Invoke Operator & Operator Overloading in Kotlin - operator-overloading

Question

I get to know about the Invoke operator that,
a() is equivalent to a.invoke()
Is there anything more regarding Invoke operator than please explain. Also, I did not get any example of Invoke operator overloading.
Is Invoke operator overloading possible? If possible then can anyone please explain about the Invoke operator overloading with an example? I did not get anything regarding this.
Thanks in advance.

Answer 1

Yes, you can overload invoke. Here's an example:
class Greeter(val greeting: String) {
operator fun invoke(target: String) = println("$greeting $target!")
}
val hello = Greeter("Hello")
hello("world") // Prints "Hello world!"
In addition to what #holi-java said, overriding invoke is useful for any class where there is a clear action, optionally taking parameters. It's also great as an extension function to Java library classes with such a method.
For example, say you have the following Java class
public class ThingParser {
public Thing parse(File file) {
// Parse the file
}
}
You can then define an extension on ThingParser from Kotlin like so:
operator fun ThingParser.invoke(file: File) = parse(file)
And use it like so
val parser = ThingParser()
val file = File("path/to/file")
val thing = parser(file) // Calls ThingParser.invoke extension function

Answer 2

The most way to use a invoke operator is use it as a Factory Method, for example:
// v--- call the invoke(String) operator
val data1 = Data("1")
// v--- call the invoke() operator
val default = Data()
// v-- call the constructor
val data2 = Data(2)
This is because the companion object is a special object in Kotlin. Indeed, the code Data("1") above is translated to the code as below:
val factory:Data.Companion = Data
// v-- the invoke operator is used here
val data1:Data = factory.invoke("1")
class Data(val value: Int) {
companion object {
const val DEFAULT =-1
// v--- factory method
operator fun invoke(value: String): Data = Data(value.toInt())
// v--- overloading invoke operator
operator fun invoke(): Data = Data(DEFAULT)
}
}

Answer 3

Operator Function invoke()
Kotlin provides an interesting function called invoke, which is an operator function. Specifying an invoke operator on a class allows it to be called on any instances of the class without a method name.
Let’s see this in action:
class Greeter(val greeting: String) {
operator fun invoke(name: String) {
println("$greeting $name")
}
}
fun main(args: Array<String>) {
val greeter = Greeter(greeting = "Welcome")
greeter(name = "Kotlin")
//this calls the invoke function which takes String as a parameter
}
A few things to note about invoke() here. It:
Is an operator function.
Can take parameters.
Can be overloaded.
Is being called on the instance of a Greeter class without method name.

Answer 4

In addition to the other answers:
It's possible to define a class extending an anonymous function.
class SpecialFunction : () -> Unit {}
In such case, the operator invoke is already defined, so it needs to be overriden:
class MyFunction : () -> Unit {
override fun invoke() { println("Hi Mom") }
}
One more thing about syntax repercussions:
If such "functor" is called right after constructing it, you end up with double parentheses:
MyFunction()()
And, if such functor returns another functor, you may see some obscurities like
MyFunction()()()()()...
perhaps including parameters. This will not surprise anyone coming from the JavaScript world, though.

Answer 5

If you have some Python background,
you can think invoke in Kotlin as __call__ in Python.
By using this, you can "call" your object as if it's a function.
One difference is: you can overload invoke, but there is no official way to overload methods in Python.