CODEWITHSUNDEEP
genericskotlin
Duncan McGregor
Duncan McGregor

Reputation: 18177

How do I constrain a Kotlin extension function parameter to be the same as the extended type?

I want to write an extension method on a generic type T, where the matched type constrains a method parameter.

I want this to compile:

"Hello".thing("world")

But not this, as 42 is not a String:

"Hello".thing(42)

This definition doesn’t work, because T is satisfied by Any

fun <T> T.thing(p: T) {}

Upvotes: 19

Views: 1389

Answers (3)

hotkey
hotkey

Reputation: 148179

Improving @miensol's workaround and making it visually identical to function call:

val <T> T.foo: (T) -> SomeType get() = { other -> ... }

This is an extension property that provides a lambda, which can be immediately called with an argument of the same type T like this:

"abc".foo(1) // Fail
"abc".foo("def") // OK

Unfortunately, there seems to be a bug in the compiler which prevents you from writing "abc".thing("abc"), but either of "abc".thing.invoke("abc") and ("abc".thing)("abc) work well and filter out calls with non-strings.

Upvotes: 5

miensol
miensol

Reputation: 41678

As mentioned by @Alexander Udalov it's not possible to do directly but there's a workaround where you define the extension method on another type like so:

data class Wrapper<T>(val value: T)

val <T> T.ext: Wrapper<T> get() = Wrapper(this)

fun <T> Wrapper<T>.thing(p: T) {
    println("value = $value, param = $p")
}

With the above the following compiles:

"abc".ext.thing("A")

but the next fails 

"abc".ext.thing(2)

with:

Kotlin: Type inference failed: Cannot infer type parameter T in fun <T> Wrapper<T>.thing(p: T): Unit
None of the following substitutions
receiver: Wrapper<String>  arguments: (String)
receiver: Wrapper<Int>  arguments: (Int)
can be applied to
receiver: Wrapper<String>  arguments: (Int)

As suggested by @hotkey it seems that it should be possible to avoid the need for explicit Wrapper type with the following extension property:

val <T> T.thing: (T) -> Any? get() = { println("extension body") }

And then use it as "abc".thing("A") but it also fails. Surprisingly the following does compile "abc".thing.invoke("A")

Upvotes: 13

Alexander Udalov
Alexander Udalov

Reputation: 32826

As far as I know, this is not possible in Kotlin 1.0. There are several issues in the tracker (first, second) about a similar use case, and the solution proposed in the first one is likely going to help here in the future as well.

Upvotes: 10

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