``` fun lt F Foo gt foo1 fooRef KProperty1 lt Bar Foo gt = fun lt F Foo gt foo2 foo Foo = foo1 WHAT GOES HERE ``` Is there a simple way to do this `Bar something with foo `

so let s say `fooRef KProperty1 lt Bar String gt ` e g `Bar name` How could you engineer `KProperty1 lt Bar String gt ` from `name`

``` fun lt F Foo gt foo1 fooRef KProperty1 lt Bar Foo gt = f kotlinlang #getting-started

Hullaballoonatic

06/13/2019, 9:38 PM

fun <F: Foo> foo1(fooRef: KProperty1<Bar, Foo>) = ...

fun <F: Foo> foo2(foo: Foo) = foo1(WHAT GOES HERE?)

Is there a simple way to do this?

Bar::(something with foo)

karelpeeters

06/13/2019, 10:23 PM

foo1

takes a property of

Bar

Foo

, and all you've got is an instance or

Foo

. How could that possibly converted to the latter?

Hullaballoonatic

06/13/2019, 10:27 PM

so let's say

fooRef: KProperty1<Bar, String>

Bar::name

. How could you engineer

KProperty1<Bar, String>

name

karelpeeters

06/13/2019, 10:31 PM

Do you happen to be confused about how variables work? If I have a class

class Bar(val name: String)

and then I do

foo2(bar.name)

there's no connection to

bar

bar.name

is just an ordinary

String

instance out there with everything else

You can't engineer a property from a random instance, it may not even be a property,

foo("test")

, it may be a "property" of multiple classes, it may be in a list somewhere, ...

Hullaballoonatic

06/13/2019, 10:32 PM

Yeah, so you can provide the

Bar

part yourself again.

alright. i'll try to think of a different solution

karelpeeters

06/13/2019, 10:33 PM

I don't get what you want to happen.