Equals method for data class in kotlin

问题:

I have the following data class

data class PuzzleBoard(val board: IntArray) {
    val dimension by lazy { Math.sqrt(board.size.toDouble()).toInt() }
}

I read that data classes in Kotlin get equals()/hashcode() method for free.

I instantiated two objects.

val board1 = PuzzleBoard(intArrayOf(1,2,3,4,5,6,7,8,0))
val board2 = PuzzleBoard(intArrayOf(1,2,3,4,5,6,7,8,0))

But still the following statements return false.

board1 == board2
board1.equals(board2)

回答1:

In Kotlin data classes equality check, arrays, just like other classes, are compared using equals(...), which compares the arrays references, not the content. This behavior is described here:

So, whenever you say

• arr1 == arr2

• DataClass(arr1) == DataClass(arr2)

• ...

you get the arrays compared through equals(), i.e. referentially.

Given that,

val arr1 = intArrayOf(1, 2, 3)
val arr2 = intArrayOf(1, 2, 3)

println(arr1 == arr2) // false is expected here
println(PuzzleBoard(arr1) == PuzzleBoard(arr2)) // false too

---

To override this and have the arrays compared structurally, you can implement equals(...)+hashCode() in your data class using Arrays.equals(...) and Arrays.hashCode(...):

override fun equals(other: Any?): Boolean{
    if (this === other) return true
    if (other?.javaClass != javaClass) return false

    other as PuzzleBoard

    if (!Arrays.equals(board, other.board)) return false

    return true
}

override fun hashCode(): Int{
    return Arrays.hashCode(board)
}

^{This code is what IntelliJ IDEA can automatically generate for non-data classes.}

Another solution is to use List<Int> instead of IntArray. Lists are compared structurally, so that you won't need to override anything.

回答2:

For Data classes in Kotlin, hashcode() method will generate and return the same integer if parameters values are same for both objects.

val user = User("Alex", 1)
val secondUser = User("Alex", 1)
val thirdUser = User("Max", 2)

println(user.hashCode().equals(secondUser.hashCode()))
println(user.hashCode().equals(thirdUser.hashCode()))

Running this code will return True and False as when we created secondUser object we have passed same argument as object user, so hashCode() integer generated for both of them will be same.

also if you will check this:

println(user.equals(thirdUser))

It will return false.

As per hashCode() method docs

open fun hashCode(): Int (source)

Returns a hash code value for the object. The general contract of hashCode is:

Whenever it is invoked on the same object more than once, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified.

If two objects are equal according to the equals() method, then calling the hashCode method on each of the two objects must produce the same integer result.

For more details see this discussion here

回答3:

In Kotlin, equals() behaves differently between List and Array, as you can see from code below:

val list1 = listOf(1, 2, 3)
val list2 = listOf(1, 2, 3)

val array1 = arrayOf(1, 2, 3)
val array2 = arrayOf(1, 2, 3)

//Side note: using a==b is the same as a.equals(b)

val areListsEqual = list1 == list2// true
val areArraysEqual = array1 == array2// false

List.equals() checks whether the two lists have the same size and contain the same elements in the same order.

Array.equals() simply does an instance reference check. Since we created two arrays, they point to different objects in memory, thus not considered equal.

Since Kotlin 1.1, to achieve the same behavior as with List, you can use Array.contentEquals().

Source: Array.contentEquals() docs ; List.equals() docs