Scala placeholder syntax

Question

There is something that I can't quite understand hope someone can shed some light.. I have Seq[String]

val strDeps: Seq[String] = ...

and I tried to sort it on the reverse of the using the sortWith method and I get the following error.

scala> print(strDeps.sortWith(_.reverse.compareTo(_.reverse) < 0) mkString ("\n"))
<console>:15: error: wrong number of parameters; expected = 2
              print(strDeps.sortWith(_.reverse.compareTo(_.reverse) < 0) mkString ("\n"))
                                                                    ^

But when I try sort it without doing a reverse it works fine.

scala> print(strDeps.sortWith(_.compareTo(_) < 0) mkString ("\n"))
// this is fine

Also it works fine without the placeholder syntax

scala> print(strDeps.sortWith((a,b) => a.reverse.compareTo(b.reverse) < 0) mkString ("\n"))
// this works fine too

See also: [Underscore in List.filter](http://stackoverflow.com/questions/7695270). The same problem. — Tomasz Nurkiewicz, Nov 24 '11 at 17:05

score 11 · Accepted Answer · answered Nov 24 '11 at 17:03

11

_ expands only to the smallest possible scope.

The inner _.reverse part is already interpreted as x => x.reverse therefore the parameter is missing inside sortWith.

answered Nov 24 '11 at 17:03

Debilski

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score 8 · Answer 2 · answered Nov 25 '11 at 02:01

compareTo(_)

Is a partially applied method. It just means "compareTo, but without applying the first parameter". Note that _ is not a parameter. Rather, it indicates the absence of a parameter.

compareTo(_.reverse)

Is a method taking an anonymous function as parameter, the parameter being _.reverse. That translates to x => x.reverse.

Scala placeholder syntax

2 Answers2

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