I found the following code in the Kotlin forum and it works fine.
sealed class JsonValue<out T>(val value: T) {
class JsonString(value: String) : JsonValue<String>(value)
class JsonBoolean(value: Boolean) : JsonValue<Boolean>(value)
class JsonNumber(value: Number) : JsonValue<Number>(value)
object JsonNull : JsonValue<Nothing?>(null)
class JsonArray<V>(value: Array<V>) : JsonValue<Array<V>>(value)
class JsonObject(value: Map<String, Any?>) : JsonValue<Map<String, Any?>>(value)
override fun toString(): String = value.toString()
}
fun main() {
var pi: JsonValue<Any?>
pi = JsonValue.JsonString("pi"); println (pi)
pi = JsonValue.JsonNumber(3.14); println (pi)
pi = JsonValue.JsonNull; println (pi)
}
But I do not understand why it uses out T.
An answer to a question about out in general states:
out T [...] means functions can return T but they can't take T as arguments.
in T [...] means functions can take T as arguments but they can't return T.
If I take a look at the above code, I can see many constructors (functions), which take T (the value) as an argument. And I see no function which returns T. So my inital impression was: this must be a typo, it should be in T. But it does not even compile with in T.
Why is it necessary to use out T, although the type goes into the constructor?
The constructor doesn't really count :) Only instance members matter - things that you can do to instances of JsonValue.
As explained in the linked answer, the whole idea of (declaration-site) covariance is that you are allowed to implicitly convert an instance of e.g. JsonValue<String> to JsonValue<Any?> if the type JsonValue<T> satisfies some requirements. One of the requirements is that JsonValue<T> should not have any functions that take in any Ts*, because if it did, weird things like this would happen:
val x: JsonValue<Any?> = JsonString("foo")
x.giveMeSomeT(123)
x at runtime holds an instance of JsonString, but the giveMeSomeT method in JsonString would expect a String, not an Int, but as far as the compiler is concerned, x is a JsonValue<Any?>, so this should compile, and bad things would happen at runtime.
So this is why having a function that takes in Ts stops you from marking JsonValue as out T. However, having a constructor that takes in a T is not problematic at all, since situations like the above cannot happen with just a constructor.
And I see no function which returns T
In fact, the getter of value returns T. Also note that you do not need something that returns T to in order to say out T. You just need to to have nothing that takes in Ts. This is vacuously valid for example:
class Foo<out T>
* More accurately and generally, whenever I say "take in any Ts", it should be "have T in an 'in' position", and whenever I say "return a T", it should be "have T in an 'out' position". This is to account for Ts being used as the type argument of other generic types.
Related
I do not fully understand how variance in Generics work. In the code below the classes are as follows Any -> Mammals -> Cats. Any is the supertype, there is a parameter called from in the copy function
From what I understand about the out and in keywords, out allows reference to any of it's subtype, can only be produced not consumed.
in allows reference to any of it's supertype, can only be consumed not produced.
However in the copytest function we are instantiating the function copy. I gave it a catlist1 argument in the from parameter. Since the parameter has an out keyword wouldn't it mean that we can only input parameters that are a subtype of catlist2?
To top of my confusion I have seen many conflicting definitions, for instance , In Kotlin, we can use the out keyword on the generic type which means we can assign this reference to any of its supertypes.
Now I am really confused could anybody guide me on how all of these works? Preferably from scratch, thanks!
class list2<ITEM>{
val data = mutableListOf<ITEM>()
fun get(n:Int):ITEM = data[n]
fun add(Item:ITEM){data.add(Item)}
}
fun <T> Copy(from: list2<out T>, to:list2<T>){
}
fun copytest(){
val catlist1 = list2<Cat>()
val catlist2 = list2<Cat>()
val mammallist = list2<Mammal>()
Copy(catlist1,mammallist)
}
I think maybe you're mixing up class-declaration-site generics and use-site generics.
Class-declaration-site generics
Defined at the class declaration site with covariant out, it is true you cannot use the generic type as the type of a function parameter for any functions in the class.
class MyList<out T>(
private val items: Array<T>
) {
fun pullRandomItem(): T { // allowed
return items.random()
}
fun addItem(item: T) { // Not allowed by compiler!
