Lambda implementation of interface in kotlin

Question

What would be the equivalent of that code in kotlin, nothing seems to work of what I try:

public interface AnInterface {
    void doSmth(MyClass inst, int num);
}

init:

AnInterface impl = (inst, num) -> {
    //...
}

For future readers, you may also want to look at stackoverflow.com/a/58725404/48136 — bric3
– bric3, Commented Nov 20, 2023 at 17:36

bric3 · Accepted Answer · 2023-11-20 17:35:01Z

41

If AnInterface is Java, it can work with SAM conversion:

val impl = AnInterface { inst, num -> 
     //...
}

Otherwise, if the interface is Kotlin:

Since Kotlin 1.4, it's possible to write functional interfaces :

fun interface AnInterface {
     fun doSmth(inst: MyClass, num: Int)
}
val impl = AnInterface { inst, num -> ... }

Otherwise, if the interface is not functional

interface AnInterface {
     fun doSmth(inst: MyClass, num: Int)
}

you can use the object syntax for implementing it anonymously:

val impl = object : AnInterface {
    override fun doSmth(inst:, num: Int) {
        //...
    }
}

edited Nov 20, 2023 at 17:35

bric3

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answered Jan 16, 2018 at 15:56

s1m0nw1

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3 Comments

abitcode Over a year ago

Why Kotlin interfaces doesn't have a similar SAM conversion to that of Java? Any way can this AnInterface { inst, num -> } apply to Kotlin interface as well?

Re'em Over a year ago

well you can upvote this issue here: discuss.kotlinlang.org/t/sam-conversion-on-kotlin-interfaces/…

MiguelMunoz Over a year ago

What if I wanted to pass it to a function that takes AnInterface as a parameter. Do I write this? myFun(object : AnInterface { override fun doSmth(ints:, num: Int) { ... } }) There has to be a better way!

Marko Topolnik · Accepted Answer · 2018-01-16 17:16:06Z

22

If you're rewriting both the interface and its implementations to Kotlin, then you should just delete the interface and use a functional type:

val impl: (MyClass, Int) -> Unit = { inst, num -> ... }

answered Jan 16, 2018 at 17:16

Marko Topolnik

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3 Comments

Sira Lam Over a year ago

This is the true answer... the other 2 does not address the lambda implementation. However if I still want to use my own declared interface (for readability reason), does it mean I cannot use lambda?

TheWanderer Over a year ago

@SiraLam you can declare a default variable name for a functional type: val listener: (name: Class, other: OtherClass) -> Unit

Matt Klein Over a year ago

@SiraLam If you want the interface for readability reasons then you're probably suffering from Primitive Obsession. tl;dr - don't use primitives as arguments. What does a function with this signature do? (String) -> String Without a name, you can't reason about it. Now, what does a function with this signature do? (UserId) -> PhoneNumber - Well, the only reasonable implementation would be something that looks up user data by ID, then pulls out the phone number and returns it. If your functions don't take primitives then you don't usually need to wrap them in an interface for readability.

hudsonb · Accepted Answer · 2018-01-16 16:02:31Z

5

You can use an object expression

So it would look something like this:

val impl = object : AnInterface {
    override fun(doSmth: Any, num: Int) {
        TODO()
    }
}

edited Jan 16, 2018 at 16:02

answered Jan 16, 2018 at 15:56

hudsonb

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Comments

Kai Karren · Accepted Answer · 2022-05-26 16:33:20Z

1

To anyone who is reading this in 2022, Kotlin now has Functional (SAM) Interfaces. Please see https://kotlinlang.org/docs/fun-interfaces.html

Maybe it will save others some time depending on your use case.

answered May 26, 2022 at 16:33

Kai Karren

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1 Comment

Péter Veres Over a year ago

Thank you, this should be the selected answer

Vinujan.S · Accepted Answer · 2022-09-26 06:59:58Z

1

you can also do something like this, using @lambda label


interface AnInterface {
     fun doSmth(inst: MyClass, num: Int)
}

val impl = AnInterface lambda@{inst, num ->
    //..
}

edited Sep 26, 2022 at 6:59

answered Jul 13, 2021 at 15:33

Vinujan.S

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