Kotlin 父类的init块中的null值
我正在创建一个非常简单的kotlin程序,看到父类有一个奇怪的行为 代码是:Kotlin 父类的init块中的null值,kotlin,Kotlin,我正在创建一个非常简单的kotlin程序,看到父类有一个奇怪的行为 代码是: fun makeSalt(name:String) = Spice(name, "non-spicy") fun main(args: Array<String>) { var salt : Spice = Spice("salt", "non-spicy") println("Salt heat = ${salt.heat}") val spicelist = listOf&l
fun makeSalt(name:String) = Spice(name, "non-spicy")
fun main(args: Array<String>) {
var salt : Spice = Spice("salt", "non-spicy")
println("Salt heat = ${salt.heat}")
val spicelist = listOf<Spice>(
Spice("salt", "non-spicy"),
Spice("turmeric", "mild"),
Spice("Pepper", "hot"),
Spice("Chilli", "hot"),
Spice("Sugar", "non-spicy")
)
val mildSpices = spicelist.filter{it.heat <=5}
val salt2 = makeSalt("rock salt")
val bhoot : SubSpice = SubSpice("bhoot", "hot")
}
open class Spice(open var name:String, open var spiciness:String = "mild" ){
var heat : Int = 5
get() = when (spiciness){
"mild"->5
"hot"->10
"non-spicy"->1
else -> 0
}
init{
if(spiciness === null) {println("spiciness is null")}
else println("Spiciness of ${name} = ${spiciness}; heat = ${heat}")
}
}
class SubSpice(override var name:String, override var spiciness:String = "hot") : Spice(name, spiciness){
}
如您所见,当我创建子类的对象时,父类的变量
spiciness
变为null。有人能解释为什么会这样吗?我希望它为null,因为它还有默认参数“mild”
当您不重写任何getter/setter方法时,您正在使用openvar
您引入了奇怪的初始化冲突,因为Spice.init
(父类构造函数)在子空间.init
之前被调用,并且通过重写字段,它们不再与父构造函数一起初始化-相反,一旦构建了子空间
,它们将可用
从父类中的变量中删除
open
关键字,并在子构造函数中删除override var
,这样,字段将在Spice
类中正确初始化,其init
块将成功运行。当您不重写任何getter/setter方法时,您正在使用openvar
您引入了奇怪的初始化冲突,因为Spice.init
(父类构造函数)在子空间.init
之前被调用,并且通过重写字段,它们不再与父构造函数一起初始化-相反,一旦构建了子空间
,它们将可用
从父类中的变量中删除
open
关键字,并在子构造函数中重写var,这样,字段将在Spice
类中正确初始化,其init
块将成功运行。要为Pawel的答案添加额外信息:
如果使用open
重写属性,则基类中的init块将在初始化派生属性之前运行
见文件:
因此,如果我们像Pawel建议的那样更改代码,我们将得到如下结果:
fun main(args: Array<String>) {
Spice("salt", "non-spicy")
Spice(name="spice")
SubSpice(name = "bhoot", spiciness = "hot")
SubSpice(name="subspice")
}
open class Spice(var name:String, var spiciness : String = "mild" ) {
var heat : Int = 5
get() = when (spiciness){
"mild"->5
"hot"->10
"non-spicy"->1
else -> 0
}
init{
if(spiciness === null) {println("spiciness is null")}
else println("Spiciness of ${name} = ${spiciness}; heat = ${heat}")
}
}
class SubSpice(name: String, spiciness : String = "hot") : Spice(name, spiciness) {
}
要在Pawel的回答中添加额外信息: 如果使用
open
重写属性,则基类中的init块将在初始化派生属性之前运行
见文件:
因此,如果我们像Pawel建议的那样更改代码,我们将得到如下结果:
fun main(args: Array<String>) {
Spice("salt", "non-spicy")
Spice(name="spice")
SubSpice(name = "bhoot", spiciness = "hot")
SubSpice(name="subspice")
}
open class Spice(var name:String, var spiciness : String = "mild" ) {
var heat : Int = 5
get() = when (spiciness){
"mild"->5
"hot"->10
"non-spicy"->1
else -> 0
}
init{
if(spiciness === null) {println("spiciness is null")}
else println("Spiciness of ${name} = ${spiciness}; heat = ${heat}")
}
}
class SubSpice(name: String, spiciness : String = "hot") : Spice(name, spiciness) {
}
谢谢由于我不能将你们的两个答案都标为“已接受”,我已将波维尔的答案标为“已接受的一方”。由于我不能将你们的两个答案都标为“已接受”,所以我将Pawel的答案标为“已接受”
Spiciness of salt = non-spicy; heat = 1
Spiciness of spice = mild; heat = 5
Spiciness of bhoot = hot; heat = 10
Spiciness of subspice = hot; heat = 10