Kotlin asterisk operator before variable name or Spread Operator in Kotlin

Question

I want to know what exactly an asterisk does before a variable name in Kotlin. I saw this (*args) in a Spring boot Kotlin example:

@SpringBootApplication
open class Application {

    @Bean
    open fun init(repository: CustomerRepository) = CommandLineRunner {
        repository.save(Customer("Jack", "Bauer"))
        repository.save(Customer("Chloe", "O'Brian"))
        repository.save(Customer("Kim", "Bauer"))
        repository.save(Customer("David", "Palmer"))
        repository.save(Customer("Michelle", "Dessler"))
    }
}

fun main(args: Array<String>) {
    SpringApplication.run(Application::class.java, *args)
}

Answer 1

The * operator is known as the Spread Operator in Kotlin.

From the Kotlin Reference...

When we call a vararg-function, we can pass arguments one-by-one, e.g. asList(1, 2, 3), or, if we already have an array and want to pass its contents to the function, we use the spread operator (prefix the array with *):

It can be applied to an Array before passing it into a function that accepts varargs.

For Example...

If you have a function that accepts a varied number of arguments...

fun sumOfNumbers(vararg numbers: Int): Int {
    return numbers.sum()
}

Use the spread operator to pass an array's elements as the arguments:

val numbers = intArrayOf(2, 3, 4)
val sum = sumOfNumbers(*numbers)
println(sum) // Prints '9'

---

Notes:

• The * operator is also the multiplication operator (of course).
• The operator can only be used when passing arguments to a function. The result of the operation cannot be stored since it yields no value (it is purely syntactic sugar).
• The operator may confuse some C/C++ programmers at first because it looks like a pointer is being de-referenced. It isn't; Kotlin has no notion of pointers.
• The operator can be used in-between other arguments when calling a vararg function. This is demonstrated in the example here.
• The operator is similar to the apply function in various functional programming languages.

Answer 2

In addition to the answers that were directly towards "what is this thing!?!", you often have the case where you have a List and want to pass it to a function that is expecting a vararg. For this, the conversion is:

someFunc(x, y, *myList.toTypedArray())

Assuming that last parameter of someFunc is vararg of the same type as the elements in the list.

Answer 3

As described in the documentation this is a spread operator:

When we call a vararg-function, we can pass arguments one-by-one, e.g. asList(1, 2, 3), or, if we already have an array and want to pass its contents to the function, we use the spread operator (prefix the array with *):

val a = arrayOf(1, 2, 3) 
val list = asList(-1, 0, *a, 4)

Answer 4

In Java you can pass an array as is but an advantage of unpacking an array with spread operator * is that spread operator lets you combine the values from an array and some fixed values in a single call. Java doesn't support this.

Answer 5

If a function which accept a vararg(Variable number of arguments) parameter like:

fun sum(vararg data:Int)
{
   // function body here         
}

Now to call this method, we can do:

sum(1,2,3,4,5)

But what if we have these value in an array, like:

val array= intArrayOf(1,2,3,4,5)

then, to call this method we have to use spread operator, like:

 sum(*array)

Here, *(spread operator) will pass all content of that array.

*array is equivalent to 1,2,3,4,5

But wait a minute, what if we call it like this: sum(array) it will give us Type Mismatch compile time error:

Type mismatch.
Required:Int
Found:IntArray

The problem is sum function accept a vararg Int parameter(which accept value like: 1,2,3,4,5) and if we pass array, it will be passed as IntArray.