How to get a function reference using reflection in Kotlin

Question

Let's suppose I have an instance of a class X with a method Y of which we will know the name only at runtime, how can I get a reference to it using reflection?

Something like:

class X{

  fun Y(){
    
  }

}

I want to able to store the method Y in a variable and call it when needed.

I tried with X::class.java::getMethod('Y').kotlinFunction but then I will need to have an instance of such method to be able to call it, so it doesn't make any sense

You mean you want a reference to a function that is bound to a specific instance of X? — Tenfour04
– Tenfour04, Commented Jan 19, 2021 at 14:28

Mohsen · Accepted Answer · 2021-01-19 14:43:07Z

1

first, you need to find the function looping through class members, then call it with the required instance. if a function needs other arguments you need to pass it in order, but the first arg always needs to be the instance

class X {
    fun y() { println("I got called") }
}

fun main() {
    val x = X()
    x::class.members.find { it.name == "y" }
        ?.call(x)
}

Performance:

I ran the following code and got the following results:

    var start = System.nanoTime()
    val y = x::class.members.find { it.name == "y" }
    y?.call(x)
    var stop = System.nanoTime()
    println(stop - start)
    println()

    start = System.nanoTime()
    y?.call(x)
    stop = System.nanoTime()
    println(stop - start)
    println()


    start = System.nanoTime()
    x.y()
    stop = System.nanoTime()
    println(stop - start)
    println()



I got called
381566500 // with loop and reflection

I got called
28000 // reflection call

I got called
12100 // direct call

edited Jan 19, 2021 at 14:43

answered Jan 19, 2021 at 14:29

Mohsen

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9 Comments

Tito Tigi Over a year ago

How slow is this way compared to a direct reference like X::Y?

Todd Over a year ago

If you don't want to deal with nullables and know your method exists, you could call first instead of find.

Mohsen Over a year ago

@Todd if you have other members first does not work

Todd Over a year ago

Not if you use the lambda version - .first { it.name == "y" }.

Tito Tigi Over a year ago

Btw, I just tested the code and I get a type mismatched exception. The method call() is expecting a X.Y instance, not an X instance

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