I am experimenting with Generic Constraints. When declaring a constraint on a class as so:
public class DocumentPrinter<T> where T : IFooDoc
I am able to access method declared by IFooDoc within the methods of the DocumentPrinter class. However, if I make DocumentPrinter implement an interface that declares the contraint, eg:
public interface IDocumentPrinter<T> where T : IFooDoc
{
void Add(T fooDoc);
void PrintFoos();
}
and then declare DocumentPrinter as so:
public class DocumentPrinter<T>: IDocumentPrinter<T>
properties/methods of IFooDoc instances are no longer available within the methods of the Document printer. It seems that I must explicitly declare an interface constraint upon the class itself if I am to access members declared by the type.
Do I understand this correctly or is it possible to declare the constraint on the interface and to have that constraint realized by the class?
Do I understand this correctly or is it possible to declare the constraint on the interface and to have that constraint realized by the class?
Correct. You have1 to declare the constraint on the type parameters for the generic class as well. Just because you named the type parameter in DocumentPrinter<T> to have the same name as the type parameter in IDocumentPrinter<T> does not mean that they are the same types. When you declare
public class DocumentPrinter<T> : IDocumentPrinter<T>
you are in fact saying to use T that parameterizes DocumentPrinter<T> to parameterize IDocumentPrinter<T> and now they are the same types. But for that to be legal, T from DocumentPrinter<T> has to satisfy all constraints on the type parameter for IDocumentPrinter<T>. Thus, you must explicitly say that T satisfies T : IFooDoc.
1: Apparently I need to state this explicitly. If you don't do something the language specification requires you to do, your code won't compile. The language specification requires that when you parameterize a generic type, the type that you parameterize it with must satisfy all the constraints on that type parameter for that generic type. If you do not, your code won't compile.
properties/methods of IFooDoc instances are no longer available within the methods of the Document printer
Well, it's kind of irrelevant what IntelliSense tells you when your code doesn't compile.
If you want to implement IDocumentPrinter<T>, you have to satisfy its constraint. If you don't, your code won't compile.
Related
I've see generic methods with this where keyword like
public static void DoStuff<T>() where T :
where seems to restrict the type of T in a specific way.
Question: how does this where work? Is it only possible to ristrict to interfaces like where T : IComparable<T> or are there different ways aswell?
It's called a generic type parameter constraint
You can constraint T to inherit from a base class, implement an interface, be a reference type, be a value type, have a parameterless constructor...
The compiler checks that the type parameter meets these requirements.
In your case, the constraint allows to call IComparable<T> methods in the DoStuff method.
I'm looking at the source code for the MvcContrib Grid and see the class declared as:
public class Grid<T> : IGrid<T> where T : class
What does the where T : class bit do?
It is a generic type constraint.
In this case it means that the generic type (T) must be a reference type, that is class, interface, delegate, or array type.
Other constraints are listed here.
You can also constrain the generic type to inherit from a specific type (base class or interface)
Another examples would be
public A<T> where T : AnInterface
where AnInterface is a interface class. It means then, that T must implement this interface.
These constraints are important, so that the compiler knows the operations which are valid for the type. For example you can not call functions of T without telling the compiler what functions the type provides.
From the Docs http://msdn.microsoft.com/en-us/library/d5x73970.aspx
where T : class
The type argument must be a reference type; this applies also to any class, interface, delegate, or array type.
It is a constraint on the type argument which says that T can either be a class or an interface but not an enum or a struct. So T must be a reference type and not a value type.
Best Regards,
Oliver Hanappi
It restricts T to be a reference type, including any class, interface, delegate, or array type.
It's a generic type constraint. It specifies that the type T has to be a reference type, i.e. a class and not a structure.
you can apply restrictions to the kinds of types that client code can use for type arguments when it instantiates your class are called as Constraints on Type Parameters
E.g : where T : class
Here where T is the Type , The type argument must be a reference type; this applies also to any class, interface, delegate, or array type.
