Declaring same property in Child class as in Parent

Question

I have stared Kotlin recently and have experience in Java. In Java, we can declare the field with same name as it is in parent. When I do same in Kotlin it gives error name hides member of Super type User. What am I missing here?

Kotlin

open class User(protected var name: String)

class TwitterUser(var name: String) : User(name)

Same concept for Java

public class A {
    protected String name;
}

public class B extends A {
    String name;
}

Techically doing that is Java *also* hides the member of the super type, Java's compiler just doesn't error out you when that happens... but it probably tosses a warning at you. — Powerlord, Dec 30 '17 at 10:40
[Unlike Java](https://kotlinlang.org/docs/reference/classes.html#overriding-methods), Kotlin requires explicit annotations for overridable members and for overrides. — BakaWaii, Dec 30 '17 at 11:46

Amit Vaghela · Answer 1 · 2017-12-30T12:14:41.297

1

Change it like,

open class ClassParent(name: String) {

}

class ClassChild(name: String) : ClassParent(name) {

}

you can use init block, check details

edited Dec 30 '17 at 12:14

answered Dec 30 '17 at 10:20

Amit Vaghela

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I know it will work like this. But why the above code gives error? – mallaudin Dec 30 '17 at 10:36
it doest override member of parent and you need to override because you are extening parent class so it show this error – Amit Vaghela Dec 30 '17 at 10:41
@mallaudin The initial error is because a data class can't have parameters in its primary constructor other than val or var properties. Removing the data keyword lifts this restriction. – Amit Vaghela Dec 30 '17 at 10:42
@mallaudin please check https://stackoverflow.com/a/44129826/2826147 and https://stackoverflow.com/questions/44419997/extending-data-class-from-a-sealed-class-in-kotlin – Amit Vaghela Dec 30 '17 at 10:42

tynn · Accepted Answer · 2017-12-30T11:00:46.523

1

The issue is, that you're hiding the implementation of the original property. But there's one possibility to do so anyhow:

open class User(protected open var name: String)

class TwitterUser(override var name: String) : User(name)

You just have to consider, that this mainly changes the implementation of the property. You will not be able to access User.name or TwitterUser.name separately. It's just the same.

edited Dec 30 '17 at 11:00

answered Dec 30 '17 at 10:38

tynn

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With override it says name property is final in User class – mallaudin Dec 30 '17 at 10:42
Ohh right. You need to make the `name` open as well. – tynn Dec 30 '17 at 11:01

s1m0nw1 · Answer 3 · 2017-12-30T17:41:50.963

If your superclass already has a name property, any subclass will have it automatically, too. Why would you then define it again in that child? It's better to simply define it as a parameter of the constructor without making it another property:

open class User(protected var name: String)
//name is not a val/val! simply passed to the constructor as an argument
class TwitterUser(name: String) : User(name)

Otherwise, if you really need to override that property, make it open in the parent and override in the child:

open class User(protected open var name: String)
class TwitterUser(override var name: String) : User(name)

Declaring same property in Child class as in Parent

3 Answers3