How do I send an accumulation function to fold in Scala, the below example will say (Int,Int,Int) does not take parameters . My questions are: Why do I get this error? How is the idomatic way to do what I want in the code below? def a(xs: List[(Int,Int)]): Int = { def logic(acc: (Int,Int, Int), post: (Int,Int)): (Int,Int,Int) = { // do some logic (1, 2, 3) } val to = xs foldLeft((0,0,0))(logic _) to._3 }

Reputation: 7394

Sending functions to fold in Scala

How do I send an accumulation function to fold in Scala, the below example will say (Int,Int,Int) does not take parameters.

My questions are:

Why do I get this error?

How is the idomatic way to do what I want in the code below?

def a(xs: List[(Int,Int)]): Int = {

def logic(acc: (Int,Int, Int), post: (Int,Int)): (Int,Int,Int) = {
    // do some logic

    (1, 2, 3)
}

val to = xs foldLeft((0,0,0))(logic _)

to._3
}

Upvotes: 0

Answers (4)

sebszyller

Reputation: 853

foldLeft is a curried function so it requires a . - go for xs.foldLeft and it will work.

Upvotes: 1

Chobeat

Reputation: 3525

The problem is here:

xs foldLeft((0,0,0))(logic _)

Never go for the dotless notation unless it's for an operator. This way it works.

xs.foldLeft((0,0,0))(logic _)

Without a context I believe this is idiomatic enough.

Upvotes: 2

talex

Reputation: 20544

Try this

xs.foldLeft((0,0,0), logic)

Upvotes: 1

Nyavro

Reputation: 8866

I don't get the goal of your code but the problem you described can be solved this way:

val to = xs.foldLeft((0,0,0))(logic)

Upvotes: 1

Sending functions to fold in Scala

Answers (4)

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