What is out keyword in kotlin

Question

I am not able to understand and I couldn\'t find the meaning of out keyword in kotlin.

You can check example here:

List


        

Answer 1

List<out T> is like List<? extends T> in Java

and

List<in T> is like List<? super T> in Java

For example in Kotlin you can do things like

 val value : List<Any> = listOf(1,2,3)
//since List signature is List<out T> in Kotlin

Answer 2

Refer to thie manual of kotlin

The Kotlin List<out T> type is an interface that provides read only operations like size, get and so on. Like in Java, it inherits from Collection<T> and that in turn inherits from Iterable<T>. Methods that change the list are added by the MutableList<T> interface. This pattern holds also for Set<out T>/MutableSet<T> and Map<K, out V>/MutableMap<K, V>

And this,

In Kotlin, there is a way to explain this sort of thing to the compiler. This is called declaration-site variance: we can annotate the type parameter T of Source to make sure that it is only returned (produced) from members of Source<T>, and never consumed. To do this we provide the out modifier:

> abstract class Source<out T> {
>     abstract fun nextT(): T }
> 
> fun demo(strs: Source<String>) {
>     val objects: Source<Any> = strs // This is OK, since T is an out-parameter
>     // ... }

The general rule is: when a type parameter T of a class C is declared out, it may occur only in out-position in the members of C, but in return C<Base> can safely be a supertype of C<Derived>.

In "clever words" they say that the class C is covariant in the parameter T, or that T is a covariant type parameter. You can think of C as being a producer of T's, and NOT a consumer of T's. The out modifier is called a variance annotation, and since it is provided at the type parameter declaration site, we talk about declaration-site variance. This is in contrast with Java's use-site variance where wildcards in the type usages make the types covariant.

Answer 3

The variance modifiers out and in allow us to make our generic types less restrictive and more reusable by allowing subtyping.

Let's understand this with the help of contrasting examples. We'll use examples of cases as containers of various weapons. Assume that we have the following type hierarchy:

open class Weapon
open class Rifle : Weapon()
class SniperRifle : Rifle()

---

out produces T and preserves subtyping

When you declare a generic type with an out modifier, it's called covariant. A covariant is a producer of T, that means functions can return T but they can't take T as arguments:

class Case<out T> {
    private val contents = mutableListOf<T>()
    fun produce(): T = contents.last()         // Producer: OK
    fun consume(item: T) = contents.add(item)  // Consumer: Error
}

The Case declared with the out modifier produces T and its subtypes:

fun useProducer(case: Case<Rifle>) {
    // Produces Rifle and its subtypes
    val rifle = case.produce()
}

With the out modifier, the subtyping is preserved, so the Case<SniperRifle> is a subtype of Case<Rifle> when SniperRifle is a subtype of Rifle. As a result, the useProducer() function can be called with Case<SniperRifle> too:

useProducer(Case<SniperRifle>())               // OK
useProducer(Case<Rifle>)                       // OK
useProducer(Case<Weapon>())                    // Error

This is less restrictive and more reusable while producing but our class becomes read only.

---

in consumes T and reverses subtyping

When you declare a generic type with an in modifier, it's called contravariant. A contravariant is a consumer of T, that means functions can take T as arguments but they can't return T:

class Case<in T> {
    private val contents = mutableListOf<T>()
    fun produce(): T = contents.last()         // Producer: Error
    fun consume(item: T) = contents.add(item)  // Consumer: OK
}

The Case declared with the in modifier consumes T and its subtypes:

fun useConsumer(case: Case<Rifle>) {
    // Consumes Rifle and its subtypes
    case.consume(SniperRifle())
}

With the in modifier, the subtyping is reversed, so now the Case<Weapon> is a subtype of Case<Rifle> when Rifle is a subtype of Weapon. As a result, the useConsumer() function can be called with Case<Weapon> too:

useConsumer(Case<SniperRifle>())               // Error          
useConsumer(Case<Rifle>())                     // OK
useConsumer(Case<Weapon>())                    // OK

This is less restrictive and more reusable while consuming but our class becomes write only.

---

Invariant produces and consumes T, disallows subtyping

When you declare a generic type without any variance modifier, it's called invariant. An invariant is a producer as well as a consumer of T, that means functions can take T as arguments and can also return T:

class Case<T> {
    private val contents = mutableListOf<T>()
    fun produce(): T = contents.last()         // Producer: OK
    fun consume(item: T) = contents.add(item)  // Consumer: OK
}

The Case declared without in or out modifier produces and consumes T and its subtypes:

fun useProducerConsumer(case: Case<Rifle>) {
    // Produces Rifle and its subtypes
    case.produce()
    // Consumes Rifle and its subtypes
    case.consume(SniperRifle())
}

Without the in or out modifier, the subtyping is disallowed, so now neither Case<Weapon> nor Case<SniperRifle> is a subtype of Case<Rifle>. As a result, the useProducerConsumer() function can only be called with Case<Rifle>:

useProducerConsumer(Case<SniperRifle>())       // Error
useProducerConsumer(Case<Rifle>())             // OK
useProducerConsumer(Case<Weapon>())            // Error

This is more restrictive and less reusable while producing and consuming but we can read and write.

---

Conclusion

The List in Kotlin is a producer only. Because it's declared using the out modifier: List<out T>. This means you cannot add elements to it as the add(element: T) is a consumer function. Whenever you want to be able to get() as well as add() elements, use the invariant version MutableList<T>.

That's it! Hopefully that helps understand the ins and outs of the variance!

Answer 4

Remember like this:

in is "for input" - you wanna put(write) something into it (so it's a "consumer")

out is "for output" - you wanna take(read) something out of it (so it's a "producer")

If you're from Java,

<in T> is for input, so it's like <? super T> (consumer)

<out T> is for output, so it's like <? extends T> (producer)

Answer 5

With this signature:

List<out T>

you can do this:

val doubleList: List<Double> = listOf(1.0, 2.0)
val numberList: List<Number> = doubleList

which means T is covariant:

when a type parameter T of a class C is declared out, C<Base> can safely be a supertype of C<Derived>.

This is contrast with in, e.g.

Comparable<in T>

you can do this:

fun foo(numberComparable: Comparable<Number>) {
  val doubleComparable: Comparable<Double> = numberComparable
  // ...
}

which means T is contravariant:

when a type parameter T of a class C is declared in, C<Derived> can safely be a supertype of C<Base>.

Another way to remember it:

Consumer in, Producer out.

see Kotlin Generics Variance

-----------------updated on 4 Jan 2019-----------------

For the "Consumer in, Producer out", we only read from Producer - call method to get result of type T; and only write to Consumer - call method by passing in parameter of type T.

In the example for List<out T>, it is obvious that we can do this:

val n1: Number = numberList[0]
val n2: Number = doubleList[0]

So it is safe to provide List<Double> when List<Number> is expected, hence List<Number> is super type of List<Double>, but not vice versa.

In the example for Comparable<in T>:

val double: Double = 1.0
doubleComparable.compareTo(double)
numberComparable.compareTo(double)

So it is safe to provide Comparable<Number> when Comparable<Double> is expected, hence Comparable<Double> is super type of Comparable<Number>, but not vice versa.