// ...
}
}
// Reason:
val cowList = MyList<Cow>(arrayOf(Cow()))
// The declaration site out covariance allows us to up-cast to a more general type.
// It makes logical sense, any cow you pull out of the original list qualifies as an animal.
val animalList: MyList<Animal> = cowList
// If it let us put an item in, though:
animalList.addItem(Horse())
// Now there's a horse in the cow list. That doesn't make logical sense
cowList.pullRandomItem() // Might return a Horse, impossible!
It is not logical to say, "I'm going to put a horse in a list that may have the requirement that all items retrieved from it must be cows."
Use-site generics
This has nothing to do with the class level restriction. It's only describing what kind of input the function gets. It is perfectly logical to say, "my function does something with a container that I'm going to pull something out of".
// Given a class with no declaration-site covariance of contravariance:
class Bag<T: Any>(var contents: T?)
// This function will take any bag of food as a parameter. Inside the function, it will
// only get things out of the bag. It won't put things in it. This makes it possible
// to pass a Bag of Chips or a Bag of Pretzels
fun eatBagContents(bagOfAnything: Bag<out Food>) {
eat(bagOfAnything.contents) // we know the contents are food so this is OK
bagOfAnything.contents = myChips // Not allowed! we don't know what kind of stuff
// this bag is permitted to contain
}
// If we didn't define the function with "out"
fun eatBagContentsAndPutInSomething(bagOfAnything: Bag<Food>) {
eat(bagOfAnything.contents) // this is fine, we know it's food
bagOfAnything.contents = myChips // this is fine, the bag can hold any kind of Food
}
// but now you cannot do this
val myBagOfPretzels: Bag<Pretzels> = Bag(somePretzels)
eatBagContentsAndPutInSomething(myBagOfPretzels) // Not allowed! This function would
// try to put chips in this pretzels-only bag.
Combining both
What could be confusing to you is if you saw an example that combines both of the above. You can have a class where T is a declaration site type, but the class has functions where there are input parameters where T is part of the definition of what parameters the function can take. For example:
abstract class ComplicatedCopier<T> {
abstract fun createCopy(item: T): T
fun createCopiesFromBagToAnother(copyFrom: Bag<out T>, copyTo: Bag<in T>) {
val originalItem = copyFrom.contents
val copiedItem = createCopy(originalItem)
copyTo.contents = copiedItem
}
}
This logically makes sense since the class generic type has no variance restriction at the declaration site. This function has one bag that it's allowed to take items out of, and one bag that it's allowed to put items into. These in and out keywords make it more permissive of what types of bags you can pass to it, but it limits what you're allowed to do with each of those bags inside the function.
I'm exploring the Substitution principal and from what I've understood about the principal is that a sub type of any super type should be passable into a function/class. Using this idea in a new section of code that I'm writing, I wanted to implement a abstract interface for a Filter like so
interface Filter {
fun filter(): Boolean
}
I would then imagine that this creates the contract for all classes that inherit this interface that they must implement the function filter and return a boolean output. Now my interpretation of this is that the input doesn't need to be specified. I would like it that way as I want a filter interface that guarantee the implementation of a filter method with a guarantee of a return type boolean. Does this concept even exists in Kotlin? I would then expect to implement this interface like so
class LocationFilter {
companion object : Filter {
override fun filter(coord1: Coordinate, coord2: Coordinate): Boolean {
TODO("Some business logic here")
}
}
}
But in reality this doesn't work. I could remove remove the filter method from the interface but that just defeats the point of the whole exercise. I have tried using varargs but again that's not resolving the issue as each override must implement varargs which is just not helpful. I know this may seem redundant, but is there a possibility to have the type of abstraction that I'm asking for? Or am I missing a point of an Interface?
Let's think about it a little. The main point of abstraction is that we can use Filter no matter what is the implementation. We don't need to know implementations, we only need to know interfaces. But how could we use Filter if we don't know what data has to be provided to filter? We would need to use LocationFilter directly which also defeats the point of creating an interface.
