Are these two curry function implementations equal?

Question

  def curry[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => f(a, _)




  def curry[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => (b: B) => f(a, b) 

What I was thinking was that in the first implementation I would take the function f and pass in an A and an anything (but the compiler will type check that the second param is a B) to get a C out.

Answer 1

Yes, they are identical. If you compile:

object Test {
  def curry[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => f(a, _)

  def curry2[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => (b: B) => f(a, b)
}

with -Xprint:typer, you get the intermediate abstract syntax tree:

[[syntax trees at end of                     typer]]
package <empty> {
  object Test extends scala.AnyRef {
    def <init>(): Test.type = {
      Test.super.<init>();
      ()
    };
    def curry[A, B, C](f: (A, B) => C): A => (B => C) = ((a: A) => ((x$1: B) => f.apply(a, x$1)));
    def curry2[A, B, C](f: (A, B) => C): A => (B => C) = ((a: A) => ((b: B) => f.apply(a, b)))
  }
}

During the "typer" stage, when the compiler assigns types to everything, it realizes that the _ (now named x$1) must be type B.