fun main() {
val highScores = listOf(4000, 2000, 10200, 12000, 9030)
}
.withIndex() .forEachIndexed() .forEach() .forIndexes() abstract class Aircraft {
init { println("Aircraft = ${getName()}") }
abstract fun getName(): String
}
class Airplane(private val name: String) : Aircraft() {
override fun getName(): String = name
}
fun add(a: Int, b: Int): Int {
return a + b
}
println(add(5,10).invoke()) println(::add.invoke(5, 10)) println(::add.invoke{5, 10}) println(add.invoke(5,10)) fun static main(){} fun main(){} fun Main(){} public static void main(){} fun showHashCode(obj: Any){
println("${obj.hashCode()}")
}
fun main() {
showHashCode(1)
}
.toClass() 方法 val task = launch { // long running job }
task.join() task.abort() job.stop() task.cancel() Long 分配给 Int,反之亦然 Int 到 Long 没有隐式转换 Long 类型 望显示一条特殊消息。使用 Kotlin 范围,您应该添加哪些代码?
when (die) { 1 -> println("die is 1") 2 -> println("die is 2") ___ -> println("die is between 3 and 6") else -> println("die is unknown") }
3,4,5,6 in 3..6 3 : 6 {3,4,5,6} instanceof is typeof as firstName?.let { println("Greeting $firstname!") }
firstName 等于 null firstName 等于空字符串 firstName 等于布尔值 false simple()。您在该函数上添加了 inline 前缀。这对代码有什么影响?inline fun simple(x: Int): Int{
return x * x
}
fun main() {
for(count in 1..1000) {
simple(count)
}
}
for (_____) { println("There are $count butterflies.") }
count in 1..10 count in 2..10 step 2 count in 1..10 % 2 var count=2; count <= 10; count+=2 val set = setOf("apple", "pear", "orange", "apple") println(set.count())
println(b!!.length ?: 0) println(b?.length ?: 0) println(b?.length ?? 0) println(b == null? 0: b.length) removeBadValues。在下面的代码行中,什么是调用 filter 中函数的正确方式?val list2 = (80..100).toList().filter(_____)
::removeBadValues GlobalScope.removeBadValues() Mainkt.removeBadValues removeBadValues for(z in 1..7) println("$z ") for(z in 1..6) print("$z ") for(z in 1 to 6) print("$z ") for(z in 1..7) print("$z ")