I need some help understanding the following code as I am complete Kotlin newbie. This is from a kotlin post I found online
typealias genericContext<T> = Demo<T>.() -> Unit
class Demo<T> {
infix fun doThis(block: genericContext<T>) = block()
fun say(obj: T) = println(obj.toString())
}
fun main(args: Array<String>)
{
val demo = Demo<String>()
demo doThis { say("generic alias") }
}
So I understand that because of the infix we can skip the usual method call syntax i.e. demo.doThis and do demo doThis.
But I don't understand the following:
typealias genericContext<T> = Demo<T>.() -> Unit
This seems to associate the string genericContext<T> with something that looks like a lambda but I don't get the .() part. That extends Demo with a function ()? I am confused on how this works. Could someone shed some light?
typealias genericContext<T> = Demo<T>.() -> Unit is a type alias. It is simply giving a new name to the type on the right hand side. This means that declaration of doThis in Demo is equivalent to this:
infix fun doThis(block: Demo<T>.() -> Unit) = block()
Now for the type Demo<T>.() -> Unit:
This is a function type. A function of this type takes a Demo as it's receiver argument, and returns Unit. It is therefore the type of all functions defined either in the Demo class or as an extension on the Demo class.
When you provide a lambda of this type (for example when you call the doThis function), then this will point to a Demo-object inside the lambda. For example:
someDemo.doThis {
/* "this" is an object of type `Demo`.
* In this case it's actually "someDemo", because the implementation of "doThis"
* calls "block" on the implicit "this".
*/
this.say("Hey!")
}
Related
Introduction
In Kotlin I have a generic conversion extension function that simplifies conversion of this object of type C to an object of another type T (declared as the receiver) with additional conversion action that treats receiver as this and also provides access to original object:
inline fun <C, T, R> C.convertTo(receiver: T, action: T.(C) -> R) = receiver.apply {
action(this#convertTo)
}
It is used like this:
val source: Source = Source()
val result = source.convertTo(Result()) {
resultValue = it.sourceValue
// and so on...
}
I noticed I often use this function on receivers that are created by parameterless constructors and thought it would be nice to simplify it even more by creating additional version of convertTo() that automates construction of the receiver based on its type, like this:
inline fun <reified T, C, R> C.convertTo(action: T.(C) -> R) = with(T::class.constructors.first().call()) {
convertTo(this, action) // calling the first version of convertTo()
}
Unfortunately, I cannot call it like this:
source.convertTo<Result>() {}
because Kotlin expects three type parameters provided.
Question
Given above context, is it possible in Kotlin to create a generic function with multiple type parameters that accepts providing just one type parameter while other types are determined from the call-site?
Additional examples (by #broot)
Imagine there is no filterIsInstance() in stdlib and we would like to implement it (or we are the developer of stdlib). Assume we have access to #Exact as this is important for our example. It would be probably the best to declare it as:
inline fun <T, reified V : T> Iterable<#Exact T>.filterTyped(): List<V>
Now, it would be most convenient to use it like this:
val dogs = animals.filterTyped<Dog>() // compile error
Unfortunately, we have to use one of workarounds:
val dogs = animals.filterTyped<Animal, Dog>()
val dogs: List<Dog> = animals.filterTyped()
The last one isn't that bad.
Now, we would like to create a function that looks for items of a specific type and maps them:
inline fun <T, reified V : T, R> Iterable<T>.filterTypedAndMap(transform: (V) -> R): List<R>
Again, it would be nice to use it just like this:
animals.filterTypedAndMap<Dog> { it.barkingVolume } // compile error
Instead, we have this:
animals.filterTypedAndMap<Animal, Dog, Int> { it.barkingVolume }
animals.filterTypedAndMap { dog: Dog -> dog.barkingVolume }
This is still not that bad, but the example is intentionally relatively simple to make it easy to understand. In reality the function would be more complicated, would have more typed params, lambda would receive more arguments, etc. and then it would become hard to use. After receiving the error about type inference, the user would have to read the definition of the function thoroughly to understand, what is missing and where to provide explicit types.
As a side note: isn't it strange that Kotlin disallows code like this: cat is Dog, but allows this: cats.filterIsInstance<Dog>()? Our own filterTyped() would not allow this. So maybe (but just maybe), filterIsInstance() was designed like this exactly because of the problem described in this question (it uses * instead of additional T).
