I'm writing an interface and I want to declare a property that returns a generic collection. The elements of the collection should implement an interface. Is this possible and, if so, what is the syntax.
This doesn't compile, what's the right way to do this?
interface IHouse
{
IEnumerable<T> Bedrooms { get; } where T : IRoom
}
Thanks
Why use generics? Just do:
interface IHouse
{
IEnumerable<IRoom> Bedrooms { get; }
}
This is cleaner, and since you're already restricting to the interface, it will act nearly identically.
You have to mark the interface as generic as well:
interface IHouse<T> where T : IRoom
{
IEnumerable<T> Bedrooms { get; }
}
There are no good reasons why Microsoft chose to add this restriction. Under the hood properties are just a pair of methods, and generic methods are already supported.
I have hit this restriction a couple of times, and had to resort to adding explicit get and set methods instead:
interface IHouse
{
IEnumerable<T> GetBedrooms<T>() where T : IRoom;
}
Generic is for classes and methods, not properties. If a class/interface is generic, then you can use the type in the property. I agree with Reed's solution.
Related
I want to do something like this:
public class MyAttribute : Attribute
{
public Type<BaseAllowedType>[] handledTypes { get; set; }
//...elided...
}
That would then only allow someone adding the attribute with the types that they handle if those types inherit from BaseAllowedType.
Example:
[MyAttribute(handledTypes = SubType)]
public class MyClass
{
}
public class SubType : BaseAllowedType
{
}
Is something like this possible in C#? (It is in Java, but not sure if C# allows it)
In Java it would be public Class<? extends BaseAllowedType>[] handledTypes
First off, that is a property, not a field.
Second, properties cannot be genericized in C#, and neither can attributes. (And neither can fields, for that matter. Only classes, structs, interfaces, delegates and methods can be genericized.)
That said, we can take a stab at your question.
There are two ways to put a constraint on a generic type parameter. The first is to put a constraint on the declaration of a generic class or method:
class Foo<T> where T : Animal {}
Now Foo<Giraffe> or Foo<Animal> or Foo<Turtle> are legal, but Foo<string> is not.
Second, on interfaces and delegates you can put a covariance annotation:
interface IFoo<out T>
{
T SomeProperty { get; }
}
But T must be used only in output positions. With this constraint you can assign an IFoo<Giraffe> to a variable of type IFoo<Animal>, but not vice-versa.
Between those two features it is usually possible to build a reasonable type constraint in C#.
I know the title may be a little unclear, so I'll explain what I'm trying to do.
Note, this is more for educational reasons around the language's capabilities. In other words if this is possible, not should this be the way to go about it.
Consider the following generic class:
public class Foo<TId>
{
TId Id { get; set; }
}
Now consider concrete subclasses based on the above. Here's two examples using 'int' and 'string'...
public class IntFoo : Foo<int>
{
}
public class StrFoo : Foo<string>
{
}
And finally a generic that takes a Foo as a type parameter, and inherits from a Laa which takes its type parameter from Foo.
public class BaseClass<TFoo, TFooId> : Laa<TFooId>
{
}
public class Laa<TFooId>
{
}
Here's how you'd do the one based on an int and string, but note in addition to IntFoo and StrFoo, I have to also define int and foo explicitly...
public class IntFinal : BaseClass<IntFoo, int>
{
char somePropSpecificToIntFinal{ get; set; }
}
public class StrFinal : BaseClass<StrFoo, string>
{
char somePropSpecificToStrFinal{ get; set; }
}
Note that these 'final' classes are concrete types with their own properties which can't be reduced to a generic that takes a type (i.e. using a generic with the single type T, that then subclasses another generic that takes Foo and T as its arguments.
I'm wondering is if there's a way to have that type inferred so it can be written like so...
public class IntFinal : BaseClass<IntFoo>
{
}
public class StrFinal : BaseClass<StrFoo>
{
}
...and have the type for Laa implied from the generic specified on Foo. Here's a pseudo-code example of what I want.
public class BaseClass<TFoo> : Laa<TFoo.IdType>
{
}
So is that possible in C#?
Note, if this can't be done with classes, can it be done with interfaces?