I'm trying to call a method with a definition similar to the following (simplified to avoid confusion):
public static void Register<T>(T value) where T : BaseClass, IInterface
This works fine so long as I have a class instance that defines both of those values. The problem occurs when I pass a `BaseClass' into a method and then try to use that instance in the above declaration. For example:
public class MyClass
{
public MyClass(BaseClass value)
{
Register(value);
}
}
I can pass and instance of a class that implements both BaseClass and IInterface into the constructor, but when I try to use that value in the Register method I get a compilation error stating:
The type 'BaseClass' cannot be used as type parameter 'T' in the generic type or method 'Register(T)'. There is no implicit reference conversion from 'BaseClass' to 'IInterface'.
If I change the type in the constructor like so:
public class MyClass
{
public MyClass(IInterface value)
{
Register(value);
}
}
I get an error stating:
The type 'IInterface' cannot be used as type parameter 'T' in the generic type or method 'Register(T)'. There is no implicit reference conversion from 'IInterface' to 'BaseClass'.
This seems like a bit of a catch-22. Is there a way that I can define the parameter to indicate that it must implement both BaseClass and IInterface?
As I was writing the question I came up with the answer and thought I would post it instead of deleting the question.
I just need to redefine the class:
public class MyClass<T> where T : BaseClass, IInterface
{
public MyClass(T value)
{
Register(value);
}
}
The solution given by Matt is an easy answer for situations where it is not necessary to store the passed-in object in a field or collection. If you need to persist the passed-in object, things get much harder. It's easy for a SomeClass<T> where T meets multiple constraints, to store items of class T and pass them as generics to routines with such constraints, but if a class has a method SomeMethod<TParam>(TParam thing) where TParam:IFoo,BaseBar, it won't have any field of type TParam. It could store thing into a field of type IFoo or BaseBar, but unless BaseBar implements IFoo, or all passed-in instances are going to derive from one particular BaseBar derivative which implements IFoo, there's no way to specify that a field's type will meet both constraints (if every instance does derive from one particular BaseBar derivative which implements IFoo, one could simply use that type as a single constraint, or for that matter not bother with generics at all—just use that as the parameter type).
There are ways of getting around these issues, either using Reflection, an interface pattern I call ISelf<T>, or some tricky nested callback interfaces. In some cases, though, it may be better to provide alternatives to methods that take double-constrained parameters (have the methods take a parameter of one constraint type, cast it to the other type, and accept the lack of compile-time safety).
In C#, given a generic type such as this:
interface IGenericType<T> where T : new()
And a descendant type, such as:
class GenericTypeImplementation<U> : IGenericType<U>
Why do we need to explicitly restrict the generic type U with all the restrictions of the parent type?
class GenericTypeImplementation<U> : IGenericType<U> where U : new()
Am I right in inferring that the issue is in the compiler computing the union of restrictions?
interface IGenericType<T> where T : new()
interface IGenericType2<T> where T : SomeOtherType
class GenericTypeImplementation<U> : IGenericType<U>, IGenericType2<U>
/* Hypothesis: Compiler can't infer U must be "SomeOtherType + new()" */
In my opinion, the compiler could be smart enough to infer the restrictions theoretically. But it shouldn't be so smart, because a too-smart compiler is sometimes dangerous. Developers always need a clear/explicit definition of everything. See this scenario:
(1) there is an interface IFoo<T> where T : new()
(2) a class Foo<T> : IFoo<T> and the new() constraint is added automatically by the compiler(brilliant!)
(3) the class Foo<T> is a very base class in the whole project, class A<T> : Foo<T>, and then class B<T> : A<T>...
(4) Now another developer can hardly realize there is such a constraint by looking into the definition of the class, he will get weird compiling errors(well that's acceptable). But what if they are invoked by reflection? Sometimes the program is correct, because the data meets the restriction by accident.
The compiler is able to to infer that U must be convertible to SomeOtherType and must have a default constructor. It will generate a compiler error for each constraint:
Error 1 The type 'U' must have a public parameterless constructor in order to use it as parameter 'T' in the generic type or method '....IGenericType<T>'
Error 2 The type 'U' must be convertible to '....SomeOtherType' in order to use it as parameter 'T' in the generic type or method '....IGenericType2<T>'
This will also happen with just one of those interfaces implemented as well. The class must successfully implement both interfaces in order to be compiled:
class GenericTypeImplementation<U> : IGenericType<U>, IGenericType2<U>
where U : SomeOtherType, new()
{...}
or as a non-generic type:
class GenericTypeImplementation : IGenericType<SomeType>, IGenericType2<SomeOtherType>
{...}
To mark a class as implementing an interface is not a way of specifying constraints on the generic type parameters of a class; it is a way of requiring that those constraints exist on a new type parameter or that they be satisfied by a supplied type.
