A meet a strange problem at compimation saying that a class does not implement an interface.
Lets say a v got a class:
public Class MyClass
{
...
}
And an interace:
public Interface IMyInterface
{
MyClass PropertyOfMyClass {get;}
}
and now a generic class:
public class MyGeneric<T> where T:MyClass
{
T PropertyOfMyClass
{
get{return ...;}
}
}
Until here everythings fine and compiles right.
But this will break at compilation time:
public class MyGeneric<T>:IMyInterace where T:MyClass
{
T PropertyOfMyClass
{
get{return ...;}
}
}
Saying that MyGeneric does not implements method of IMyInterface. But obviously it does, not ?
You can't implement properties (or methods) from interfaces with variance. This doesn't just affect generics. For example:
public interface IFoo
{
object Bar();
}
public class Foo : IFoo
{
// This won't compile
string Bar() { return "hello"; }
}
Now you can get round this with explicit interface implementation:
public class Foo : IFoo
{
// Make the interface implementation call the more strongly-typed method
object IFoo.Bar() { return Bar(); }
string Bar() { return "hello"; }
}
That may be an answer for you - or it may not. We'd need to know exactly why you wanted to declare the property as being of type T
rather than just MyClass
to know for sure.
Another workaround would be to make the interface generic itself :
public interface IMyInterface<T> where T : MyClass
{
T PropertyOfMyClass { get; }
}
You can then use it on a class :
public class MyGenericClass<T> : IMyInterface<T> where T : MyClass
{
T PropertyOfMyClass
{
get { ... }
}
}
Note that using this implementation, the constraint on T in the generic class can be different from the one on the interface, as long as it ensures the interface constraint is respected :
public class MyOtherClass : MyClass
{
}
public class MyOtherGenericClass<T> : IMyInterface<T> where T : MyOtherClass
{
T PropertyOfMyClass
{
get { ... }
}
}
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