Can someone please explain why `mkDivider1` compiles but `mkDivider2` doesn t ` this is not defined in this context` ``` fun mkDivider1 x Int Int gt Int = this x fun mkDivider2 x Int Int gt Int = if x == 0 null else this x ``` By the way this one does compile ``` fun mkDivider3 x Int = if x == 0 null else fun Int Int return this x ```

Shouldn t your second example be like this `fun mkDivider2 x Int Int gt Int = if x == 0 null else this x `

So actually not ` Int gt Int ` but ` Int gt Int ` is the return type I m going for

But you gave me an interesting idea ``` fun mkDivider2 x Int Int gt Int = if x == 0 null else this x ```

WOW yeah that seems to do what I intended ``` fun mkDivider2 x Int Int gt Int = if x == 0 null else this x val f Int gt Int = mkDivider2 0 println f invoke 20 ``` It prints null as I hoped

Can someone please explain why `mkDivider1` compiles but `mk kotlinlang #announcements

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Title

lifter

09/06/2018, 9:29 PM

Can someone please explain why

mkDivider1

compiles but

mkDivider2

doesn't? (

'this' is not defined in this context

)

fun mkDivider1(x: Int): (Int.() -> Int) = { this / x }
fun mkDivider2(x: Int): (Int.() -> Int)? = if (x == 0) null else { this / x }

By the way, this one does compile:

fun mkDivider3(x: Int) = if (x == 0) null else fun Int.(): Int { return this / x }

Ivan

09/06/2018, 9:37 PM

Shouldn’t your second example be like this?

fun mkDivider2(x: Int): (Int.() -> Int?) = {if (x == 0) {null} else { this / x }}

lifter

09/06/2018, 9:41 PM

Wait a sec... woah.

Ivan

09/06/2018, 9:42 PM

If I understood your intent correctly, you need another pair of curly braces to make it a lambda and a question mark after Int (lambda’s output) to be able to return null from this lambda

lifter

09/06/2018, 9:43 PM

Yeah what I'm trying to do with

mkDivider2

is exactly what I managed to do with

mkDivider3

But while I'm using an anonymous function in

mkDivider3

, I was hoping to use a lambda instead in

mkDivider2

So actually not

(Int.() -> Int?)

but

(Int.() -> Int)?

is the return type I'm going for.

But you gave me an interesting idea:

fun mkDivider2(x: Int): (Int.() -> Int)? = if (x == 0) null else { { this / x } }

WOW, yeah that seems to do what I intended:

fun mkDivider2(x: Int): (Int.() -> Int)? = if (x == 0) null else { { this / x } }
    
val f: (Int.() -> Int)? = mkDivider2(0)
println(f?.invoke(20))

It prints "null" as I hoped.

Ivan

09/06/2018, 9:51 PM

Well that’s great! So you see, the problem with your first attempt was that you weren’t returning a lambda from if expression, but a result of divide opeation.

lifter

09/06/2018, 9:51 PM

Right!

Thanks!

Ivan

09/06/2018, 9:53 PM

You’re welcome

lifter

09/06/2018, 9:55 PM

I think I'd be correct to say that this particular lambda (

{ this / x }

) is an example of a "lambda with a receiver" because altho it's a lambda it satisfies the type

(Int.() -> Int)?

- an extension function type - and I can use

this

inside this particular lambda.

Int

is the receiver type and

this

is the receiver object.

Please correct me if I'm wrong. 🙂

Ivan

09/06/2018, 10:00 PM

Well, as long as the compiler agrees with you, you’re good 🙂

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