Here is a piece of my code:
public interface IA<in TInput>
{
void Method(IB<TInput> entities);
}
public interface IB<in T> { }
I can't figure out why I get following compile error:
"Parameter must be input-safe. Invalid variance: The type parameter |TInput| must be contravariantly valid on "IB< in T>".
Any help will be appreciated.
The designator of contravariance in C# (i.e. in) is intuitive only at the immediate level, when you make a method that "takes in" a parameter of generic type. Internally, however, contravariance means an inversion of a relation (Q&A with an explanation) so using in inside IA makes it incompatible with IB.
The problem is best illustrated with an example. Consider class Animal and its derived class Tiger. Let's also assume that IB<T> has a method void MethodB(T input), which is called from IA's Method:
class A_Impl<T> : IA<T> {
T data;
public void Method(IB<TInput> entities) {
entities.MethodB(data);
}
}
Declaring IA<in TInput> and IB<in TInput> means that you can do
IA<Animal> aForAnimals = new A_Impl<Animal>();
IA<Tiger> aForTigers = aForAnimals;
IA<in TInput> has a method that takes IB<TInput>, which we can call like this:
aForTigers.Method(new B_Impl<Tiger>());
This is a problem, because now A_Impl<Animal> passes an Animal to MethodB of an interface that expects a Tiger.
You would have no problem with IB<out T>, though - both with covariance and contravariance:
public interface IB<out T> {
// ^^^
}
// This works
public interface IA<in TInput> {
void Method(IB<TInput> x);
}
// This works too
public interface IC<out TInput> {
void Method(IB<TInput> x);
}
Related
This question already has an answer here:
Parameter must be input-safe error
(1 answer)
Closed 4 years ago.
I have two interfaces that are both covariant, with both being passed in to each other like so:
public interface Perfomer<in T>
{
void Perform(T t, Tracer<T> tracer);
}
public interface Tracer<in T>
{
void Notify();
}
However even though both interfaces are marked covariant, and T is only ever being used as input, I'm still getting the error:
"Invalid variance: The type parameter 'T' must be covariantly valid on
'Perfomer<T>.Do(T, Tracer<T>)'. 'T' is contravariant. [_Console].
Any ideas why having covariant interface parameter using the same type makes T contravariant?
Edit
(Sorry, I am new to StackOverflow, based on the answers I realize I should've been more exact in my question, I had just tried to eliminate as much noise as possible to a single error).
The code actually has two interfaces with generally similar interfaces:
public interface Performer<in T>
{
bool Perform(T t, Tracer<T> tracer = null);
}
public interface Tracer<in T>
{
void Notify(Performer<T> performer, T t, ref bool success);
}
It's purpose is to allow the an optional "tracer" to see things happen/modify the results of a performer.
When you declare that Performer is contravariant, you are declaring that anything a Performer does to a T can also be done to a more specific version of T. For example, an action that acts on a object can be given a string, and it'll just act as if that string is an object.
So for example you could do this, because all streams support Length:
class MyClass : Performer<Stream>
{
void Perform(Stream t)
{
Console.WriteLine(t.Length)
}
}
Performer<FileStream> p = new MyClass();
p.Perform(new FileStream());
But you can't do this, because you gave it a class that doesn't support IsAsync:
class MyClass : Performer<FileStream>
{
void Perform(Stream t)
{
Console.WriteLine(t.IsAsync)
}
}
Performer<Stream> p = new MyClass();
p.Perform(new Stream()); //Stream isn't good enough; it has to be a FileStream, since it needs IsAsync
So far so good. Now let's add in that second parameter:
class MyClass : Performer<Stream>
{
void Perform(Stream t, Tracer<Stream> tracer)
{
Console.WriteLine(tracer.Notify())
}
}
In order for this to work, the contravariance has to work. If the contravariance works, it means that Perform can store a Tracer<FileStream> (which you pass in) in a variable that is typed as a Tracer<Stream> (which is how it is implemented). That means that Tracer must be covariant with respect to its type argument.
So you can fix your code by changing in to out, like so:
public interface Performer<in T>
{
void Perform(T t, Tracer<T> tracer);
}
public interface Tracer<out T> //out instead of in
{
void Notify();
}
From what you've provided I'd avoid the issue all together, Modify the Tracer interface to remove the T because it's not needed:
public interface INotify
{
void Notify();
}
Then just take in an the new interface in your performer
public interface Perfomer<in T>
{
void Perform(T t, INotify entity);
}
PS: there might be a type in your interface name Perfomer => Performer
Just modifyTracer<in T> to Tracer (non-generic) and define void Perform(T t, Tracer tracer);.
Your code was not using T in Tracer anyways.
Since you edited your question with new details, the alternative fix is to remove in from generics definition. You don't need it. Another way to achieve what you want is following:
public interface Performer<T>
{
bool Perform(T t, Tracer tracer = null);
}
public interface Tracer
{
bool Notify<T>(Performer<T> performer);
}
Note: drop ref bool and return bool instead
This question already has answers here:
Implementing multiple generic interfaces - type error
(5 answers)
Inheritance from multiple interfaces with the same method name
(9 answers)
Closed 4 years ago.
