Transform the value of an argument

[GitAntoinee]

Hello, Is there a way to transform the value of an argument like:

val parser = ArgParser(args)
val file: File by parser.option(ArgType.String, transform = { File(it) })

instead of

val parser = ArgParser(args)
val filename: String by parser.option(ArgType.String)
val file: File = File(filename)

[GitAntoinee]

Or another way with a function:

val parser = ArgParser(args)
val file: File by parser.option(ArgType.String).transform { File(it) }

[LepilkinaElena]

Now you can't do so. And I have doubts that you really need transformation in current example. It seems that you need new ArgType.File (unfortunately it isn't supported yet). Or could you provide another example where you need exactly transformation?

[GitAntoinee]

Hello, thanks for the reply. I do not think ArgType.File is supported because the File class used in the examples above is java.io.File, which is not in Kotlin's standard library.

The transformation can be very useful when you need to create an instance from a primitive (Int, Double, String, etc).

Here is another example:

val parser = ArgParser(args)
val repository: UserRepository by parser.option(ArgType.String, transform = { userId: String -> UserRepository(host) })

[LepilkinaElena]

Ok, I unsertood. That was an idea that user can describe his own custom ArgType

You can implement this now using

data class UserRepository(val value: kotlin.String)
object UserRepositoryArgType : ArgType<UserRepository>(true) {
    override val description: kotlin.String
        get() = "{ UserRepository }"

    // You can add checks in case if not all values are valid.
    override fun convert(value: kotlin.String, name: kotlin.String): UserRepository = UserRepository(value)
}

val parser = ArgParser(args)
val repository by parser.option(UserRepositoryArgType)

Is this enough?

[GitAntoinee]

Hello, I know I can do it in this manner, but I do not want to create a new ArgType for each class. If I am not too bored, I will create a pull request