Kotlin - How to reproduce the effect of "lateinit" on a "var" of an interface?

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Kotlin - How to reproduce the effect of "lateinit" on a "var" of an interface?

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#1 by (1 votes)

1

I have an interface called UserManager which has a var of type User

interface UserManager {

    var user:User

    /* ... */
}

and a class named UserManagerImpl that implements UserManager

class UserManagerImpl : UserManager {

    override var user: User // = uma instancia default de User deve ser fornecida

    /* ... */
}

My question is as follows

How to allow a class to set a User in UserManager() at any time (* ie * I would not provide a User default ), but would leave that another class would create and select a User when needed)?

Notes

Interfaces do not have lateinit properties

I would like the User class to be not null , so I did not want null to be the default value for user ie doing var user: User? = null )

I would like to access the fields directly rather than using interface methods ( i.e ) to call userManager.user to return a user, rather than userManager.getUser ())

interface kotlin propriedade

asked by anonymous 21.04.2016 / 00:32

1 answer

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score 1 · Accepted Answer

You can not really set a lateinit property on an interface.

But you can set this property to lateinit in the subclass.

It would look something like:

interface UserManager {
    var user: User
}

class UserManager Impl: UserManager {
    override lateinit var user: User

    fun initUser(): UserManager = apply { user = User() }
}

println("${UserManagerImpl().initUser().user}")

It will generate the result you want.

Another alternative would be to use delegated properties with lazy , but this will force the property to be immutable, but it may serve you in another context:

interface UserManager {
    val user: User
}

class UserManager Impl: UserManager {
    override val user: User by lazy {
        User()
    }
}

println("${UserManagerImpl().user}")