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

Question

I have an interface called UserManager who owns a var of the kind User

interface UserManager {

    var user:User

    /* ... */
}

and a class called UserManagerImpl implementing UserManager

class UserManagerImpl : UserManager {

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

    /* ... */
}

My question is this:

How to allow a class to set one User in the UserManager() at any time (* ie * I would not provide a User default, but let another class create and setasse one User when necessary) ?

Remarks

1. Interfaces have no properties lateinit
2. I’d like the class User were nonzero, so I didn’t want null is the default value for user (i and. making var user: User? = null )
3. I would like to access the fields directly instead of using interface methods (i and. call userManager.user to return a user, rather than userManager.getUser() )

Answer 1

It is really not possible to define a property lateinit in an interface.

But it is possible to define this property as 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}")

Will generate the result you want.

---

Another alternative would be to use properties delegated with lazy, but this will require 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}")