Consider this...
interface IFoo
{
Type FoosType { get; }
}
public class Foo<TId> : Foo
{
TId Id { get; set; }
Type FoosType { get{ return TId } }
}
Then do this...
public class BaseClass<TFoo> : Laa<TFoo.FoosType>
where TFoo : Foo
{
}
(Note: FoosType would have to be static technically, and you can't inherit using statics so again, this is pseudo-code.)
If you constrained TFoo to IFoo, could you then use 'FoosType' as the type specifier when defining Laa?
You can't do that based on the C# specification. Type inference currently works for methods only and doesn't work for types (classes like your case)
Second rule that breaks your needed result is that you can't specify one generic type argument and infer the other, it is Provide all or Infer all case for methods.
C# specification:
1.6.3 Type parameters
When the generic class is used, type arguments must be provided for each of the type parameters
Your question is not very specific, and it's not clear what the actual constraints and requirements are. That said…
Type inference only occurs for generic methods, not generic types. So taking your question literally, the answer is no, there is no way to infer the type.
What might work for you is to use Foo<TId> in the class definition instead of IntFoo:
class BaseClass<TFooId> : Laa<TFooId>
{
public Foo<TFooId> Method() { ... }
}
(Of course, you can apply the type anywhere appropriate: fields, property types, etc.)
I.e. instead of coding the BaseClass type with two type parameters, just use the one that uniquely defines the interesting/useful elements of the Foo<TFooId> base class you're using, and then use that base type instead of the more-derived IntFoo
In your example, you have no constraints for the TFoo class, so it's not like BaseClass<TFoo, TFooId> was going to be able to use even the base type class members from Foo<TId> anyway. But even if you did mean to constrain TFoo to Foo<TFooId>, it seems likely you wouldn't really need to specify that type anyway.
If the above is not useful, then you need to add a lot more detail to your question, to explain precisely what is needed. Consider also that it's likely people have already gone down this road, and that if you express your question less about the mechanics of the implementation you think you need, you instead phrase it at a higher level, you might likely find existing questions on Stack Overflow or articles elsewhere that already address that broader question.
At the very least, if you are unable to find such references yourself, expressing your question that way may yield better answers faster.
See also XY Problem.
For example, suppose I want an ICar interface and that all implementations will contain the field Year. Does this mean that every implementation has to separately declare Year? Wouldn't it be nicer to simply define this in the interface?
Though many of the other answers are correct at the semantic level, I find it interesting to also approach these sorts of questions from the implementation details level.
An interface can be thought of as a collection of slots, which contain methods. When a class implements an interface, the class is required to tell the runtime how to fill in all the required slots. When you say
interface IFoo { void M(); }
class Foo : IFoo { public void M() { ... } }
the class says "when you create an instance of me, stuff a reference to Foo.M in the slot for IFoo.M.
Then when you do a call:
IFoo ifoo = new Foo();
ifoo.M();
the compiler generates code that says "ask the object what method is in the slot for IFoo.M, and call that method.
If an interface is a collection of slots that contain methods, then some of those slots can also contain the get and set methods of a property, the get and set methods of an indexer, and the add and remove methods of an event. But a field is not a method. There's no "slot" associated with a field that you can then "fill in" with a reference to the field location. And therefore, interfaces can define methods, properties, indexers and events, but not fields.
Interfaces in C# are intended to define the contract that a class will adhere to - not a particular implementation.
In that spirit, C# interfaces do allow properties to be defined - which the caller must supply an implementation for:
interface ICar
{
int Year { get; set; }
}
Implementing classes can use auto-properties to simplify implementation, if there's no special logic associated with the property:
class Automobile : ICar
{
public int Year { get; set; } // automatically implemented
}
Declare it as a property:
interface ICar {
int Year { get; set; }
}
Eric Lippert nailed it, I'll use a different way to say what he said. All of the members of an interface are virtual and they all need to be overridden by a class that inherits the interface. You don't explicitly write the virtual keyword in the interface declaration, nor use the override keyword in the class, they are implied.
The virtual keyword is implemented in .NET with methods and a so-called v-table, an array of method pointers. The override keyword fills the v-table slot with a different method pointer, overwriting the one produced by the base class. Properties, events and indexers are implemented as methods under the hood. But fields are not. Interfaces can therefore not contain fields.
