Partially applied functions in Scala

Question

Wondering if you can comment on why following two scenarios behave differently:

The following works:

var la= List(12, 13 , 14 ,15); var func = (x:Int) => println(x) la.foreach(func) // 1 la.foreach(func(_)) // 2

But the following does not:

var la= List(12, 13 , 14 ,15); var func1 = (x:Int) => { for (i <- 0 to x) yield i*2 } mkString la.foreach(println(func1)) // similar to 1 above la.foreach(println(func1(_))) // similar to 2 above

error: type mismatch; found : Unit required: Int => ? la.foreach(println(func1(_)))

tightest non-degenerate scope : see my comment at stackoverflow.com/a/5259946/562716 — Peter Schmitz
– Peter Schmitz, Commented Nov 23, 2012 at 9:48

pedrofurla · Accepted Answer · 2012-11-22 20:49:08Z

5

This case is desugared

la.foreach(println(func1(_)))

to

la.foreach(println(x => func1(x)))

So you passing the function to println, print return type is Unit and foreach requires some Int => ? function.

In contrasting, the first sample in both cases you are feeding foreach with Int => Unit, while in the 2nd sample in both cases you are feeding foreach with Unit.

edited Nov 22, 2012 at 20:49

answered Nov 22, 2012 at 19:08

pedrofurla

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2 Comments

Shalab Over a year ago

I understand that by explicitly specify the argument (x) we can make it work.. but why it does not work with partially supplied argument.? In both your code bits, println is still returning Unit.. so how does it change the behavior for foreach

pedrofurla Over a year ago

It doesn't, it only doesn't type check. See how the desugaring of func1 is different from func

sepp2k · Accepted Answer · 2012-11-22 19:07:54Z

3

In the second code snippet, you're calling println with a function as its argument and then trying to pass the result of that call as an argument to foreach. Since println does not return a function, but foreach wants one, that does not work.

answered Nov 22, 2012 at 19:07

sepp2k

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2 Comments

Shalab Over a year ago

Thanks! how can we fix this, but still using partially supplied argument?

sepp2k Over a year ago

@Shalab To the best of my knowledge, you can't. You'll have to use a lambda with explicit arguments.

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