Can someone please explain why the second

println(f())

prints “Yes” instead of “No”?

fun main(args: Array<String>) {
    var (f, x) = g()
    println(f()) // Prints "yes"
    x = 1
    println(f()) // Prints "yes"
}

fun g(): Pair<() -> Char, Int> {
    var x = 0
    var f = fun(): Char { return if (x == 0) { println("yes"); 'Y' } else { println("no"); 'N' } }
    return f to x
}

x is returned by value. So when you set x to 1. The x referenced by f doesn't change.

Is it? I’m confused because this prints “true”:

fun main(args: Array<String>) {
    val x = 0
    val y = f(x)
    println(x === y)
}

fun f(x: Int): Int {
    return x
}

===

is the same as

==

for "primitives".

That’s interesting, I thought everything in Kotlin was an object and there were no primitives.

Although it may return false for

Int?

instead of

Int

.

So it sounds like it’s impossible to compare `Int`s for equality of reference? Even though they are objects?

===

is the same as

==

for values that are represented as primitive types at runtime.

Wow that’s a little unfortunate.

No problem.

Everything is by-value-of reference, and there is nothing else in Java/Kotlin.

I think I get what you are saying @karelpeeters. Tho I have to admit this, to me, seems like an exception to the statement “Everything is by-value-of reference”:

fun main(args: Array<String>) {
    val x = 0
    val y = 0
    println(x === y) // Prints "true"
}

Right, pretend like primitives do interning like strings do. Replace

0

by

"hello"

and it probably still prints

true

.

Yeah, it does

But that stuff only happens for primitives and strings, both immutable types, where the distinction between reference and value is less relevant.

Either way, setting x in your very first example won't work, if its a primitive or object. You change the reference, so in the end x points to a different object.

exactly, your

x = 1

assigns a new value to the local

x

, you aren't changing the old value