int, bools or whatever - but can it extend beyond that?
Can I pass an entire function/method to it just for the sake of it? Or is it strictly for return types if that makes sense?
Generic parameters, specified using the <> notation indicate the type. It is up to the defining class whether or not to limit the types that can be supplied. For example you could have a base class
public class Animal
{}
and a several derived classes
public class Dog : Animal
{}
public class Cat : Animal
{}
You could then make human a class like this
public class Human
{
public void AddPet<T>() where T: Animal
{}
}
In this case the T is constrained to be only things that inherit from Animal
So, you could do this
var me = new Human();
me.AddPet<Dog>();
but not this
var me = new Human();
me.AddPet<Human>();
This is a way of declaring new types/classes that can use another type inside initially not defined. But whenever you instantiate new object of that class, you can to specify that type and it will only be used for that object.
In short each of object of that class might have different type for that value on instantiation.
public class Test<T>
{
public T a;
}
This is a generic class where the type of a isn't known at this stage, but let's see example objects instantiations of that class:
var obj1 = new Test<int>(); // a will be of type int
var obj2 = new Test<List<string>>(); // a will be a list of strings
Type is dependent on object instantiation...
Update: Check docs for more details about generics https://learn.microsoft.com/en-us/dotnet/csharp/programming-guide/generics/generic-classes
T represents generic type. It could be any type (user/system defined).
https://learn.microsoft.com/en-us/dotnet/api/system.collections.generic?view=net-6.0
Please mark this answer, if it is useful to you.
Item class
public class Item
{
public bool Check(int value) { ... }
}
Base abstract class with generic type constraint
public abstract class ClassBase<TItem>
where TItem : Item
{
protected IList<TItem> items;
public ClassBase(IEnumerable<TItem> items)
{
this.items = items.ToList();
}
public abstract bool CheckAll(int value);
}
Inherited class without constraints
public class MyClass<TItem> : ClassBase<TItem>
{
public override bool CheckAll(int value)
{
bool result = true;
foreach(TItem item in this.items)
{
if (!item.Check(value)) // this doesn't work
{
result = false;
break;
}
}
return result;
}
}
I would like to know why aren't generic type constraints inheritable? Because if my inherited class inherits from base class and passes over its generic type which has a constraint on the base class it automatically means that generic type in inherited class should have the same constraint without explicitly defining it. Shouldn't it?
Am I doing something wrong, understanding it wrong or is it really that generic type constraint aren't inheritable? If the latter is true, why in the world is that?
A bit of additional explanation
Why do I think that generic type constraints defined on a class should be inherited or enforced on child classes? Let me give you some additional code to make it bit less obvious.
Suppose that we have all three classes as per above. Then we also have this class:
public class DanteItem
{
public string ConvertHellLevel(int value) { ... }
}
As we can see this class does not inherit from Item so it can't be used as a concrete class as ClassBase<DanteItem> (forget the fact that ClassBase is abstract for now. It could as well be a regular class). Since MyClass doesn't define any constraints for its generic type it seems perfectly valid to have MyClass<DanteItem>...
But. This is why I think generic type constraints should be inherited/enforced on inherited classes just as with member generic type constraints because if we look at definition of MyClass it says:
MyClass<T> : ClassBase<T>
When T is DanteItem we can see that it automatically can't be used with MyClass because it's inherited from ClassBase<T> and DanteItem doesn't fulfill its generic type constraint. I could say that **generic type on MyClass depends on ClassBase generic type constraints because otherwise MyClass could be instantiated with any type. But we know it can't be.
It would be of course different when I would have MyClass defined as:
public class MyClass<T> : ClassBase<Item>
in this case T doesn't have anything to to with base class' generic type so it's independent from it.
This is all a bit long explanation/reasoning. I could simply sum it up by:
If we don't provide generic type constraint on MyClass it implicitly implies that we can instantiate MyClass with any concrete type. But we know that's not possible, since MyClass is inherited from ClassBase and that one has a generic type constraint.
I hope this makes much more sense now.
ANOTHER UPDATE:
This question was the subject of my blog in July 2013. Thanks for the great question!
UPDATE:
I've given this some more thought and I think the problem is that you don't want inheritance at all. Rather, what you want is for all constraints that must be placed on a type parameter in order for that type parameter to be used as a type argument in another type to be automatically deduced and invisibly added to the declaration of the type parameter. Yes?
Some simplified examples:
class B<T> where T:C {}
class D<U> : B<U> {}
U is a type parameter that is used in a context where it must be C. Therefore in your opinion the compiler should deduce that and automatically put a constraint of C on U.
What about this?
class B<T, U> where T : X where U : Y {}
class D<V> : B<V, V> {}
Now V is a type parameter used in a context where it must be both X and Y. Therefore in your opinion the compiler should deduce that and automatically put a constraint of X and Y on V. Yes?
