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Title
a

Ayden

07/30/2018, 7:14 AM
val parameterObject = element?.asJsonObject
val parameters: Map<String, List<String>> = HashMap()
for (entry in parameterObject?.entrySet().orEmpty()) {
    val key = entry.key
    val value = entry.value.asString
    parameters.put(key, value)
}
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Lucas Ł

07/30/2018, 7:16 AM
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Ayden

07/30/2018, 7:17 AM
Thanks @Lucas Ł!
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arekolek

07/30/2018, 7:21 AM
your map is mutable, but then you narrow the type down to immutable
either do
val parameters = HashMap<String, List<String>>()
or
val parameters: MutableMap<String, List<String>> = HashMap()
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Lucas Ł

07/30/2018, 7:22 AM
or
val parameters = mutableMapOf<String, List<String>>()
a

arekolek

07/30/2018, 7:24 AM
I assumed he cared about the implementing type, but if you don't care about it, that's the best one to use
a

Ayden

07/30/2018, 7:32 AM
Thanks guys.
I managed to implement it.
I though if I don't defined anything.
I would be MutableList by default.
a

arekolek

07/30/2018, 7:41 AM
only if it comes from Java
t

thymecypher

07/30/2018, 11:38 PM
Let's not get into the ways we could do this. This is Kotlin, we'd be here all year :p