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
Related
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 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.
I saw this code example and was wondering what the purpose of the new() constraint was:
public class Client<T> : IClient where T : IClientFactory, new()
{
public Client(int UserID){ }
}
That's called a "'new' constraint". Here's the documentation on it.
The new constraint specifies that any type argument in a generic class declaration must have a public parameterless constructor. To use the new constraint, the type cannot be abstract.
(Emphasis mine)
Basically, you need it whenever you're creating a new T somewhere in the class, to ensure that you're only able to pass in things which the compiler can create a new instance of.
Client is a collection of T objects, and those T objects must implement the IClientFactory interface and have a public parameterless constructor.
new() means
The type argument must have a public parameterless constructor. When
used together with other constraints, the new() constraint must be
specified last.
Ref Generic Constraints on MSDN
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.
Is it possible for a generic interface's type to be based on a specific parent class?
For example:
public interface IGenericFace<T : BaseClass>
{
}
Obviously the above code doesn't work but if it did, what I'm trying to tell the compiler is that T must be a sub-class of BaseClass. Can that be done, are there plans for it, etc.?
I think it would be useful in terms of a specific project, making sure a generic interface/class isn't used with unintended type(s) at compile time. Or also to sort of self-document: show what kind of type is intended.
public interface IGenericFace<T> where T : SomeBaseClass
What your are referring to is called "Generic Constraints". There are numerous constraints that can be put on a generic type.
Some basic examples are as follows:
where T: struct - The type argument must be a value type. Any value type except Nullable - can be specified. See Using Nullable Types (C# Programming Guide) for more information.
where T : class - The type argument must be a reference type; this applies also to any class, interface, delegate, or array type.
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.
where T : <base class name> - The type argument must be or derive from the specified base class.
where T : <interface name> - The type argument must be or implement the specified interface. Multiple interface constraints can be specified. The constraining interface can also be generic.
where T : U - The type argument supplied for T must be or derive from the argument supplied for U. This is called a naked type constraint.
These can also be linked together like this:
C#
public class TestClass<T> where T : MyBaseClass, INotifyPropertyChanged, new() { }
public interface IGenericFace<T> where T : SomeBaseClass
VB
Public Class TestClass(Of T As {MyBaseClass, INotifyPropertyChanged, New})
Public Interface IGenericInterface(Of T As SomeBaseClass)
yes.
public interface IGenericFace<T>
where T : BaseClass
{
}