What about this?
class B<T> where T : C<T> {}
class C<U> : B<D<U>> where U : IY<C<U>> {}
class D<V> : C<B<V>> where V : IZ<V> {}
I just made that up, but I assure you that it is a perfectly legal type hierarchy. Please describe a clear and consistent rule that does not go into infinite loops for determining what all the constraints are on T, U and V. Don't forget to handle the cases where type parameters are known to be reference types and the interface constraints have covariance or contravariance annotations! Also, the algorithm must have the property that it gives exactly the same results no matter what order B, C and D appear in source code.
If inference of constraints is the feature you want then the compiler has to be able to handle cases like this and give clear error messages when it cannot.
What is so special about base types? Why not actually implement the feature all the way?
class B<T> where T : X {}
class D<V> { B<V> bv; }
V is a type parameter used in a context where it must be convertible to X; therefore the compiler should deduce this fact and put a constraint of X on V. Yes? Or no?
Why are fields special? What about this:
class B<T> { static public void M<U>(ref U u) where U : T {} }
class D<V> : B<int> { static V v; static public void Q() { M(ref v); } }
V is a type parameter used in a context where it can only be int. Therefore the C# compiler should deduce this fact and automatically put a constraint of int on V.
Yes? No?
You see where this is going? Where does it stop? In order to implement your desired feature properly the compiler must do whole-program analysis.
The compiler does not do this level of analysis because that is putting the cart before the horse. When you construct a generic, you are required to prove to the compiler that you've satisfied the constraint. It's not the compiler's job to figure out what you meant to say and work out what further set of constraints satisfy the original constraint.
For similar reasons, the compiler also does not attempt to automatically infer variance annotations in interfaces on your behalf. See my article on that subject for details.
http://blogs.msdn.com/b/ericlippert/archive/2007/10/29/covariance-and-contravariance-in-c-part-seven-why-do-we-need-a-syntax-at-all.aspx
Original answer:
I would like to know why aren't generic type constraints inheritable?
Only members are inherited. A constraint is not a member.
if my inherited class inherits from base class and passes over its generic type which has a constraint on the base class it automatically means that generic type in inherited class should have the same constraint without explicitly defining it. Shouldn't it?
You're just asserting how something should be, without providing any explanation of why it should be that way. Explain to us why you believe that the world should be that way; what are the benefits and what are the drawbacks and what are the costs?
Am I doing something wrong, understanding it wrong or is it really that generic type constraint aren't inheritable?
Generic constraints are not inherited.
If the latter is true, why in the world is that?
Features are "not implemented" by default. We don't have to provide a reason why a feature is not implemented! Every feature is not implemented until someone spends the money to implement it.
Now, I hasten to note that generic type constraints are inherited on methods. Methods are members, members are inherited, and the constraint is a part of the method (though not part of its signature). So the constraint comes along with the method when it is inherited. When you say:
class B<T>
{
public virtual void M<U>() where U : T {}
}
class D<V> : B<IEnumerable<V>>
{
public override void M<U>() {}
}
Then D<V>.M<U> inherits the constraint and substitutes IEnumerable<V> for T; thus the constraint is that U must be convertible to IEnumerable<V>. Note that C# does not allow you to restate the constraint. This is in my opinion a misfeature; I would like to be able to restate the constraint for clarity.
But D does not inherit any kind of constraint on T from B; I don't understand how it possibly could. M is a member of B, and is inherited by D along with its constraint. But T is not a member of B in the first place, so what is there to inherit?
I'm really not understanding at all what feature it is that you want here. Can you explain with more details?
Below is a scenario where the implicit nature of this behavior causes different behavior than expected:
I recognize that this scenario may seem extravagant in the amount of setup, but this is just one example of where this behavior might cause a problem. Software applications can be complicated, so even though this scenario may seem complicated, I wouldn't say that this can't happen.
In this example there is an Operator class that implements two similar interfaces: IMonitor and IProcessor. Both have a start method and an IsStarted property, but the behavior for each interface within the Operator class is separate. I.e. there is a _MonitorStarted variable and a _ProcessorStarted variable within the Operator class.
MyClass<T> derives from ClassBase<T>. ClassBase has a type constraint on T that it must implement the IProcessor interface, and according to the suggested behavior MyClass inherits that type constraint.
MyClass<T> has a Check method, which is built with the assumption that it can get the value of the IProcessor.IsStarted property from the inner IProcessor object.
