Scala function definition vs application

Question

Why does f in the following code snippet gives the value of 1. I was expected f() to be 1. How can I obtain a reference to the function f:()=> Int

var y = 0
def f():Int = {y + 1}

f

Somethings in scala drive me nuts.

One of the many [differences between methods and functions](https://stackoverflow.com/questions/2529184/difference-between-method-and-function-in-scala). — jwvh, Dec 05 '18 at 00:33
You cannot obtain a reference to the function `f`, because there is no function `f` in your code. There is only a method `f`, which is something completely different. — Jörg W Mittag, Dec 08 '18 at 07:31

Aki · Accepted Answer · 2018-12-12T22:53:31.077

5

If you're calling a function that has no parameters, then you can drop the brackets. That's why f evaluates to 1.

The exact same expression can also evaluate into a function reference if the compiler knows that you're expecting a value of that type.

val foo: () => Int = f

edited Dec 12 '18 at 22:53

answered Dec 04 '18 at 23:29

Aki

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score 3 · Answer 2 · edited Dec 11 '18 at 15:15

3

You can obtain so using _ :

  var y = 0
  def m:Int = {y + 1}
  val result = m _  // type of result is an instance of Function0 "() => Int"

Use _ when compiler is not expecting Function object.

edited Dec 11 '18 at 15:15

Rpant

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answered Dec 05 '18 at 06:17

mukesh210

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I get a Eta-expansion of zero-argument method values is deprecated with this. – Rpant Dec 05 '18 at 18:45
@Rpant Which Scala version are you using? – mukesh210 Dec 06 '18 at 01:53
@Rpant I am not getting any such error or warning with Scala "2.12.7". – mukesh210 Dec 06 '18 at 02:00
@Rpant I tried and it is not giving any warning/error at https://scastie.scala-lang.org/ – mukesh210 Dec 07 '18 at 08:01
I still get the error , but going by the warning i mentioned above , i removed the empty parameter and the eta expansion worked. `var y = 0; def m = {y + 1}; val resultingFunction = m _; resultingFunction`; – Rpant Dec 11 '18 at 15:04

score 2 · Answer 3 · answered Dec 05 '18 at 00:17

If you want f to be an expression of type () => Int that evaluates to { y + 1 }, then just define it as such:

var y = 0
val f: () => Int = () => { y + 1 }

Now

does nothing (it just gives back the lambda of type () => Int), but

f()

gives 1.

You don't really need the type ascription, this works too:

val f = () => { y + 1 }

Scala function definition vs application

3 Answers3