Unresolved reference: Curry - kotlin

Question

I want to print out labels of spices in SpiceContainer but I get
Here is my code. How do I solve this?
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

You are trying to create an instance of a class Curry you just don't have. Instead, you have a class Spice with a name and a spiciness. If you want an instance of, say Red Curry, you would have to instantiate a Spice and give it the desired name: Spice("Red Curry", "kinda hot").
A good practice for that spiciness (at least a better one than just a String) would be an enum class Spiciness...
Here's a simple example:
// 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) }
}
and run the fun jack() like this
fun main() {
jack()
}
in order to get an output of
Yellow Curry
Red Curry
Green Curry
Pepper
I don't really get why you need or want the SpiceContainer, but that's beyond the scope of this answer (and question, most likely).