Suppose someone changes the implementation of ClassBase to remove the type constraint of IProcessor on the generic parameter T and replace it with a type contraint of IMonitor. This code will silently work, but will produce different behavior. The reason is because the Check method in MyClass<T> is now calling the IMonitor.IsStarted property instead of the IProcessor.IsStarted property, even though the code for MyClass<T> hasn't changed at all.
public interface IMonitor
{
void Start();
bool IsStarted { get; }
}
public interface IProcessor
{
void Start();
bool IsStarted { get; }
}
public class Operator : IMonitor, IProcessor
{
#region IMonitor Members
bool _MonitorStarted;
void IMonitor.Start()
{
Console.WriteLine("IMonitor.Start");
_MonitorStarted = true;
}
bool IMonitor.IsStarted
{
get { return _MonitorStarted; }
}
#endregion
#region IProcessor Members
bool _ProcessorStarted;
void IProcessor.Start()
{
Console.WriteLine("IProcessor.Start");
_ProcessorStarted = true;
}
bool IProcessor.IsStarted
{
get { return _ProcessorStarted; }
}
#endregion
}
public class ClassBase<T>
where T : IProcessor
{
protected T Inner { get; private set; }
public ClassBase(T inner)
{
this.Inner = inner;
}
public void Start()
{
this.Inner.Start();
}
}
public class MyClass<T> : ClassBase<T>
//where T : IProcessor
{
public MyClass(T inner) : base(inner) { }
public bool Check()
{
// this code was written assuming that it is calling IProcessor.IsStarted
return this.Inner.IsStarted;
}
}
public static class Extensions
{
public static void StartMonitoring(this IMonitor monitor)
{
monitor.Start();
}
public static void StartProcessing(this IProcessor processor)
{
processor.Start();
}
}
class Program
{
static void Main(string[] args)
{
var #operator = new Operator();
#operator.StartMonitoring();
var myClass = new MyClass<Operator>(#operator);
var result = myClass.Check();
// the value of result will be false if the type constraint on T in ClassBase<T> is where T : IProcessor
// the value of result will be true if the type constraint on T in ClassBase<T> is where T : IMonitor
}
}
I think you're confused becuase you're declaring you derived class with TItem as well.
If you think about it if you were using Q instead so.
public class MyClass<Q> : BaseClass<Q>
{
...
}
Then how is it to be determined that Q is of the type item?
You need to add the constraint to the derived classes Generic Type as well so
public class MyClass<Q> : BaseClass<Q> were Q : Item { ... }
Because the ClassBase has a constraint on his template (should by typeof Item), you have to add this constraint to MyClass too.
If you don't do this, you could create a new instance of MyClass, where the template isn't a type of Item. When creating the base class, it will fail.
[edit]
Hmm now a re-read your question, and I see your code does compile? Ok.
Well, im MyClass you don't know the basetype of this.items, so you can't call the Check method.
this.items is of the type IList, and in your class, TItem isn't specified, thats why the class doesn't understand the Check method.
Let me counter your question, why don't you want to add the constraint to your MyClass class? Given any other class type as template to this class, would result in an error. Why not prevent this errors by adding a constraint so it will fail compiletime.
I'm working on a small class library at work, and it naturally involves using generics for this task. But there is this thing that I don't really understand with generics:
Why would I need to use generic type parameters, and then constrain the the type parameter to a specific base class or interface.
Here's an example to what I mean:
public class MyGenericClass<T> where T : SomeBaseClass
{
private T data;
}
And here's the implementation without generics
public class MyClass
{
private SomeBaseClass data;
}
Are these two definitions the same (if yes, then i don't see the advatage of using generics here)?
If not, what do we benefit from using generics here?
As with almost all uses of generics, the benefit comes to the consumer. Constraining the type gives you the same advantages that you get by strongly typing your parameter (or you can do other things like ensure that there's a public parameterless constructor or ensure that it's either a value or reference type) while still retaining the niceties of generics for the consumer of your class or function.
Using generics also, for example, allows you to obtain the actual type that was specified, if that's of any particular value.
This example is a little contrived, but look at this:
public class BaseClass
{
public void FunctionYouNeed();
}
public class Derived : BaseClass
{
public void OtherFunction();
}
public class MyGenericClass<T> where T: BaseClass
{
public MyGenericClass(T wrappedValue)
{
WrappedValue = wrappedValue;
}
public T WrappedValue { get; set; }
public void Foo()
{
WrappedValue.FunctionYouNeed();
}
}
...
var MyGenericClass bar = new MyGenericClass<Derived>(new Derived());
bar.Foo();
bar.WrappedValue.OtherFunction();
The difference is that the former defines the new class as a specific type; the latter simply defines a plain class with a field of that type.
It's all about type safety. Using generics you can return a concrete type (T) instead of some base type which defines the API you need in your generic class. Therefore, the caller of your method won't have to cast the result to the concrete type (which is an error-prone operation).
The main difference is in usage. In the first case, the usage can have:
MyGenericClass<SomeDerivedClass> Variable
Variable.data.SomeDerivedProperty = X
And so that when you use that class, you can still access anything from SomeDerivedClass without casting back to it.
The second example will not allow this.