Why not just have a Year property, which is perfectly fine?
Interfaces don't contain fields because fields represent a specific implementation of data representation, and exposing them would break encapsulation. Thus having an interface with a field would effectively be coding to an implementation instead of an interface, which is a curious paradox for an interface to have!
For instance, part of your Year specification might require that it be invalid for ICar implementers to allow assignment to a Year which is later than the current year + 1 or before 1900. There's no way to say that if you had exposed Year fields -- far better to use properties instead to do the work here.
The short answer is yes, every implementing type will have to create its own backing variable. This is because an interface is analogous to a contract. All it can do is specify particular publicly accessible pieces of code that an implementing type must make available; it cannot contain any code itself.
Consider this scenario using what you suggest:
public interface InterfaceOne
{
int myBackingVariable;
int MyProperty { get { return myBackingVariable; } }
}
public interface InterfaceTwo
{
int myBackingVariable;
int MyProperty { get { return myBackingVariable; } }
}
public class MyClass : InterfaceOne, InterfaceTwo { }
We have a couple of problems here:
Because all members of an interface are--by definition--public, our backing variable is now exposed to anyone using the interface
Which myBackingVariable will MyClass use?
The most common approach taken is to declare the interface and a barebones abstract class that implements it. This allows you the flexibility of either inheriting from the abstract class and getting the implementation for free, or explicitly implementing the interface and being allowed to inherit from another class. It works something like this:
public interface IMyInterface
{
int MyProperty { get; set; }
}
public abstract class MyInterfaceBase : IMyInterface
{
int myProperty;
public int MyProperty
{
get { return myProperty; }
set { myProperty = value; }
}
}
Others have given the 'Why', so I'll just add that your interface can define a Control; if you wrap it in a property:
public interface IView {
Control Year { get; }
}
public Form : IView {
public Control Year { get { return uxYear; } } //numeric text box or whatever
}
A lot has been said already, but to make it simple, here's my take.
Interfaces are intended to have method contracts to be implemented by the consumers or classes and not to have fields to store values.
You may argue that then why properties are allowed? So the simple answer is - properties are internally defined as methods only.
Interfaces do not contain any implementation.
Define an interface with a property.
Further you can implement that interface in any class and use this class going forward.
If required you can have this property defined as virtual in the class so that you can modify its behaviour.
Beginning with C# 8.0, an interface may define a default implementation for members, including properties. Defining a default implementation for a property in an interface is rare because interfaces may not define instance data fields.
https://learn.microsoft.com/en-us/dotnet/csharp/programming-guide/classes-and-structs/interface-properties
interface IEmployee
{
string Name
{
get;
set;
}
int Counter
{
get;
}
}
public class Employee : IEmployee
{
public static int numberOfEmployees;
private string _name;
public string Name // read-write instance property
{
get => _name;
set => _name = value;
}
private int _counter;
public int Counter // read-only instance property
{
get => _counter;
}
// constructor
public Employee() => _counter = ++numberOfEmployees;
}
For this you can have a Car base class that implement the year field, and all other implementations can inheritance from it.
An interface defines public instance properties and methods. Fields are typically private, or at the most protected, internal or protected internal (the term "field" is typically not used for anything public).
As stated by other replies you can define a base class and define a protected property which will be accessible by all inheritors.
One oddity is that an interface can in fact be defined as internal but it limits the usefulness of the interface, and it is typically used to define internal functionality that is not used by other external code.
I haven't found my use case in the existing questions so here I go.
I have an abstract class that has generics like this :
public abstract class aParameter<T>
{
private T _parameter;
public T Parameter
{
get { return _parameter;}
}
...
}
Then I have my "Type" classes like this :
public class IntParameter : aParameter<Int32>
{
public IntParameter(Int32 value)
{
_parameter = value;
}
}
public class TextParameter : aParameter<String>
{
public TextParameter(String value)
{
_parameter = value;
}
}
Now in my application code, I'd like to create a dictionary with the abstract generic class without specifying the generic type. Like this :
...
private Dictionary<Int32, aParameter> _paramDict = new Dictionary<Int32, aParameter>();
...
