如何在Scala中的匿名函数中指定元组类型

Question

I have a function called map_tree as follows:

def fold_tree[A,B](f1: A => B) (f2: (A,B,B) => B) (t: Tree[A]) : B = t match {
    case Leaf(value) => f1(value)
    case Node(value , l, r) => f2 (value, fold_tree (f1) (f2) (l), fold_tree (f1) (f2) (r) )
  }

and I need to implement a function called right_most that takes a Tree[A] and returns A . Here's my attempt at it:

def right_most [A](t:Tree[A]) : A =
    fold_tree ((x: A) => x) ((v: (A, A, A)) => v._3) (t)

But I get the following errors:

 found   : ((A, A, A)) => A
 required: (A, A, A) => A
    fold_tree ((x: A) => x) ((v: (A, A, A)) => v._3) (t)
                                            ^
one error found

Looks to me like found and required are the same. What's the error then? Additionally, how do we specify the tuple type in anonymous functions? And why do I need to specify the tuple type in the function signature. Can't scala infer it?

Answer 1

Scala compiler can infer lot of things for you. so, do this

def right_most [A](t:Tree[A]) : A =
    fold_tree[A, A](_) ((_, _, c) => c) (t)

You got compilation error because you were using f2 params like triplet (tuples). Instead you need function params like this ( (a, b, c) => c )