Say I have an interface WorksWithType<T> and class MyClass that implements both WorksWithType<TypeA> and WorksWithType<TypeB>.
If my interface looks like
public interface WorksWithType<T> {
void DoSomething(T foo);
void DoSomethingElse();
}
it is easy to implement two different DoSomething method overloads in MyClass.
public class MyClass : WorksWithType<TypeA>, WorksWithType<TypeB> {
{
public void DoSomething(TypeA fooA) { ... }
public void DoSomething(TypeB fooB) { ... }
...
}
However, there doesn't seem to be a way to implement overloads of DoSomethingElse. In my mind I feel as though I should be able to change the signature on the interface to be
void DoSomethingElse<T>();
and then overload the class with
public void DoSomethingElse<TypeA>() { ... }
public void DoSomethingElse<TypeB>() { ... }
What is the correct approach here, if there is one?
Assuming you want the two implementations of DoSomethingElse to be different , you need to use explicit interface implementation to distinguish between the calls:
public class TypeA {}
public class TypeB {}
public interface IWorksWithType<T>
{
void DoSomething(T foo);
void DoSomethingElse();
}
public class MyClass : IWorksWithType<TypeA>, IWorksWithType<TypeB>
{
public void DoSomething(TypeA fooA) {}
public void DoSomething(TypeB fooB) {}
// Note the syntax here - this indicates which interface
// method you're implementing
void IWorksWithType<TypeA>.DoSomethingElse() {}
void IWorksWithType<TypeB>.DoSomethingElse() {}
}
You don't have to make both of them use explicit interface implementation. For example:
public class MyClass : IWorksWithType<TypeA>, IWorksWithType<TypeB>
{
public void DoSomething(TypeA fooA) {}
public void DoSomething(TypeB fooB) {}
// Explicit interface implementation
void IWorksWithType<TypeA>.DoSomethingElse() {}
// Implicit interface implementation
public void DoSomethingElse() {}
}
If you don't need the implementations to be different, you can just have three methods of course:
public class MyClass : IWorksWithType<TypeA>, IWorksWithType<TypeB>
{
public void DoSomething(TypeA fooA) {}
public void DoSomething(TypeB fooB) {}
// Implementation of both IWorksWithType<TypeA>.DoSomethingElse()
// and IWorksWithType<TypeB>.DoSomethingElse()
public void DoSomethingElse() {}
}
That's assuming you to want the type parameter to be on the interface. You could put it on the method instead, but that really represents a very different interface - and you wouldn't be able to say that MyClass can only call DoSomethingElse for types TypeA and TypeB, for example.
By the current C# specifications (draft 6) https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/language-specification/introduction
The signature of a method must be unique in the class in which the method is declared. The signature of a method consists of the name of the method, the number of type parameters and the number, modifiers, and types of its parameters. The signature of a method does not include the return type
(emphasis mine)
So, unfortunately,
void MyMethod<TypeA>()
and
void MyMethod<TypeB>()
would not have different signatures, so you cannot, per the specs, define both of them : they differ only by the type of their type parameter, but not by the number of their type parameter
Other answers have already pointed out how you could work around this (the explicit interface implementation is a good idiomatic option in my opinion)
Let's say we have an interface like
public interface IEnumerable<out T>
{ /*...*/ }
that is co-variant in T.
Then we have another interface and a class implementing it:
public interface ISomeInterface {}
public class SomeClass : ISomeInterface
{}
Now the co-variance allows us to do the following
IEnumerable<ISomeInterface> e = Enumerable.Empty<SomeClass>();
So a IEnumerable<SomeClass> is assignable to a variable (or method parameter) of type IEnumerable<ISomeInterface>.
But if we try this in a generic method:
public void GenericMethod<T>(IEnumerable<T> p) where T : ISomeInterface
{
IEnumerable<ISomeInterface> e = p;
// or
TestMethod(p);
}
public void TestMethod(IEnumerable<ISomeInterface> x) {}
we get the compiler error CS0266 telling us that an IEnumerable<T> cannot be converted to an IEnumerable<ISomeInterface>.
The constraint clearly states the T is derived from ISomeInterface, and since IEnumerable<T> is co-variant in T, this assignment should work (as shown above).
Is there any technical reason why this cannot work in a generic method? Or anything I missed that makes it too expensive for the compiler to figure it out?
Change your GenericMethod and add generic constraint class:
public void GenericMethod<T>(IEnumerable<T> p) where T : class, ISomeInterface
{
IEnumerable<ISomeInterface> e = p;
// or
TestMethod(p);
}
Covariance does not support structs, so we need to tell that we want to use classes only.
I've two interfaces:
public interface IAmA
{
}
public interface IAmB<T> where T : IAmA
{
}
And two classes implementing these interfaces like this:
public class ClassA : IAmA
{
}
public class ClassB : IAmB<ClassA>
{
}
When trying to use these classes as shown:
public class Foo
{
public void Bar()
{
var list = new List<IAmB<IAmA>>();
list.Add(new ClassB());
}
}
I get this compiler error:
cannot convert from 'ClassB' to 'IAmB<IAmA>'
I know I can make the compiler happy using:
public class ClassB : IAmB<IAmA>
{
}
But I need to be able to be the Type parameter for IAmB<> in ClassB an implementation of IAmA.
