Partially applied function doesn't work in Scala

Question

Why I can't invoke g2 with the second parameter in the line g2(1, _: Int)(9)?

g2(1, _: Int) returns a function <function1> and then I invoke that <function1> with parameter. At the same time in the example above I can invoke f3(9) however it is the same as g2(1, _: Int)(9) IMHO.

// OK

def f1: (Int, Int) => Int   =   _ + _
val f2: (Int, Int) => Int   =   f1
val f3: (Int) => Int        =   f2(1, _: Int)

println {
  f3(9)
}

// FAIL

def g1: (Int, Int) => Int   =   _ + _
val g2: (Int, Int) => Int   =   g1

println {
  g2(1, _: Int)(9) // Error: 'g2' does not take parameters 
}

To get it work I need to add parentheses: `(g2(1, _: Int))(9)` which is weird. — Finkelson, Oct 08 '15 at 12:53
does `(g2(1, _: Int))(9)` work? scala has multiple parameter lists, so `f(x)(y)` might parse differently from `(f(x))(y)`. I'm not sure how that interacts with the wildcard (`f(x, _)`), though — Silly Freak, Oct 08 '15 at 12:53

score 2 · Answer 1 · edited May 23 '17 at 11:44

2

It's the issue with the scope of underscore in anonymous functions. See here for the rules: What are the rules to govern underscore to define anonymous function?

According to the rule 1 from the link above g2(1, _: Int)(9) is interpreted as (x: Int) => g2(1, x)(9). So you need to add parentheses, for the rule 2 to start working: (g2(1, _: Int))(9)

edited May 23 '17 at 11:44

Community

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answered Oct 08 '15 at 12:59

Kolmar

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Wouldn't that be precedence, not scope? – Silly Freak Oct 08 '15 at 13:03

score 1 · Answer 2 · answered Oct 08 '15 at 12:59

As you add in a comment, (g2(1, _: Int))(9) works.

Scala has multiple parameter lists, so f(x)(y) parses differently from (f(x))(y). As g takes only one parameter list, it can't take the second parameter list, hence the explicit grouping is necessary.

g2(1, _: Int)(9) means: take the function g2(x, y)(z), and apply parameters x = 1, z = 9.

(g2(1, _: Int))(9) means: take the function g2(x, y), and apply parameter x = 1. Then apply y = 9 to the resulting function.

Partially applied function doesn't work in Scala

2 Answers2