Another example, utilizing already existing reduce() function. We have function like this:
operator fun Animal.plus(other: Animal): Animal
(Don't ask, it doesn't make sense)
Now, reducing a list of dogs seems pretty straightforward:
dogs.reduce { acc, item -> acc + item } // compile error
Unfortunately, this is not possible, because compiler does not know how to properly infer S to Animal. We can't easily provide S only and even providing the return type does not help here:
val animal: Animal = dogs.reduce { acc, item -> acc + item } // compile error
We need to use some awkward workarounds:
dogs.reduce<Animal, Dog> { acc, item -> acc + item }
(dogs as List<Animal>).reduce { acc, item -> acc + item }
dogs.reduce { acc: Animal, item: Animal -> acc + item }
The type parameter R is not necessary:
inline fun <C, T> C.convertTo(receiver: T, action: T.(C) -> Unit) = receiver.apply {
action(this#convertTo)
}
inline fun <reified T, C> C.convertTo(action: T.(C) -> Unit) = with(T::class.constructors.first().call()) {
convertTo(this, action) // calling the first version of convertTo()
}
If you use Unit, even if the function passed in has a non-Unit return type, the compiler still allows you to pass that function.
And there are other ways to help the compiler infer the type parameters, not only by directly specifying them in <>. You can also annotate the variable's result type:
val result: Result = source.convertTo { ... }
You can also change the name of convertTo to something like convert to make it more readable.
Another option is:
inline fun <T: Any, C> C.convertTo(resultType: KClass<T>, action: T.(C) -> Unit) = with(resultType.constructors.first().call()) {
convertTo(this, action)
}
val result = source.convertTo(Result::class) { ... }
However, this will conflict with the first overload. So you have to resolve it somehow. You can rename the first overload, but I can't think of any good names off the top of my head. I would suggest that you specify the parameter name like this
source.convertTo(resultType = Result::class) { ... }
Side note: I'm not sure if the parameterless constructor is always the first in the constructors list. I suggest that you actually find the parameterless constructor.
This answer does not solve the stated problem but incorporates input from #Sweeper to provide a workaround at least simplifying result object instantiation.
First of all, the main stated problem can be somewhat mitigated if we explicitly state variable's result type (i.e. val result: Result = source.convertTo {}) but it's not enough to solve the problem in cases described by #broot.
Secondly, using KClass<T> as result parameter type provides ability to use KClass<T>.createInstance() making sure we find a parameterless constructor (if there's any – if there is none, then result-instantiating convertTo() is not eligible for use). We can also benefit from Kotlin's default parameter values to make result parameter type omittable from calls, we just need to take into account that action might be provided as lambda (last parameter of call) or function reference – this will require two versions of result-instantiating convertTo().
So, taking all the above into account, I've come up with this implementation(s) of convertTo():
// version A: basic, expects explicitly provided instance of `receiver`
inline fun <C, T> C.convertTo(receiver: T, action: T.(C) -> Unit) = receiver.apply {
action(this#convertTo)
}
// version B: can instantiate result of type `T`, supports calls where `action` is a last lambda
inline fun <C, reified T : Any> C.convertTo(resultType: KClass<T> = T::class, action: T.(C) -> Unit) = with(resultType.createInstance()) {
(this#convertTo).convertTo(this#with, action)
}
// version C: can instantiate result of type `T`, supports calls where `action` is passed by reference
inline fun <C, reified T : Any> C.convertTo(action: T.(C) -> Unit, resultType: KClass<T> = T::class) = with(resultType.createInstance()) {
(this#convertTo).convertTo(T::class, action)
}
All three versions work together depending on a specific use case. Below is a set of examples explaining what version is used in what case.
class Source { var sourceId = "" }
class Result { var resultId = "" }
val source = Source()
fun convertX(result: Result, source: Source) {
result.resultId = source.sourceId
}
fun convertY(result: Result, source: Source) = true
fun Source.toResultX(): Result = convertTo { resultId = it.sourceId }
fun Source.toResultY(): Result = convertTo(::convertX)
val result0 = source.convertTo(Result()) { resultId = it.sourceId } // uses version A of convertTo()
val result1: Result = source.convertTo { resultId = it.sourceId } // uses version B of convertTo()
val result2: Result = source.convertTo(::convertX) // uses version C of convertTo()
val result3: Result = source.convertTo(::convertY) // uses version C of convertTo()
val result4: Result = source.toResultX() // uses version B of convertTo()
val result5: Result = source.toResultY() // uses version C of convertTo()
P.S.: As #Sweeper notices, convertTo might not be a good name for the result-instantiating versions (as it's not as readable as with basic version) but that's a secondary problem.
I am writing a sample DSL to create an infra-as-code library. The basic structure would be as below:
class Employee internal constructor (private val init: Employee.() -> Unit) {
var name:String = ""
var address:String = ""
infix fun showMessage(msg:String) =
println("${this.name} resides at ${this.address} and wishes you $msg")
internal fun describe(){
init()
//initialization
}
}
fun employee(process:Employee.()->Unit) = Employee(process).describe()
fun main(){
employee {
name="John Doe"
address="Amsterdam"
this showMessage "happy new year"
// showMessage("This works")
}
}
I am thinking that showMessage infix function should work as other infix inside the Employee context but i need to use this to make it work as an infix.
