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
}
}
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