本文介绍: MutableSet的默认实现是LinkedHashSet,LinkedHashSet是有保留元素插入的顺序。MutableSet怎么能调用indexOf去获取它的插入顺序呢?翻不了墙的看下面 (

MutableSet怎么能调用indexOf去获取它的插入顺序呢?参考官网Collections overview | Kotlin Documentationicon-default.png?t=N7T8https://kotlinlang.org/docs/collections-overview.html#set翻不了墙的看下面 (MutableSet的默认实现是LinkedHashSet,LinkedHashSet是有保留元素插入的顺序

Set

Set<T> stores unique elements; their order is generally undefined. null elements are unique as well: a Set can contain only one null. Two sets are equal if they have the same size, and for each element of a set there is an equal element in the other set.()

val numbers = setOf(1, 2, 3, 4)
println("Number of elements: ${numbers.size}")
if (numbers.contains(1)) println("1 is in the set")

val numbersBackwards = setOf(4, 3, 2, 1)
println("The sets are equal: ${numbers == numbersBackwards}")


Number of elements: 4
1 is in the set
The sets are equal: true

MutableSet is a Set with write operations from MutableCollection.

The default implementation of MutableSet – LinkedHashSet – preserves the order of elements insertion. Hence, the functions that rely on the order, such as first() or last(), return predictable results on such sets.

val numbers = setOf(1, 2, 3, 4)  // LinkedHashSet is the default implementation
val numbersBackwards = setOf(4, 3, 2, 1)

println(numbers.first() == numbersBackwards.first())
println(numbers.first() == numbersBackwards.last())
println(numbersBackwards.indexOf(2))


false
true
2

An alternative implementation – HashSet – says nothing about the elements order, so calling such functions on it returns unpredictable results. However, HashSet requires less memory to store the same number of elements.

原文地址:https://blog.csdn.net/qq_50675668/article/details/135403859

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