未解决的参考：咖喱

Question

我想在 SpiceContainer 中打印出香料标签，但我得到了 这是我的代码。我该如何解决？

package Spice
fun main(args: Array<String> ) {
    jack()
}

class Spice(){
    val name:String=("Curry")
    val spiciness:String=("mild")
}
data class SpiceContainer(var spice: Spice){
    val label=spice.name
}

fun jack(){
    val spiceCabinet = listOf(SpiceContainer(Curry("Yellow Curry", "mild")),
            SpiceContainer(Curry("Red Curry", "medium")),
            SpiceContainer(Curry("Green Curry", "spicy")))
    for(element in spiceCabinet) println(element.label)
}

Answer 1

您正在尝试创建一个您没有的class Curry 实例。相反，您有一个class Spice 和一个name 和一个spiciness。如果你想要一个实例，比如说 Red Curry，你必须实例化一个 Spice 并给它所需的 name：Spice("Red Curry", "kinda hot")。
spiciness（至少比 String 更好）的一个好的做法是 enum class Spiciness...

这是一个简单的例子：

// a spice consisting of a name and some degree of spiciness
data class Spice(val name: String, val spiciness: Spiciness)

// the spice container that gets the label of the spice contained
data class SpiceContainer(val spice: Spice) {
    val label: String = spice.name
}

// enumerate possible degrees of spiciness
enum class Spiciness {
    NONE_AT_ALL,
    MILD,
    MODERATE,
    HOT,
    DANGEROUSLY_HOT,
    LIFE_THREATENING,
    UNKNOWN_TRY_YOURSELF
}

fun jack() {
    // create a list of spices and add some arbitrary ones
    val spiceCabinet = listOf(SpiceContainer(Spice("Yellow Curry", Spiciness.MODERATE)),
                              SpiceContainer(Spice("Red Curry", Spiciness.HOT)),
                              SpiceContainer(Spice("Green Curry", Spiciness.UNKNOWN)),
                              SpiceContainer(Spice("Pepper", Spiciness.MILD)))
     // print each item's label
     spiceCabinet.forEach{ it -> println(it.label) }
                              
}

然后像这样运行fun jack()

fun main() {
    jack()
}

为了得到输出

Yellow Curry
Red Curry
Green Curry
Pepper

我真的不明白你为什么需要或想要SpiceContainer，但这超出了这个答案的范围（很可能是问题）。