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

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遇见更好的自我
遇见更好的自我 2020-12-18 19:25

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\".thin         


        
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  • 2020-12-18 19:56

    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")

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  • 2020-12-18 20:05

    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.

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  • 2020-12-18 20:15

    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.

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