Below fn2 failed to compile,
def fn(x: Int)(y: Int) = x + y
def fn2(f: ((Int)(Int)) => Int) = f
fn2(fn)(1)(2) // expected = 3
How to define fn2 to accept fn?
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Jacek Laskowski
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3Please don't just say "failed to compile" but also give the compiler error message. – The Archetypal Paul Oct 07 '14 at 16:12
1 Answers
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It should be as follows:
scala> def fn2(f: Int => Int => Int) = f
fn2: (f: Int => (Int => Int))Int => (Int => Int)
scala> fn2(fn)(1)(2)
res5: Int = 3
(Int)(Int) => Int is incorrect - you should use Int => Int => Int (like in Haskell), instead. Actually, the curried function takes Int and returns Int => Int function.
P.S. You could also use fn2(fn _)(1)(2), as passing fn in previous example is just a short form of eta-expansion, see The differences between underscore usage in these scala's methods.
