Suppose I have a list of lambdas like: `List<(I...

# getting-started

[Mark]

Suppose I have a list of lambdas like: List<(Int) -> Unit> but I’d rather be more explicit so define an interface MyFun: (Int) -> Unit so that my list now looks like List<MyFun>. However, now, to create such a MyFun, I need to do something like object: MyFun { override invoke(val: Int) { //some stuff } }. Does this mean I’m creating an extra object compared to with not using the interface? If so, is there a better way?

[Ruckus]

You can use typealias MyFun = (Int) -> Unit

[Mark]

Yep, I just tried that, thanks!

[diesieben07]

Just to answer your original question: No, lambdas are also just objects, internally implementing an interface, at least on the JVM.

[Alowaniak]

Also doesn't the new type inference enable kotlin SAM conversion?

[Ruckus]

No, but there will be a new fun interface construct.