How to initialize a list / collection option from a set?

[sschuberth]

If in the following code PackageManager.ALL is a set:

private val packageManagers by option(
    "--package-managers", "-m",
    help = "The comma-separated package managers to activate, any of ${allPackageManagersByName.keys}."
).convert { name ->
    allPackageManagersByName[name]
        ?: throw BadParameterValue("Package managers must be one or more of ${allPackageManagersByName.keys}.")
}.split(",").default(PackageManager.ALL)

It does not compile as default() expects to get a list. I tried to play with unique(), but failed to find a syntax that compiles as wanted. In the end, I'd also like the packageManagers option to be a set.

Any hints?

[ajalt]

split turns the option value into a list, so default needs a list:

val packageManagers by option().enum<PackageManager>()
        .split(",")
        .default(listOf(PackageManager.ALL)) // or ALL.toList(), if all is already a collection
        .unique()

The type contraints on unique currently prevent it from being used this way, but they can be loosened to allow it.

fun <T, EachT, ValueT> OptionWithValues<List<T>, EachT, ValueT>.unique(): OptionWithValues<Set<T>, EachT, ValueT> {
    return copy(transformValue, transformEach, { transformAll(it).toSet() }, {})
}

[sschuberth]

Thanks, I ended up using PackageManager.ALL.toList(). Not super nice, but I guess it's not worth to implement a splitTo() that would take a mutable set to solve this.

[ajalt]

You should also be able to do split(",").convert{it.toSet()}.default(ALL) if you prefer.

[sschuberth]

No, thanks, I don't want to add another convert call. I just wanted to check that I'm not overlooking an existing elegant solution.