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Why does this result in CS0695?

public interface PipelineElement<in TIn, out TOut>
{
    IEnumerable<TOut> Run(IEnumerable<TIn> input, Action<Error> errorReporter);
}

public interface Stage
{
}

public abstract class PipelineElementBase<TIn, TOut> : PipelineElement<object, object>,
    PipelineElement<TIn, TOut> where TIn : Stage where TOut : Stage
{
    IEnumerable<object> PipelineElement<object, object>.Run(IEnumerable<object> input, Action<Error> errorReporter)
    {
        return this.Run(input.Cast<TIn>(), errorReporter).Cast<object>();
    }

    public abstract IEnumerable<TOut> Run(IEnumerable<TIn> input, Action<Error> errorReporter);
}

object doesn't implement Stage, therefore neither TIn nor TOut could ever be object, right? So why does the compiler think that PipelineElement<object, object> and PipelineElement<TIn, TOut> can become identical?

EDIT: Yes, it is perfectly possible to implement the same generic interface multiple times:

public interface MyInterface<A> { }
public class MyClass: MyInterface<string>, MyInterface<int> { }
like image 480
main-- Avatar asked Mar 09 '13 22:03

main--


1 Answers

From Compiler Error CS0695

'generic type' cannot implement both 'generic interface' and 'generic interface' because they may unify for some type parameter substitutions.

This error occurs when a generic class implements more than one parameterization of the same generic interface, and there exists a type parameter substitution which would make the two interfaces identical. To avoid this error, implement only one of the interfaces, or change the type parameters to avoid the conflict.

You can't implement both PipelineElementBase<TIn, TOut> and PipelineElement<object, object> interfaces to your abstract class.

As the error page said, you should;

  • Implement only one of these or
  • Change the type parameters to avoid the conflict.

From C# 5.0 Language Specification

13.4.2 Uniqueness of implemented interfaces

The interfaces implemented by a generic type declaration must remain unique for all possible constructed types. Without this rule, it would be impossible to determine the correct method to call for certain constructed types. For example, suppose a generic class declaration were permitted to be written as follows:

interface I<T>
{
    void F();
}
class X<U,V>: I<U>, I<V>
{
    void I<U>.F() {...}
    void I<V>.F() {...}
}

Were this permitted, it would be impossible to determine which code to execute in the following case:

I<int> x = new X<int,int>();
x.F();

To determine if the interface list of a generic type declaration is valid, the following steps are performed:

  • Let L be the list of interfaces directly specified in a generic class, struct, or interface declaration C.

  • Add to L any base interfaces of the interfaces already in L.

  • Remove any duplicates from L.

  • If any possible constructed type created from C would, after type arguments are substituted into L, cause two interfaces in L to be identical, then the declaration of C is invalid. Constraint declarations are not considered when determining all possible constructed types.

In the class declaration X above, the interface list L consists of I<U> and I<V>. The declaration is invalid because any constructed type with U and V being the same type would cause these two interfaces to be identical types.

It is possible for interfaces specified at different inheritance levels to unify:

interface I<T>
{
  void F();
}
class Base<U>: I<U>
{
  void I<U>.F() {…}
}
class Derived<U,V>: Base<U>, I<V> // Ok
{
  void I<V>.F() {…}
}

This code is valid even though Derived<U,V> implements both I<U> and I<V>. The code

I<int> x = new Derived<int,int>();
x.F();

invokes the method in Derived, since Derived<int,int> effectively re-implements I<int>(§13.4.6).

[Emphasis by the SO editor.]

like image 78
Soner Gönül Avatar answered Nov 03 '22 08:11

Soner Gönül