CODEWITHSUNDEEP
kotlin
Zorgan
Zorgan

Reputation: 9173

Why do Kotlin Lambda functions not execute when called?

In the following code I have 2 functions - first one wrapped in a lambda body and the other without.

fun first() = { println("first")}
fun second() = println("second")

first()
second()

Only second() prints - why is this?

Upvotes: 7

Views: 3887

Answers (4)

Cililing
Cililing

Reputation: 4773

It's very simple. Check types of those:

fun first(): () -> Unit = { println("first") }
fun second(): Unit = println("second")

So, when you call first you get lamda expression. To invoke this function use .invoke() (or simply ()): 

first().invoke()
// or
first()()

Second is obvious - it's executed on call.

Upvotes: 5

Steven Spungin
Steven Spungin

Reputation: 29209

Try

first()()

or

first().invoke()

first returns a function, it does not invoke it.

Upvotes: 0

Fred
Fred

Reputation: 17095

The first one is a function that returns a function. What happens in fact is that first() returns another function that prints "first", but doesn't execute it.

To do this you have to call it by adding another set of of parenthesis:

first()()
// Or
val resultOfFirst = first()
resultOfFirst()

This happens because the = sign for functions is analogous to a return statement and when you wrap things in {} you're in fact creating a lambda. Hence, first returns a lambda, but doesn't execute it

Upvotes: 9

DVarga
DVarga

Reputation: 21829

The function first returns a function { println("first")}.

Calling first() does nothing - not even its return argument is catched.

Without a lambda an equivalent would be, maybe it is easier to understand it in this form:

fun firstWithoutLambda() = fun() { println("first w/o lambda")}

To call them:

first().invoke()
firstWithoutLambda().invoke()

which will print the messages that you would expect.

From the original Kotlin tutorial a good article: https://kotlinlang.org/docs/reference/lambdas.html

Upvotes: 2

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