I am attempting to create the following class name signature:
public class MyClass<T> where T : struct, MyBase
(I am using struct to constrain to Enums)
I am getting error
The class type constraint 'MyBase' must come before any other constraints
I understand the message, however, rearranging the code I cannot get past that or some other syntax error. How can I fix that line if at all?
If I have to, I will remove struct.
Thank you
Constraints are "and-ed" together; all the constraints must be satisfied.
Under what circumstances can T be both a non-nullable value type and also be implicitly convertible to the class MyBase via identity, boxing or reference conversion?
There are no such circumstances, so there is no possible type argument that will satisfy T's constraints. Rather than letting you define a set of constraints that cannot be met, the compiler simply disallows it. You cannot state that you require both the struct constraint and a class type constraint.
I am using struct to constrain to Enums
That illustrates my point. Since there are no enums that inherit from MyBase, the constraint could not possibly be met.
Can you describe for me what you thought this meant? For example, did you think that it meant "any non-nullable value type or any type that is convertible to MyBase"? I am interested to learn why people believe false things about C# so that I can try to improve it.
UPDATE: Ah, I see -- MyBase is intended to be the base class of MyClass<T>, not the base class of T. In C#, it goes:
class [class name] < [generic type parameters] >
: [base classes and interfaces]
where [type parameter] : [constraints]
You have to put the base classes and interfaces before the constraints, otherwise the compiler thinks that they are the constraints.
Did you mean class MyClass<T> : MyBase where T : struct?
You're defining <T> as two different types.
struct is a value type where as MyBase is a class referring to a reference type.
It's not something that is interchangeable.
In this case it would be either:
public class MyClass<T> where T : struct
or
public class MyClass<T> where T : MyBase
Here is so more information regarding generics and how to use them.
If T must be a struct, it can't inherit from any other type... Value types don't support inheritance.
Not 100% sure on this, but a quick check of MSDN comes up with this where (generic type constraint) (C# Reference):
public class MyClass<T, U> where T : MyBase where U : struct
Not sure that's what you're looking for though.
Related
In C#, one is allowed to write:
public class Foo<T> where T : class {
}
And according to the C# specifications, this means that:
The type argument must be a reference type; this applies also to any class, interface, delegate, or array type.
Some are of the opinion that the literal statement T : class is confusing, one could see it as.
The type argument must be a class type. So interfaces and delegates are not allowed. (wrong)
Where for instances interfaces and delegates are not allowed. I'm wondering if the C# design team considered this and why they didn't introduce a constraint like:
where T : reference
(Or another keyword that is more precise). What was the rationale to use : class instead?
They wanted to stick to a keyword that was already present in the language. Surely reference or referencetype would be more precise.
Similarly, it is called where T : struct, not valuetype (or nonnullablevaluetype), even if both structs and enums are value types that can be used for T in that case. Also note that the special struct Nullable<> is not allowed if the constraint where T : struct was used.
where T: class
reads
where T: <reference type>
I'm not sure why it was called class, though. Someone who was sitting through the language team's meetings on that one might have to chime in.
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 have a generic type that should be specified with an Enum type (actually, it's one of several specified enums, but I'll settle for System.Enum).
Of course the compiler balks at code like:
class Generic<T> where T : Enum {}
with a "Constraint cannot be special class 'System.Enum'" exception.
The only solution I've been able to come up so far is using the static type initializer to inspect the type parameter and throw an exception if it is not, in fact, an Enum, like this:
class Generic<T>
{
static Generic()
{
if (typeof(T).BaseType != typeof(Enum))
throw new Exception("Invalid Generic Argument");
}
}
which at least gives me runtime security that it wont we used with a non-enum parameter. However this feels a bit hacky, so is there a better way to accomplish this, ideally with a compile-time construct?
You can use Jon Skeet's Unconstrained Melody project to do this.
Using Unconstrained Melody you would write:
class Generic<T> where T : IEnumConstraint
Which would accomplish the same thing.
More info about Unconstrained Melody with usage examples.
Unfortunately, there is no way to ensure that a type passed to a generic class is an enum at compile time. You can only add the following constraint:
class Generic<T> where T : struct { }
In order to exclude all reference types, but the runtime check will still be necessary.
I think you can use Struct:
class Generic<T> where T : Struct
enums also implement the IConvertible interface, so you can add that as a constraint, along with the struct.
class Generic<T> where T : struct, IConvertible {}
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
{
}