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:
Late answer, but it might help someone arriving to this question from search:
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
来源:https://stackoverflow.com/questions/37524422/equals-method-for-data-class-in-kotlin