5
votes

I'm stuck trying to translate some Java code that uses (bounded) wildcard generics to C#. My problem is, Java seems to allow a generic type to be both covariant and contravariant when used with a wildcard. For instance:

Java:

interface IInterf { }

class Impl implements IInterf { }

interface IGeneric1<T extends Impl> {
    void method1(IGeneric2<?> val);
    void method1WithParam(T val);
}

interface IGeneric2<T extends Impl> {
    void method2(IGeneric1<?> val);
}

abstract class Generic<T extends Impl> implements IGeneric1<T>, IGeneric2<T> {
    public void method1(IGeneric2<?> val2) {
        val2.method2(this);
    }
}

...works.

C# equivalent (?)

interface IInterf { }

class Impl : IInterf { }

interface IGeneric1<T> where T:Impl {
  //Java was: 
  //void method1(IGeneric2<?> val2);
    void method1(IGeneric2<Impl> val);
    void method1WithParam(T to);
}

interface IGeneric2<T>where T:Impl {
    void method2(IGeneric1<Impl> val);
}

abstract class Generic<T> : IGeneric1<T>, IGeneric2<T> where T : Impl
{
  //Java was: 
  //public void method1(IGeneric2<?> val2) {
    public void method1(IGeneric2<Impl> val2)
    {
         val2.method2(this); //'this': Argument type 'Generic<T>' is not 
                             //assignable to parameter type 'IGeneric1<Impl>'
    }

    public abstract void method1WithParam(T to);
    public abstract void method2(IGeneric1<Impl> val);
}

...fails to compile - see the error in the comment. Which is to be expected, since IGeneric's generic parameter is not marked 'out' for covariance.

If I change this:

interface IGeneric1<T> where T:Impl {

to this

interface IGeneric1<out T> where T:Impl 

the error goes away, but another one appears, for the declaration of the method that takes a generic parameter inside the same interface:

interface IGeneric1<T> where T:Impl {
    void method1WithParam(T val);  //Parameter must be input-safe. 
                      //Invalid variance: The type parameter 'T' must be
                      //contravariantly valid on 'IGeneric1<out T>'.

Suggestions?

[Also see the follow-up question for a somewhat harder scenario]

1
What is the meaning of the question mark inside the angle brackets in Java?Daniel Hilgarth
"Any type goes" (or in this case, any type derived from Impl goes). It's called "wildcard". docs.oracle.com/javase/tutorial/extra/generics/wildcards.htmlCristian Diaconescu

1 Answers

7
votes

You need to translate the Java wildcard generic methods to C# methods that are generic in their own right. For example, this:

interface IGeneric2<T extends Impl> {
    void method2(IGeneric1<?> val);
}

should be translated to

interface IGeneric2<T>where T:Impl {
    void method2<U>(IGeneric1<U> val) where U:Impl;
}

It is necessary to repeat the type constraint for T specified by IGeneric1<T> as the type constraint for U.

The reason for this is that in the Java version there are implicit constraints for the type arguments of the parameters of method1 and method2: if the parameter must be some kind of IGeneric1<X> then X must obviously be an Impl because otherwise it could not possibly implement IGeneric1 for that type.

In C# the constraints must be explicit, so you repeat what IGeneric1<T> and IGeneric2<T> require of T.

So the equivalent code would be:

interface IInterf { }

class Impl : IInterf { }

interface IGeneric1<T> where T:Impl {
    void method1<U>(IGeneric2<U> val) where U:Impl;
    void method1WithParam(T to);
}

interface IGeneric2<T>where T:Impl {
    void method2<U>(IGeneric1<U> val) where U:Impl;
}

abstract class Generic<T> : IGeneric1<T>, IGeneric2<T> where T : Impl
{
    public void method1<U>(IGeneric2<U> val2) where U:Impl
    {
        val2.method2(this);
    }

    public abstract void method1WithParam(T to);
    public abstract void method2<U>(IGeneric1<U> val) where U:Impl;
}