Your problem isn't really related to Kotlin, but to OOP in general. In most languages it is solved by generics/templates/parameterized types. It means that an interface/class is parameterized by another type. You use it in Kotlin like this:
interface Filter<in T> {
fun filter(value: T): Boolean
}
object LocationFilter : Filter<Coordinate> {
override fun filter(value: Coordinate): Boolean {
TODO()
}
}
fun acquireCoordinateFilter(): Filter<Coordinate> = LocationFilter
fun main() {
val coord: Coordinate = TODO()
val filter: Filter<Coordinate> = acquireCoordinateFilter()
val result = filter.filter(coord)
}
Filter is parameterized, meaning that we can have a filter for filtering strings (type is: Filter<String>), for filtering integers (Filter<Int>) or for filtering coordinates (Filter<Coordinate>). Then we can't use e.g. Filter<String> to filter integers.
Note that the code in main() does not use LocationFilter directly, it only knows how to acquire Filter<Coordinate>, but the specific implementation is abstracted from it.
Also note there is already a very similar interface in Java stdlib. It is called Predicate.
my interpretation of this is that the input doesn't need to be specified.
Where did you get that interpretation from?
You can see that it can't be correct, by looking at how the method would be called. You should be able to write code that works for any instance of Filter — and that can only happen if the number and type of argument(s) is specified in the interface. To use your example:
val f: Filter = someMethodReturningAFilterInstance()
val result = f.filter(coord1, coord2)
could only work if all implementations used two Coordinate parameters. If some used one String param, and others used nothing at all, then how would you call it safely?
There are a few workarounds you could use.
If every implementation takes the same number of parameters, then you could make the interface generic, with type parameter(s), e.g.:
interface Filter<T1, T2> {
fun filter(t1: T1, t2: T2): Boolean
}
Then it's up to the implementation to specify which types are needed. However, the calling code either needs to know the types of the particular implementation, or needs to be generic itself, or the interface needs to provide type bounds with in variance.
Or if you need a variable number of parameters, you could bundle them up into a single object and pass that. However, you'd probably need an interface for that type, in order to handle the different numbers and types of parameters, and/or make that type a type parameter on Filter — all of which smells pretty bad.
Ultimately, I suspect you need to think about how your interface is going to be used, and in particular how its method is going to be called. If you're only ever going to call it when the caller knows the implementation type, then there's probably no point trying to specify that method in the interface (and maybe no point having the interface at all). Or if you'll want to handle Filter instances without knowing their concrete type, then look at how you'll want to make those calls.
The whole this is wrong!
First, OOP is a declarative concept, but in your example the type Filter is just a procedure wrapped in an object. And this is completely wrong.
Why do you need this type Filter? I assume you need to get a collection filtered, so why not create a new object that accepts an existing collection and represents it filtered.
class Filtered<T>(private val origin: Iterable<T>) : Iterable<T> {
override fun iterator(): Iterator<T> {
TODO("Filter the original iterable and return it")
}
}
Then in your code, anywhere you can pass an Iterable and you want it to be filtered, you simply wrap this original iterable (any List, Array or Collection) with the class Filtered like so
acceptCollection(Filtered(listOf(1, 2, 3, 4)))
You can also pass a second argument into the Filtered and call it, for example, predicate, which is a lambda that accepts an element of the iterable and returns Boolean.
class Filtered<T>(private val origin: Iterable<T>, private val predicate: (T) -> Boolean) : Iterable<T> {
override fun iterator(): Iterator<T> {
TODO("Filter the original iterable and return it")
}
}
Then use it like:
val oddOnly = Filtered(
listOf(1, 2, 3, 4),
{ it % 2 == 1 }
)
Suppose I have the following function definition.
fun<T> parse(a: Any): T = when (a) {
is String -> a
else -> false
}
I guessed it should be valid. However, the IntelliJ IDEA linter shows a type mismatch error
That being said, I would change the return type of my parse function to Any, right? So that, what is the difference between using Any type and Generics in Kotlin? In which cases should use each of those?
I did read the following question but not understood at all about star-projection in Kotlin due to the fact I am quite new.
