【Kotlin】Kotlin笔记2-程序的逻辑控制(if,when,循环语句)

矫情吗；* 2022-11-16 02:47 267阅读 0赞

Kotlin笔记-程序的逻辑控制

2. 程序的逻辑控制

  • if条件语句

if

Example:

  1. fun largerNumber3(num1: Int, num2: Int): Int {
  2.     var value = 0
  3.     if (num1 > num2) {
  4.         value = num1
  5.     } else
  6.         value = num2
  8.     return value
  9. }

语法糖

  1. fun largerNumber4(num1: Int, num2: Int): Int {
  2.     val value = if (num1 > num2) {
  3.         num1
  4.     } else {
  5.         num2
  6.     }
  7.     return value
  8. }
  10. fun largerNumber5(num1: Int, num2: Int): Int {
  11.     return if (num1 > num2)
  12.         num1
  13.     else
  14.         num2
  15. }
  17. fun largerNumber6(num1: Int, num2: Int) = if (num1 > num2) {
  18.     num1
  19. } else {
  20.     num2
  21. }
  23. fun largerNumber7(num1: Int, num2: Int) = if (num1 > num2) num1 else num2
  • when条件语句

when

Example if:

  1. fun getScore(name: String) = if (name == "Tom") 86
  2. else if (name == "Jim") 77
  3. else if (name == "Jack") 95
  4. else if (name == "Lily") 100
  5. else 0

Example when:

  1. fun getScore1(name: String) = when (name) {
  2.     "Tom" -> 86
  3.     "Jim" -> 77
  4.     "Jack" -> 95
  5.     "Lily" -> 100
  6.     else -> 0
  7. }

when允许传入任意类型参数,格式:

  1. 匹配值 -> { 执行逻辑 }

类型匹配

  1. fun checkNumber(num: Number) {
  2.     when (num) {
  3.         is Int -> println("number is Int")
  4.         is Double -> println("number is Double")
  5.         else -> println("number not support")
  6.     }
  7. }

is = instanseof(Java)

不传参when

  1. fun getScore2(name: String)=when{
  2.     name=="Tom"->86
  3.     name=="Jim"->77
  4.     name=="Jack"->95
  5.     name=="Lily"->100
  6.     else->0
  7. }

特殊场景 example:

  1. name.startsWith("Tom")->86
  2.     name=="Jim"->77
  3.     name=="Jack"->95
  4.     name=="Lily"->100
  5.     else->0
  6. }
  • 循环语句

while

for ->for-in

闭区间

  1. va range = 0..10 //[0,10]
  3. fun circulation(){
  4.     for (i in 1..10){
  5.         println(i)
  6.     }
  7. }

左闭右开

until

  1. val range 0 until 10

步长

step

  1. fun circulation1(){
  2.     //val range=0 until 10
  3.     for (i in 0 until 10 step 2){
  4.         println(i)
  5.     }
  6. }

downTo

  1. fun circulation2(){
  2.     for (i in 10 downTo 0){
  3.         println(i)
  4.     }
  5. }

q(≧▽≦q):Kotlin新特性

  1. for-in无法实现特殊场景,可以改用while

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