The quick answer is that you can do what you ask by declaring the type parameter of IAmB<T> as covariant, only if the type is used as a return type:
public interface IAmB<out T> where T : IAmA
{
T SomeMethod(string someparam);
}
out T means that you can use a more specific type than then one specified in the constraints.
You won't be able to use T as a parameter. The following won't compile:
public interface IAmB<out T> where T : IAmA
{
void SomeMethod(T someparam);
}
From the documentation
You can use a covariant type parameter as the return value of a method that belongs to an interface, or as the return type of a delegate. You cannot use a covariant type parameter as a generic type constraint for interface methods.
This isn't a compiler quirk.
Assuming you could declare a covariant method parameter, your list would end up containing some objects that couldn't handle an IAmB<IAmA> parameter - they would expect an input of ClassA or more specific. Your code would compile but fail at runtime.
Which begs the question - why do you want to use IAmB<ClassA> ?
You should think about before using this though, as there may be other, more suitable ways to address your actual problem. It's unusual to use a generic interface implementing a concrete type but trying to use it as if it were implementing another interface.
You can check the MSDN documentation's section on Covariance and Contravariance as well as Eric Lippert's an Jon Skeet's answers to this SO question: Difference between Covariance and Contravariance
Fast answer : make the generic type covariant (see msdn) in your interface
public interface IAmB<out T> where T : IAmA
{
}
this will resolve the compiler problem.
But this won't answer the why asked by Panagiotis Kanavos !
The trick is making the type constraint T on IAmB<T> covariant, with the out keyword:
public interface IAmB<out T> where T : IAmA
{
}
This allows you to use a more specific type than originally specified, in this case allowing you to assign an IAmB<ClassA> to a variable of type IAmB<IAmA>.
For more information, see the documentation.
I just tell why this error reported.
if your IAmB has a method
public interface IAmB<T> where T : IAmA
{
void foo(T p);
}
public class ClassB : IAmB<ClassA>
{
void foo(ClassA p)
{
p.someIntField++;
}
}
and we have another class
public class ClassC : IAmB<ClassA2>
{
void foo(ClassA2 p)
{
p.someOtherIntField++;
}
}
and we assume List<IAmB<IAmA>>.Add(T p) implement like this
IAmA mParam = xxxx;
void Add(IAmB<IAmA>> p){
p.foo(mParam);
}
thinking all compile OK. you pass a ClassB instance to List.Add, it becomes
void Add(IAmB<IAmA>> p){
//p is ClassB now
p.foo(mParam);//COMPILER CAN NOT MAKE SURE mParam fit ClassB.foo
}
It can be solved using Contravariance and Covariance.
public interface IAmA
{
}
**public interface IAmB<out T> where T : IAmA
{
}**
public class ClassA : IAmA
{
}
public class ClassB : IAmB<ClassA>
{
}
public class Foo
{
public void Bar()
{
var list = new List<IAmB<IAmA>>();
**list.Add(new ClassB());**
}
}
Now you don't get compiler error. Compiler is happy.
The following does not compile on line fm.AddFoo(new StringFoo()); with the error message:
Argument 1: cannot convert from 'ClassLibrary2.StringFoo' to 'ClassLibrary2.IFoo'
This seems logical to me since string inherits from object.
public interface IFoo<T>
{
void Handle(T value);
}
public class StringFoo : IFoo<string>
{
public void Handle(string value)
{ }
}
public class ObjectFoo : IFoo<object>
{
public void Handle(object value)
{ }
}
public class FooManager
{
private readonly List<IFoo<object>> _foos;
public FooManager()
{
_foos = new List<IFoo<object>>();
}
public void AddFoo(IFoo<object> foo)
{
_foos.Add(foo);
}
}
public class Bad
{
public Bad()
{
var fm = new FooManager();
fm.AddFoo(new StringFoo()); \\ This does not compile
}
}
Thanks
Although it may seem like IFoo is a subclass of IFoo it is not. When you close IFoo<> to a specific type is is not creating a subclass of IFoo from IFoo, they are seperate and distinct types with no common hierarchy.
If you could make your IFoo<> interface covariant it would work, that is if you were allowed to change the declaration of it into:
public interface IFoo<out T>
(note the out). Because with covariance any IFoo<string> would also be an IFoo<object> because string is a reference type and derives from object.
But: A member of IFoo<>, the Handle method, uses the type parameter in a contravariant manner. So your interface cannot be declared covariant (out). (It could be declared contravariant (in) but that goes in the wrong direction for your example above.)
Read up on covariance and contravariance in generics.
The fundamental problem here is that your StringFoo handles only strings. Therefore it can never be used as an IFoo<object> because then you could pass for example a Giraffe instance (Giraffe derives from object, so a Giraffe is an object) into the StringFoo, and that is impossible when its Handle takes a string.