Extension functions for generic classes in Kotlin

Question

What's wrong with my extension function below

class Foo<T> {
    fun <T> Foo<T>.plus(that: Foo<T>): Foo<T> = throw Exception()

    init {
        Foo<Int>() + Foo<String>()  // A receiver of type Foo<T> is required
    }
}

Update

I wonder why it's different than regular extension functions, where T successfully gets inferred as Any and would like to achieve the same behavior, e. g. T to get inferred as Foo<Any>

class Foo {
    fun <T> T.foo(that: T): T = throw Exception()

    init {
        "str" foo 42
    }
}

Parceling

>>>"I wonder why it's different" @Parceling There is +no+ difference to regular functions. It's just a question of subtyping Foo<A> and Foo<B> where A is a subtype of B. In Java it would require wild casting. Not sure on Scala, I believe they have Foo[+T] for the same purpose – Iquique

Answer 1

The issue is at the very heart of how generics work.

class Foo {
    fun <T> T.foo(that: T): T = throw Exception()

    init {
        "str" foo 42
    }
}

This works, because the compiler can find a T that fits both the function signature and the arguments: it is Any, and the function is turned into this one:

fun Any.foo(that: Any): Any = ...

Now, String is a subtype of Any, Int is a subtype of Any, so this function is applicable to the arguments.

But in your first example:

class Foo<T> {
    fun <T> Foo<T>.plus(that: Foo<T>): Foo<T> = throw Exception()

    init {
        Foo<Int>() + Foo<String>()  // A receiver of type Foo<T> is required
    }
}

It's all different. There's no such T. Let's be naïve and try Any:

fun Foo<Any>.plus(that: Foo<Any>): Foo<Any> = ...

Now, Foo is invariant in T, so Foo<Int> is not a subtype of Foo<Any>, and in fact there's no type T other than Int that would make Foo<T> a supertype of Foo<Int>. So, T must be exactly Int, but it also must be exactly String by the same logic (because of the second argument), so there's no solution, and the function is not applicable.

You could make it work by making Foo co-variant in T:

class Foo<out T> {
    fun <T> Foo<T>.plus(that: Foo<T>): Foo<T> = throw Exception()

    init {
        Foo<Int>() + Foo<String>()  // A receiver of type Foo<T> is required
    }
}

This imposes some limitations on possible signatures of members of Foo, but if you are OK with them, it fixes your issue.

Have a look at this link for more details: http://kotlinlang.org/docs/reference/generics.html

Frizzle

Answer 2

I think the accepted answer by Andrey Breslaw is correct, but provides incorrect solution.

The compiler simply needs to be told to infer common supertype for the generic type arguments supplied, i.e., as long as the generic type arguments of Foo share a common supertype (and they always will), use it. Like:

operator fun <T, R: T, S: T> Foo<R>.plus(that: Foo<S>): Foo<T> = throw Exception()

Now the resulting generic type argument of returned Foo will be widened as necessary if the types do not match, but the operation itself is legal, without introducing covariance.

Heartsease

Answer 3

Your method plus expects the parameter to have the same generic type parameter T as the receiver. Hence, you can't add a Foo<String> to a Foo<Int>.

If you want to be able to add all types of Foo, than you need to declare your extension function like so:

operator fun <T,R> Foo<T>.plus(that: Foo<R>): Foo<T> = throw Exception()

Whipping