Kotlin - get name from lambda expression

Question

I want to get the name of a parameter passed into a lambda expression in Kotlin. In C# I would use an Expression<Func<T, ...>> to get the parameter name, but I'm not sure whether this is possible in Kotlin

import java.util.*

fun main(args: Array<String>) {
    val foo = Foo<Model>()
    foo.bar { it.age }
    // Should print "age"
}

data class Model(val id: UUID, val name: String, val age: Int)

class Foo<T> {
    fun bar(expression: (x: T) -> Any) {
        println(/*The name of the parameter*/)
    }
}

Is this possible in Kotlin?

Do you mean you want to pass a property which name shall be printed? Since age does not function as a “parameter” here — s1m0nw1
– s1m0nw1, Commented Feb 16, 2018 at 11:28
Try to pass a KProperty<*> directly (change parameter type) and use its name in bar. Passed as an argument like this: foo.bar(foo::age) — s1m0nw1
– s1m0nw1, Commented Feb 16, 2018 at 11:35
@s1m0nw1 not sure what you mean, would you mind providing an answer? — Matthew Layton
– Matthew Layton, Commented Feb 16, 2018 at 11:36

s1m0nw1 · Accepted Answer · 2018-02-16 11:44:49Z

2

If you want to pass around properties and print their names, you can do it by using KProperty:

fun main(args: Array<String>) {
    val foo = Foo<Model>()
    foo.bar(Model::age)
}

class Foo<T> {
    fun bar(p: KProperty<*>) {
        println(p.name)
    }
}

edited Feb 16, 2018 at 11:44

answered Feb 16, 2018 at 11:38

s1m0nw1

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3 Comments

Matthew Layton Over a year ago

This almost works, except you can't pass foo::age because age is a property of Model

s1m0nw1 Over a year ago

Model::age then? Well probably not what you want to achieve I’m afraid

Matthew Layton Over a year ago

Yes. It might not be exactly what I wanted, but it does the job!

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