MyClass.data = SomeDerivedClassInstance
MyClass.data.SomeDerivedProperty = X //Compile Error
((SomeDerivedClass)MyClass.data).SomeDerivedProperty = X //Ewwwww
You will have to cast back up to the SomeDerivedClass (which is unsafe) to use something specific to the derived class.
I don't think that there is a huge amount of difference except that the generic version is constraining your Class, whereas the second is just a constraint on a member of the class. If you added more members and methods to your first Class, you would have the same constraint in place.
Consider, I have the following 3 classes / interfaces:
class MyClass<T> { }
interface IMyInterface { }
class Derived : IMyInterface { }
And I want to be able to cast a MyClass<Derived> into a MyClass<IMyInterface> or visa-versa:
MyClass<Derived> a = new MyClass<Derived>();
MyClass<IMyInterface> b = (MyClass<IMyInterface>)a;
But I get compiler errors if I try:
Cannot convert type 'MyClass<Derived>' to 'MyClass<IMyInterface>'
I'm sure there is a very good reason why I cant do this, but I can't think of one.
As for why I want to do this - The scenario I'm imagining is one whereby you ideally want to work with an instance of MyClass<Derived> in order to avoid lots of nasty casts, however you need to pass your instance to an interface that accepts MyClass<IMyInterface>.
So my question is twofold:
Why can I not cast between these two types?
Is there any way of keeping the niceness of working with an instance of MyClass<Derived> while still being able to cast this into a MyClass<IMyInterface>?
This does not work because C# only supports covariance on the type parameters of interfaces and delegates. If your type parameter exists only in output positions (i.e. you only return instances of it from your class and don't accept it as an argument) you could create an interface like this:
interface IClass<out T> { }
class MyClass<T> : IClass<T> { }
Which would allow you to do this:
IClass<Derived> a = new MyClass<Derived>();
IClass<IMyInterface> b = a;
Honestly that is about as close as you are going to get and this requires the C# 4 compiler to work.
The reason you cannot do this in general is because most classes are not simple empty examples. They have methods:
class MyClass<T>
{
static T _storage;
public void DoSomethingWith(T obj)
{
_storage = obj;
}
}
interface IMyInterface { }
class Derived : IMyInterface { }
MyClass<Derived> a = new MyClass<Derived>();
Now, a has a method DoSomethingWith that accepts a Derived and stores it in a static variable of type Derived.
MyClass<IMyInterface> b = (MyClass<IMyInterface>)a;
If that was allowed, b would now appear to have a method DoSomethingWith that accepts anything that implements IMyInterface, and would then internally attempt to store it in a static variable of type Derived, because it's still really the same object referred to by a.
So now you'd have a variable of type Derived storing... who knows what.
I have an interface with a single generic type parameter:
public interface IDriveable<T> where T : ITransmission { ... }
I also have a class that has a type parameter that needs to be of that interface type:
public class VehicleFactory<T> where T : /* ??? */
There is a problem here with this declaration. I can't put "IDriveable", because that has no type parameters and doesn't match the type signature of IDriveable. But I also don't want to put IDriveable<U> either, because then VehicleFactory has to know what kind of IDriveable it's getting. I want VehicleFactory to accept any kind of IDriveable.
The proposed solution a coworker had was to use:
public class VehicleFactory<T, U> where T : IDriveable<U>
But I don't like this, since it's redundant. I have to say the "U" type twice:
var factory = new VehicleFactory<IDriveable<AllWheelDrive>, AllWheelDrive>();
What should go in the question marks?
What is VehicleFactory going to do with T? Does it actually need the constraint in order to work, or is it just for the sake of developer sanity checking?
One common way round this is to declare a non-generic interface (IDriveable) and then make your generic one extend that:
public interface IDriveable {}
public interface IDriveable<T> : IDriveable {}
public class VehicleFactory<T> where T : IDriveable
If you do want the factory to be able to do things with T, you could put any interface members from IDriveable<T> which don't care about T into the nongeneric IDriveable.
Does this work for you?
public class VehicleFactory<T, U> where T : IDriveable<U>
This will let the factory know what types the driveables are.
You can define the VehicleFactory with 2 types generics and anign one of them to the interface. something like:
public class VehicleFactory<T1,T2> where T1 : IDriveabel<T2>
I hope that ist not what you ment with the Idriveable. I guess the U is a specific type. Like String, etc.
You can shorten often used cases like this:
interface IDriveableAllWheel : IDriveable<AllWheelDrive>
{}
var factory = new VehicleFactory<IDriveableAllWheel, AllWheelDrive>();
or even
class AllWheelFactory : VehicleFactory<IDriveableAllWheel, AllWheelDrive>
{}
also see a kvb's answer in Calling Generic Property In Generic Class From Interface Implemented By Generic Class for possible workaround.