The function invocation works well in the context without this.
Is this the behaviour of infix functions while using along with DSL or am i missing something?
This is working as designed. From the docs on infix functions:
Note that infix functions always require both the receiver and the parameter to be specified. When you're calling a method on the current receiver using the infix notation, use this explicitly. This is required to ensure unambiguous parsing.
I have an Activity that holds a EditText. Imported via:
import kotlinx.android.synthetic.main.myActivity.*
I would like to use the with function but for some reason only the other scoping function are accessible:
apply, also, run, runCatching, takeIf and takeUnless are shown, but not with (and yes, I scrolled further down and even typed it out. Its marked as unknown if I do so).
To clarify, here what actually happens:
Are there rules to when an object has these functions and when not?
It is because apply, also, run, runCatching, takeIf and takeUnless are extensions, when with is function with 2 parameters. Here is good article about this.
You can use with like this:
with(editText) {
//your code
}
But you can not call with, as you tried to do this:
editText.with() //compilation error
Update:
with function purpose is to call object methods more easy, you don't need to write something like this:
someObject.a();
someObject.b();
someObject.c();
//etc
When you use with, you can write it like this:
with(someObject) {
a();
b();
c();
//etc
}
You're not getting any suggestions about with function because, it has no source type as extension, while other extensions contains source attached to it. How?
Look at the difference below :
with function
#kotlin.internal.InlineOnly
public inline fun <T, R> with(receiver: T, block: T.() -> R): R {
...
return receiver.block()
}
let function
#kotlin.internal.InlineOnly
public inline fun <T, R> T.let(block: (T) -> R): R {
...
return block(this)
}
Here T is the source type for the extension.
The following example is perfectly legal in Kotlin 1.3.21:
fun <T> foo(bar: T): T = bar
val t: Int = foo(1) // No need to declare foo<Int>(1) explicitly
But why doesn't type inference work for higher order functions?
fun <T> foo() = fun(bar: T): T = bar
val t: Int = foo()(1) // Compile error: Type inference failed...
When using higher order functions, Kotlin forces the call site to be:
val t = foo<Int>()(1)
Even if the return type of foo is specified explicitly, type inference still fails:
fun <T> foo(): (T) -> T = fun(bar: T): T = bar
val t: Int = foo()(1) // Compile error: Type inference failed...
However, when the generic type parameter is shared with the outer function, it works!
fun <T> foo(baz: T) = fun (bar: T): T = bar
val t: Int = foo(1)(1) // Horray! But I want to write foo()(1) instead...
How do I write the function foo so that foo()(1) will compile, where bar is a generic type?
I am not an expert on how type inference works, but the basic rule is: At the point of use the compiler must know all types in the expression being used.
So from my understanding is that:
foo() <- using type information here
foo()(1) <- providing the information here
Looks like type inference doesn't work 'backward'
val foo = foo<Int>()//create function
val bar = foo(1)//call function
To put it in simple (possibly over-simplified) terms, when you call a dynamically generated function, such as the return value of a higher-order function, it's not actually a function call, it's just syntactic sugar for the invoke function.
At the syntax level, Kotlin treats objects with return types like () -> A and (A, B) -> C like they are normal functions - it allows you to call them by just attaching arguments in parenthesis. This is why you can do foo<Int>()(1) - foo<Int>() returns an object of type (Int) -> (Int), which is then called with 1 as an argument.
However, under the hood, these "function objects" aren't really functions, they are just plain objects with an invoke operator method. So for example, function objects that take 1 argument and return a value are really just instances of the special interface Function1 which looks something like this
interface Function1<A, R> {
operator fun invoke(a: A): R
}
Any class with operator fun invoke can be called like a function i.e. instead of foo.invoke(bar, baz) you can just call foo(bar, baz). Kotlin has several built-in classes like this named Function, Function1, Function2, Function<number of args> etc. used to represent function objects. So when you call foo<Int>()(1), what you are actually calling is foo<Int>().invoke(1). You can confirm this by decompiling the bytecode.
So what does this have to do with type inference? Well when you call foo()(1), you are actually calling foo().invoke(1) with a little syntactic sugar, which makes it a bit easier to see why inference fails. The right hand side of the dot operator cannot be used to infer types for the left hand side, because the left hand side has to be evaluated first. So the type for foo has to be explicitly stated as foo<Int>.
Just played around with it a bit and sharing some thoughts, basically answering the last question "How do I write the function foo so that foo()(1) will compile, where bar is a generic type?":
A simple workaround but then you give up your higher order function (or you need to wrap it) is to have an intermediary object in place, e.g.:
object FooOp {
operator fun <T> invoke(t : T) = t
}
with a foo-method similar as to follows:
fun foo() = FooOp
Of course that's not really the same, as you basically work around the first generic function. It's basically nearly the same as just having 1 function that returns the type we want and therefore it's also able to infer the type again.