Your return type it defined as T, but there is nothing assuring that T and a:Any are related. T may be more restrictive than Any, in which case you can't return a boolean or whatever you provided for a.
The following will work, by changing the return type from T to Any:
fun<T> parse(a: Any): Any = when (a) {
is String -> a
else -> false
}
Any alternate option, if you really want to return type T:
inline fun<reified T> parse(a: Any): T? = when (a) {
is T -> a
else -> null
}
Your example does not use T and thus it's nonsense to make it generic anyways.
Think about this: As a client you put something into a function, e.g. an XML-ByteArray which the function is supposed to parse into an Object. Calling the function you do not want to have it return Any (Casting sucks) but want the function return the type of the parsed object. THIS can be achieved with generics:
fun <T> parse(xml: ByteArray): T {
val ctx: JAXBContext = JAXBContext.newInstance()
val any = ctx.createUnmarshaller().unmarshal(ByteArrayInputStream(xml))
return any as T
}
val int = parse<Int>("123".toByteArray())
val string = parse<String>("123".toByteArray())
Look at the method calls: You tell with generics what type is expected to be returned. The code is not useful and only supposed to give you an idea of generics.
I guessed it should be valid
Why would it be? You return a String in one branch and a Boolean in the other. So the common type for the entire when expression is Any and that's what the compiler (and IDEA) says is "found". Your code also says it should be T (which is "required").
Your generic method should work for any T, e.g. for Int, but Any isn't a subtype of Int and so the code isn't valid.
So that, what is the difference between using Any type and Generics in Kotlin?
This is like asking "what is the difference between using numbers and files": they don't have much in common in the first place. You use generics to write code which can work with all types T (or with all types satisfying some constraint); you use Any when you want the specific type Any.
From the Kotlin documentation:
If a function does not return any useful value, its return type is Unit. Unit is a type with only one value — Unit.VALUE. This value does not have to be returned explicitly:
fun printHello(name : String?) : Unit {
if (name != null)
print("Hello, $name!")
else
print("Hi there!")
// We don't need to write 'return Unit.VALUE' or 'return', although we could
}
What is the purpose of Unit-returning in functions? Why is VALUE there? What is this VALUE?
The purpose is the same as C's or Java's void. Only Unit is a proper type, so it can be passed as a generic argument etc.
Why we don't call it "Void": because the word "void" means "nothing", and there's another type, Nothing, that means just "no value at all", i.e. the computation did not complete normally (looped forever or threw an exception). We could not afford the clash of meanings.
Why Unit has a value (i.e. is not the same as Nothing): because generic code can work smoothly then. If you pass Unit for a generic parameter T, the code written for any T will expect an object, and there must be an object, the sole value of Unit.
How to access that value of Unit: since it's a singleton object, just say Unit
The main reason why Unit exists is because of Generic reasons.
Let's use the example from the Kotlin docs.
class Box<T>(t: T) {
var value = t
}
We can have
var box = Box(Unit)
This is why Unit returns a value so the Kotlin can infer it from the type passed into class initialization. Of course, you could also explicitly write it like this,
var box = Box<Unit>(Unit)
but all the same, it must have a return value.
Now, the void keyword in Java in Kotlin is Nothing. Nothing is the last but one type in the type hierarchy in Kotlin with the last one being Nothing? (Nullable Nothing). This does not return any value at all. Because it doesn't return any value at all, we can't pass it as a type in the above code.
var box = Box(Nothing) //This will return an Error
UNIT actually contains valuable information, it basically just means "DONE". It just returns the information to the caller, that the method has been finished. This is a real piece of information so it can be seen as the return value of a method.
In Scala, you can overload a method by having methods that share a common name, but which either have different arities or different parameter types. I was wondering why this wasn't also extended to the return type of a method? Consider the following code:
class C {
def m: Int = 42
def m: String = "forty two"
}
val c = new C
val i: Int = C.m
val s: String = C.m
Is there a reason why this shouldn't work?
Thank you,
Vincent.