Perhaps you could think of it this way: an interface is a constrained set of classes and a generic class is also a constrained set of classes. A generic interface is a constrained set of generic classes. When you say that a generic class implements a generic interface, you are asking the compiler, "Is this generic class strictly within the set specified by this generic interface?" You are not merely intersecting them as a further constrained set of classes.
Because a generic type restriction is on the type parameter of the defining class (U in your example), from a CLR point of view, that is a different type from the type parameter of the interface.
The type parameter of the class need not be the actual type parameter of the interface. It need not even be a simple type, as in:
class Implementation<T> : IGenericType<List<T>> { /* ... */ }
In this case, the compiler recognizes that List<T> satisfies the constraint, and so no further specification is necessary. But without such knowledge about the generic type parameter, the compiler requires you to declare it explicitly.
It is instructive to compare this to the similar but not identical behaviour of generic methods. As with classes that implement interfaces, the type restrictions must be specified with the declaration. There is one notable exception: if the implementation is explicit. In fact, the compiler will generate an error when you try to re-impose the restrictions.
For example, given an interface
interface ISomething {
void DoIt<T>() where T : new();
}
the two correct ways to implement this interface are:
class OneThing : ISomething {
public void DoIt<T>() where T : new() { }
}
class OtherThing : ISomething {
void ISomething.DoIt<T>() { }
}
Leaving out the constraint in OneThing or iserting it in OtherThing produces a compile-time error. Why do we need the constraint in the first implementation and not in the second one? I'd say for the same reason I mentioned above for type constraints on interfaces: in the first implementation, the type T has no relation to the type parameter on the interface method, so it must be made explicit for the method to match the interface method. In the second implementation, the fact that we explicitly declare the interface means that the type parameter T is the exact same one that was used in the interface.
I tried to google this, but all I could find was documents on ordinary class declarations.
public class DataContextWrapper<T> : IDataContextWrapper where T : DataContext, new()
{
}
I see that the class implements IDataContextWrapper, inherits from DataContext and varies with type T depending on how it is instantiated.
I don't know what "where T" or the ", new()" might mean.
It's a generic constraint and restricts what types can be passed into the generic parameter.
In your case it requires that T is indentical to or derived from DataContext and has a default(argumentless) constructor(the new() constraint).
You need generic constraints to actually do something non trivial with a generic type.
The new() constraint allows you to create an instance with new T().
The DataContext constraint allows you to call the methods of DataContext on an instance of T
MSDN wrote:
where T : <base class name>
The type argument must be or derive from the specified base class.
where T : new()
The type argument must have a public parameterless constructor. When used together with other constraints, the new() constraint must be specified last.
Only allow types T that are derived from or implement DataContext, and have a public constructor that takes no arguments.
It's a generic type constraint and specifies constraint on the generic types (for example, only classes, or must implement a specific interface).
In this case, T must be a class that is either DataContext or inherits from it and must have a parameterless public constructor (the new() constraint).
It's a generic type restriction. In this case, T must inherit from DataContext and be a type with a constructor that takes no arguments.
The where keyword is used to constrain your generic type variable, in your case it means that the type T must be a DataContext and must contain a public default constructor.
where T: DataContext reads as: T must be a (or derived from a) DataContext
the ", new()" reads as: must have an parameterless constructor.
It is constraints in the types that can be used as generic. This gives you compiler checks plus the ability to do something meaningful with T.
Ie. new() tells the compiler that T has to have a parameterless constructor. This means that you can instantiate instances of T by writing new T(); and by knowing T is a DataContext as well, you can both make instances of T but also call methods on it.
It's a generics constraint.
MSDN has more information on that.
See Constraints on Type Parameters (C# Programming Guide)
Where is there to place a constraint upon the type of T. The new says that the type T must be instantiable without any parameters. ie T thing = new T();
See more here