Considering this class: ```open class KotlinPracti...

# getting-started

[Klitos Kyriacou]

Considering this class:

open class KotlinPractice(val id: Int) {
    override fun equals(other: Any?) =
        this === other ||
                other != null &&
                other::class == KotlinPractice::class &&
                (other as KotlinPractice).id == id

    override fun hashCode() = id
}

Is there a reason other is not smart-cast and instead I have to cast it to KotlinPractice manually?

[Luke]

I can't answer why there is no smartcast, but you can do this === other || (other as? KotlinPractice)?.id == id

[edrd]

It's because the compiler doesn't perform smart casts when type checks are based on ::class. You should use is instead:

open class KotlinPractice(val id: Int) {
    override fun equals(other: Any?) =
        this === other ||
                other != null &&
                other is KotlinPractice &&
                other.id == id

    override fun hashCode() = id
}

[edrd]

You could then also eliminate the other != null check as is KotlinPractice is checking that it's not a nullable type either:

open class KotlinPractice(val id: Int) {
    override fun equals(other: Any?) =
        this === other ||
                other is KotlinPractice &&
                other.id == id

    override fun hashCode() = id
}

[Klitos Kyriacou]

Thanks, Eduardo, but I deliberately compared classes to comply with the requirement for equals to be symmetric. Unfortunately, when using is, baseObj == derivedObj does not imply derivedObj == baseObj.

[edrd]

I think only if you override equals in derivedObj, otherwise it's symmetric:

open class KotlinPractice(val id: Int) {
    override fun equals(other: Any?) =
        this === other ||
                other is KotlinPractice &&
                other.id == id

    override fun hashCode() = id
}

class SubKotlinPractice(id: Int) : KotlinPractice(id)

fun main() {
    val kotlinPractice = KotlinPractice(1)
    val subKotlinPractice = SubKotlinPractice(1)
    
    println(kotlinPractice == subKotlinPractice) // true
    println(subKotlinPractice == kotlinPractice) // true
}

[edrd]

x is Y means "x is of type Y or inherits Y"

[Vampire]

Klitos is right with his concern.
As long as it is not a sealed class, anyone could subclass KotlinPractice and overwrite equals, thereby most probably violating symmetry

[Vampire]

Lukes approach is probably best as long as id stays non-nullable

[Klitos Kyriacou]

Thanks, the fact anyone could override equals and break symmetry was exactly my concern. I think Luke's approach (elegant as it looks) is unfortunately still prone to symmetry violations.

[Vampire]

Ah, yeah, you are right of course. Did get too less sleep it seems

[Vampire]

Maybe this one then:

open class KotlinPractice(val id: Int) {
    override fun equals(other: Any?) =
        this === other ||
                other is KotlinPractice &&
                other::class == KotlinPractice::class &&
                other.id == id

    override fun hashCode() = id
}

The is also does the null-check, while then provides smart-cast for later.
This is especially useful if you have more properties to check

[Klitos Kyriacou]

Thanks, that's probably the best we can do for now, although this code checks the other class twice. I was wondering if there was any technical reason that the compiler couldn't do a smart cast after comparison of an instance's class.

[edrd]

IDK either, but out of curiosity, you could use contracts to achieve smart casting for this use-case:

@OptIn(ExperimentalContracts::class)
inline fun <reified T> Any?.isOfType(): Boolean {
  contract {
    returns(true) implies(this@isOfType is T)
  }
  return this != null && this::class == T::class
}

open class KotlinPractice(val id: Int) {
  override fun equals(other: Any?) =
    this === other ||
      other.isOfType<KotlinPractice>() &&
      other.id == id

  override fun hashCode() = id
}