Actually, you can make it work by the magic of 'implicit'. As following:
scala> case class Result(i: Int,s: String)
scala> class C {
| def m: Result = Result(42,"forty two")
| }
scala> implicit def res2int(res: Result) = res.i
scala> implicit def res2str(res: Result) = res.s
scala> val c = new C
scala> val i: Int = c.m
i: Int = 42
scala> val s: String = c.m
s: String = forty two
scala>
You can of course have overloading for methods which differ by return type, just not for methods which differ only by return type. For example, this is fine:
def foo(s: String) : String = s + "Hello"
def foo(i: Int) : Int = i + 1
That aside, the answer to your question is evidently that it was a design decision: the return type is part of the method signature as anyone who has experienced an AbstractMethodError can tell you.
Consider however how allowing such overloading might work in tandem with sub-typing:
class A {
def foo: Int = 1
}
val a: A = //...lookup an A
val b = a.foo
This is perfectly valid code of course and javac would uniquely resolve the method call. But what if I subclass A as follows:
class B extends A {
def foo: String = "Hello"
}
This causes the original code's resolution of which method is being called to be broken. What should b be? I have logically broken some existing code by subtyping some existing class, even though I have not changed either that code or that class.
The main reason is complexity issues: with a "normal" compiler approach, you go inside-out (from the inner expression to the outer scope), building your binary step by step; if you add return-type-only differentiation, you need to change to a backtracking approach, which greatly increases compile time, compiler complexity (= bugs!).
Also, if you return a subtype or a type that can be automatically converted to the other, which method should you choose? You'd give ambiguity errors for perfectly valid code.
Not worth the trouble.
All in all, you can easily refactor your code to avoid return-type-only overload, for example by adding a dummy parameter of the type you want to return.
I've never used scala, so someone whack me on the head if I'm wrong here, but this is my take.
Say you have two methods whose signatures differ only by return type.
If you're calling that method, how does the compiler (interpreter?) know which method you actually want to be calling?
I'm sure in some situations it might be able to figure it out, but what if, for example, one of your return types is a subclass of the other? It's not always easy.
Java doesn't allow overloading of return types, and since scala is built on the java JVM, it's probably just a java limitation.
(Edit)
Note that Covariant returns are a different issue. When overriding a method, you can choose to return a subclass of the class you're supposed to be returning, but cannot choose an unrelated class to return.
In order to differentiate between different function with the same name and argument types, but different return types, some syntax is required, or analysis of the site of an expression.
Scala is an expression oriented language (every statement is an expression). Generally expression oriented languages prefer to have the semantics of expressions to be dependent only on the scope evaluation occurs in, not what happens to the result, so for the expression foo() in i_take_an_int( foo() ) and i_take_any_type ( foo()) and foo() as a statement all call the same version of foo().
There's also the issue that adding overloading by return type to a language with type inference will make the code completely incomprehensible - you'd have to keep an incredible amount of the system in mind in order to predict what will happen when code gets executed.
All answers that say the JVM does not allow this are straight up wrong. You can overload based on return type. Surprisingly, the JVM does allow this; it's the compilers for languages that run on the JVM that don't allow this. But there are ways to get around compiler limitations in Scala.
For example, consider the following snippet of code:
object Overload{
def foo(xs: String*) = "foo"
def foo(xs: Int*) = "bar"
}
This will throw a compiler error (Because varargs, indicated by the * after the argument type, type erase to Seq):
Error:(217, 11) double definition:
def foo(xs: String*): String at line 216 and
def foo(xs: Any*): String at line 217
have same type after erasure: (xs: Seq)String
def foo(xs: Any*) = "bar";
However, if you change value of the second foo to 3 instead of bar (that way changing the return type from String to Int) as follows:
object Overload{
def foo(xs: String*) = "foo"
def foo(xs: Int*) = 3
}
... you won't get a compiler error.
So you can do something like this:
val x: String = Overload.foo()
val y: Int = Overload.foo()
println(x)
println(y)
And it will print out:
3
foo
However, the caveat to this method is having to add varargs as the last (or only) argument for the overloaded functions, each with with their own distinct type.
Source: http://www.drmaciver.com/2008/08/a-curious-fact-about-overloading-in-scala/