Kotlin default generic parameter of function return value

Question

I have a function that returns generic type based on passed parameter, but I can't set its default value. Here's example function I have:

fun <T : BaseClass> parse(json: String, gson: Class<T> = BaseClass::class): T =
  Gson().fromJson(json, gson)

However, I get type mismatch error for default parameter: expected Class<T>, found Class<BaseClass>. I can achieve same thing using second function:

fun <T : BaseClass> parse(json: String, gson: Class<T>): T =
  Gson().fromJson(json, gson)

fun BaseClass parse(json: String) =
  parse(json, BaseClass::class)

Which doesn't look Kotlin-way. Is it possible to have default generic parameter? Thanks.

Tenfour04 · Accepted Answer · 2020-03-12 12:50:34Z

3

Your first line of code wouldn't be usable in practice. What happens if you do this?

class Derived: BaseClass()

val x = parse<Derived>(someJson)

It would be violating its own definition of the generic type.

Your solution above may be the best you can do for your use case. You might also consider using a reified type like this:

inline fun <reified T : BaseClass> parse(json: String): T =
    Gson().fromJson(json, T::class)

This doesn't provide you any default, but it allows the compiler to infer the type where possible, so in some cases you wouldn't have to specify the class:

fun someFunction(derived: Derived) {
    //...
}

someFunction(parse(someJson)) // no need to specify <Derived>

answered Mar 12, 2020 at 12:50

Tenfour04

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1 Comment

NapoleonTheCake Over a year ago

thanks, I see your point about violating definition. sad part is that I have to rewrite old Java code in Kotlin and still keep them compatible. otherwise I'd go with your solution. still, 2 functions is better than 4 overloaded we had..

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