In a previous version, I used interfaces instead of abstract class for aParameter but I wanted to migrate it to abstract so that I could simplify my "Type" classes and not repeat identical code for each "Type".
Any thoughts here?
EDIT
I forgot to mention that it will be part of a library that is meant to be distributed among the company so I really need to secure the allowed types. I can't allow for just objects to be fed into my code.
You probably still going to need the non-generic interface (or nongeneric abstract base class as Brandon says), unless you drop to working with objects.
The reason is that aParameter<String> is not the same type as aParameter<int32> so you'll not be able to get them into the one dictionary without some help.
My preferred method around this is to declare a non-generic interface and use that as the basis for the dictionary, then implement it in each of your typed generics, and also in an abstract base class. That way you can inherit from the base class where possible, or just implement the interface if one of your types needs to inherit from something else - you get maximum flexibility.
Make a non-generic abstract base class and have your generic abstract class inherit from it.
The purpose of generic types is (among other things) that you can only write code once that can be used for multiple types. However, as C# is strongly types, the compiler needs to know what types it is dealing with in the generic class. This is why you have to specify the type of the generic (i.e. the type in angled brackets)
If you don't specify the type of the generic then, in your example, the compiler wouldn't know the type of the Parameter property.
Depending on exactly what you're doing with it, there are a number of possible approaches:
Use a Dictionary<int, object>.
When reading the values, you can use a bunch of if/elses to check the specific type of the object and convert to the appropriate type. e.g.
if(obj.GetType() == typeof(TextParameter))
{
TextParameter p = obj as TextParameter
// Do stuff
}
else if obj.GetType() == typeof(IntParameter))
{
IntParameter p = obj as IntParameter
// Do stuff
}
Have multiple dictionaries. Dictionary<int, TextParameter>, Dictionary<int, IntParameter>, etc.
If there are methods/properties in aParameter that are not type dependent then move them to a lower level non-generic abstract class. This could give you at least some of your functionality without having to resort to type conversions.
I need some advice/help on this, I can't see the wood from the trees any more.
It's a straight forward series of classes implementing some interfaces using generics.
Then I'm trying to cast the concrete types for example:
MyGenericObject<SomeObject> _obj;
IMyGenericObject<ISomeObject> _genObj = (IMyGenericObject<ISomeObject>)_obj;
// Invalid cast
I've read some articles about covariance and contravariance but not too clear why this wouldn't be possible, or how to get round it?
So, in this example:
public interface IMyObject<in T> where T : IBaseObject
{
T Activity { get; set; }
}
wouldn't work...
....because, you can't get and set the Activity property.
In this example, I needed to do:
public interface IMyObject<out T> where T : IBaseObject
{
T Activity { get; }
}
hope that helps someone, and thanks to all for help!
You can only do that if you declare the interface as having a covariant (out) parameter. You can only do that if the parameter is used covariantly.
For example, if the interface IMyGenericObject<T> has a method taking a T parameter, this prevents you from declaring the parameter as covariant. Conversely, if there is a method that returns a T, that prevents you from declaring the parameter as contravariant.
EDIT
In response to your comment on SLaks's answer, I'm tempted to repeat everything Eric Lippert has ever written on co- and contravariance. See http://blogs.msdn.com/b/ericlippert/archive/tags/Covariance+and+Contravariance/ and also his answers in SO (most recently https://stackoverflow.com/a/8380213/385844)
To summarize:
You can't cast IList<string> to IList<object> because it's legal to pass a FileInfo to an IList<object>, but it is not legal to pass it to an IList<string>.
You can't cast an IList<object> to an IList<string>, because it's legal to retrieve an item from an IList<string> and assign it to a string reference, but an IList<object> might contain a FileInfo, which can't be assigned to a string reference.
EDIT 2
Since you asked for advice, it's also possible to split your interfaces into co- and contravariant parts. To continue with the list example, you could have these interfaces
public interface ICovariantList<out T>
{
T this[int index] { get; }
//...
}
public interface IContravariantList<in T>
{
T this[int index] { set; }
void Add(T item);
//...
}
public class SomeList<T> : ICovariantList<T>, IContravariantList<T>
{
//...
}
This allows you to use the class covariantly or contravariantly, depending on the context.
You need to declare the interface as having a covariant (out) generic parameter.