Variable arguments with partial applied functions

Question

I have a compilation problem on the following code.

object Main {
    def main(args:Array[String]) = {
      def collectBigger(median:Int)(values:Int*) = values.filter { _ > median }
      val passedRanks = collectBigger(5)_
      //this compiles
      println(passedRanks(Seq(5,9,5,2,1,3)))
      //this doesn't
      println(passedRanks(5,9,5,2,1,3))
    }
}

The sample is inspired from com.agical.gsl which is a scala adapter to swt. I assume it used scala features before scala 2.8.

The error is too many arguments for method apply: (v1: Seq[Int])Seq[Int] in trait Function1 and is connected on how the variable arguments are passed to a partially applied function.

Thanks for any hints that you could give.

When you partially apply the function collectBigger it appears that the return function is of type Seq[Int] => Seq[Int] = <function1>. If you do however apply both arguments without partially applying it does work as you want. — Rafael Saraiva
– Rafael Saraiva, Commented May 30, 2016 at 13:36
I would also like to know why you are experiencing that kind of behaviour — Rafael Saraiva
– Rafael Saraiva, Commented May 30, 2016 at 13:37
but your code works fine in my ide. the scala version of mine is 2.11.7, what happened? — Jerry
– Jerry, Commented May 30, 2016 at 18:48
I get the same error in scala 2.11.8 and in scala 2.10.6 in both eclipse and sbt. — raisercostin
– raisercostin, Commented May 30, 2016 at 20:46

Dima · Accepted Answer · 2016-05-30 13:43:30Z

10

To put it simply, you can have a varargs method in scala, but not a varags function. Why? Well, all functions have a FunctionN[T1..TN,R] type. In your case, it is Function1[Seq[Int], Seq[Int]].

There is simply no type for a "varargs function", so whenever you convert a method to a function, it must be desugared into the Seq.. notation.

answered May 30, 2016 at 13:43

Dima

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Comments

som-snytt · Accepted Answer · 2016-05-30 19:48:31Z

0

You used to be able to:

$ scala210 -Yeta-expand-keeps-star
Welcome to Scala version 2.10.5 (OpenJDK 64-Bit Server VM, Java 1.7.0_95).
Type in expressions to have them evaluated.
Type :help for more information.

scala> def f(vs: Int*) = vs.sum
f: (vs: Int*)Int

scala> f(1,2,3)
res0: Int = 6

scala> val g = f _
g: Int* => Int = <function1>

scala> g(1,2,3)
res1: Int = 6

But no longer.

https://issues.scala-lang.org/browse/SI-6816

answered May 30, 2016 at 19:48

som-snytt

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Comments

raisercostin · Accepted Answer · 2016-05-30 20:52:59Z

An workaround a value with apply as suggested by Rob Norris

object Main {
  def main(args: Array[String]) = {
    {
      //workaround use Seq as requested
      def collectBigger(median: Int)(values: Int*) = values.filter { _ > median }
      val passedRanks = collectBigger(5)_
      println(passedRanks(Seq(5, 9, 5, 2, 1, 3)))
      println(passedRanks(5, 9, 5, 2, 1, 3))//compile error: too many arguments for method apply: (v1: Seq[Int])Seq[Int] in trait Function1
    }

    {
      //
      def collectBigger(median: Int) = new { def apply(values: Int*) = values.filter { _ > median } }
      val passedRanks = collectBigger(5)
      import scala.language.reflectiveCalls
      println(passedRanks(Seq(5, 9, 5, 2, 1, 3)))//compile error (as expected): type mismatch; found : Seq[Int] required: Int

      //this now works
      println(passedRanks(5, 9, 5, 2, 1, 3))
    }
  }
}

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