An alternative to your problem could be the following. Just add another function that actually specifies the type:
fun <T> foo() = fun(bar: T): T = bar
#JvmName("fooInt")
fun foo() = fun(bar : Int) = bar
The following two will then succeed:
val t: Int = foo()(1)
val t2: String = foo<String>()("...")
but... (besides potentially needing lots of overloads) it isn't possible to define another function similar to the following:
#JvmName("fooString")
fun foo() = fun(bar : String) = bar
If you define that function it will give you an error similar as to follows:
Conflicting overloads: #JvmName public final fun foo(): (Int) -> Int defined in XXX, #JvmName public final fun foo(): (String) -> String defined in XXX
But maybe you are able to construct something with that?
Otherwise I do not have an answer to why it is infered and why it is not.
In an attempt to understand more about Kotlin and play around with it, I'm developing a sample Android app where I can try different things.
However, even after searching on the topic for a while, I haven't been able to find a proper answer for the following issue :
Let's declare a (dummy) extension function on View class :
fun View.isViewVisibility(v: Int): Boolean = visibility == v
Now how can I reference this function from somewhere else to later call invoke() on it?
val f: (Int) -> Boolean = View::isViewVisibility
Currently gives me :
Error:(57, 35) Type mismatch: inferred type is KFunction2 but (Int) -> Boolean was
expectedError:(57, 41) 'isViewVisibility' is a member and an extension
at the same time. References to such elements are not allowed
Is there any workaround?
Thanks !
Extensions are resolved statically, where the first parameter accepts an instance of the receiver type. isViewVisibility actually accept two parameters, View and Int. So, the correct type of it should be (View, Int) -> Boolean, like this:
val f: (View, Int) -> Boolean = View::isViewVisibility
The error message states:
'isViewVisibility' is a member and an extension at the same time. References to such elements are not allowed
It's saying that the method is both an extension function, which is what you're wanting it to be, and a member. You don't show the entire context of your definition, but it probably looks something like this:
// MyClass.kt
class MyClass {
fun String.coolStringExtension() = "Cool $this"
val bar = String::coolStringExtension
}
fun main() {
print(MyClass().bar("foo"))
}
Kotlin Playground
As you can see the coolStringExtension is defined as a member of MyClass. This is what the error is referring to. Kotlin doesn't allow you to refer to extension function that is also a member, hence the error.
You can resolve this by defining the extension function at the top level, rather than as a member. For example:
// MyClass.kt
class MyClass {
val bar = String::coolStringExtension
}
fun String.coolStringExtension() = "Cool $this"
fun main() {
print(MyClass().bar("foo"))
}
Kotlin Playground
A better fit is the extension function type View.(Int) -> Boolean:
val f: View.(Int) -> Boolean = View::isViewVisibility
But actually the extension types are mostly interchangeable (assignment-compatible) with normal function types with the receiver being the first parameter:
View.(Int) -> Boolean ↔ (View, Int) -> Boolean
I faced the same problem when I declared extension function inside another class and try to pass that extension function as parameter.
I found a workaround by passing function with same signature as extension which in turn delegates to actual extension function.
MyUtils.kt:
object MyUtils {
//extension to MyClass, signature: (Int)->Unit
fun MyClass.extend(val:Int) {
}
}
AnyClass.kt:
//importing extension from MyUtils
import MyUtils.extend
// Assume you want to pass your extension function as parameter
fun someMethodWithLambda(func: (Int)->Unit) {}
class AnyClass {
fun someMethod() {
//this line throws error
someMethodWithLambda(MyClass::extend) //member and extension at the same time
//workaround
val myClassInstance = MyClass()
// you pass a proxy lambda which will call your extension function
someMethodWithLambda { someIntegerValue ->
myClassInstance.extend(someIntegerValue)
}
}
}
As a workaround you can create a separate normal function and invoke it from an inline extension method:
inline fun View.isVisibility(v: Int): Boolean = isViewVisibility(this, v)
fun isViewVisibility(v: View, k: Int): Boolean = (v.visibility == k)
You can't call directly the extension method because you don't have the implicit this object available.
Using either a type with two parameters (the first for the implicit receiver, as #Bakawaii has already mentioned) or an extension type should both work without any warnings at all.
Let's take this function as an example:
fun String.foo(f: Int) = true
You can use assign this to a property that has a two parameter function type like this:
val prop: (String, Int) -> Boolean = String::foo
fun bar() {
prop("bar", 123)
}
Or, you can use an extension function type, that you can then call with either of these two syntaxes:
val prop2: String.(Int) -> Boolean = String::foo
fun bar2() {
prop2("bar2", 123)
"bar2".prop2(123)
}
Again, the above should all run without any errors or warnings.