Entries tagged with “foldRight”.

In my last post I reviewed the implementation of scala.List’s foldLeft and foldRight methods. That post included a couple of simple examples, but today I’d like to give you a whole lot more. The foldLeft method is extremely versatile. It can do thousands of jobs. Of course, it’s not the best tool for EVERY job, but when working on a list problem it’s a good idea to stop and think, “Should I be using foldLeft?”

Below, I’ll present a list of problem descriptions and solutions. I thought about listing all the problems first, and then the solutions, so the reader could work on his own solution and then scroll down to compare. But this would be very annoying for those who refuse, against my strenuous urging, to start up a Scala interpreter and try to write their own solution to each problem before reading my solution.


Write a function called ‘sum’ which takes a List[Int] and returns the sum of the Ints in the list. Don’t forget to use foldLeft.

def sum(list: List[Int]): Int = list.foldLeft(0)((r,c) => r+c)
def sum(list: List[Int]): Int = list.foldLeft(0)(_+_)

I’ll explain this first example in a bit more depth than the others, just to make sure we all know how foldLeft works.

These two definitions above are equivalent. Let’s examine the first one. The foldLeft method is called on the list parameter. The first parameter is 0. This is the starting value, and the value that will be returned if list is empty. The second parameter is a function literal. It takes parameters ‘r’ (for result) and ‘c’ (for current) and returns the sum of these two values. Scala is smart enough to figure out that since the first parameter (0) is an Int, the ‘r’ parameter must also be an Int. The initial value is always the same type as ‘r’. Scala can also tell that since ‘list’ is a List[Int] the ‘c’ parameter must also be an Int, so we don’t have to specify their types in the parameter list.

The foldLeft method takes that initial value, 0, and the function literal, and it begins to apply the function on each member of the list (parameter ‘c’), updating the result value (parameter ‘r’) each time. That result value that we call ‘r’ is sometimes called the accumulator, since it accumulates the results of the function calls.

In the first defintion, foldLeft’s second parameter (a function literal) uses explicitly named parameters. Notice that ‘r’ and ‘c’ are each referred to exactly once in the function literal, and in the same order as the parameter list. When function literal parameters are used in this way (once each, same order) you can use the shorthand demonstrated in the second definition. The first ‘_’ stands for ‘r’, and the second one stands for ‘c’.


Now that you’ve got the idea, try this one. Write a function that takes a List[Int], and returns the product (of multiplication) of all the Ints in the list. It will be similar to the ‘sum’ function, but with a couple of differences.

def product(list: List[Int]): Int = list.foldLeft(1)(_*_)

Did you get it? It’s the same as ‘sum’ with two exceptions. The initial value is now 1, and the function literal’s parameters are multiplied instead of added. If the initial value were 0, as in ‘sum’, then the function would always return 0.


This one’s a little different. Write a function that takes a List[Any] and returns the number of items in the list. Don’t just call list.length()! Implement it using foldLeft.

def count(list: List[Any]): Int =
  list.foldLeft(0)((sum,_) => sum + 1)

First, we pick our initial value. Remember that this is the value that will be returned for an empty list. An empty list has 0 elements, so we use 0. What function do we want to apply for every item in the list? We just want to increase the result value by one. We call that parameter ‘sum’ in this solution. We don’t care about the actual value of each list element, so we call the second parameter ‘_’, which means it should be discarded.


Here’s a fun one. Write a function that takes a List[Double] and returns the average of the list’s values. There are two ways to go about this one. You could combine two of the previous solutions, using two foldLeft calls, or you could combine them into a single foldLeft. Try to find both solutions.

def average(list: List[Double]): Double =
  list.foldLeft(0.0)(_+_) / list.foldLeft(0.0)((r,c) => r+1)

def average(list: List[Double]): Double = list match {
  case head :: tail => tail.foldLeft( (head,1.0) )((r,c) =>
    ((r._1 + (c/r._2)) * r._2 / (r._2+1), r._2+1) )._1
  case Nil => NaN

The first solution is pretty easy and combines the ‘sum’ and ‘count’ solutions. In real life, of course, you wouldn’t use foldLeft to find the length of the list. You’d just use the length() method. Other than that, though, this is a perfectly sensible solution.

The second solution is more complex. First, the list is matched against two patterns. It is either interpreted as a head item followed by a tail, or as an empty list (Nil). If it’s empty, the function returns the same thing as the first solution, NaN (Not a Number) because you can’t divide by 0.

If the list is not empty, we use a Pair as our initial value. A Pair is just an ordered pair of values. It’s a convenient way to bundle values together. We use it when we need to keep track of more than one accumulator value. In this case, we want to keep track of the average “so far” and also the number of values that the average represents. If the function literal were just passed the average so far, it wouldn’t know how to weight the next value. Members of a Pair are accessed using special methods called ‘_1’ and ‘_2’. You can have groupings longer than 2, also. These are named Tuple3, Tuple4, and so on. In fact, Pair is just an alias of Tuple2. Notice that we didn’t use the word Pair or Tuple2 anywhere in the code. If you enclose a comma-delimited series of values in parentheses, Scala converts that series into the appropriate TupleX.

After we have built up the result, it is a Pair containing the average and the number of items in the list. We only want to return the average so we call ‘_1’ on the result of foldLeft.


Whew! That one was a little tough. Here’s an easier one. Given a List[A] return the last value in the list. Again, no using List’s last() method.

def last[A](list: List[A]): A =
  list.foldLeft[A](list.head)((_, c) => c)

Easy! Mostly. You’ll notice that we’re using a type parameter, A, in this one. If you’re not familiar with type parameters, too bad. I can’t explain them here. Suffice it to say that our use of A here allows us to take a list of any type of contents, and return a result of just that type. So Scala knows that when this is called on a List[Int], it will return an Int. When it’s called on a List[String], it returns a String.

First, we pick an initial value. For the empty list the concept of a last item doesn’t make any sense, so forget that. We can use any value, so long as it’s of type A. list.head is convenient, so that’s our initial value. The function literal is the simplest we’ve seen. For each item in the list, it just returns that item itself. So when it gets to the end of the list, the accumulator holds the last item. We don’t use the accumulator value in the function literal, so it gets parameter name ‘_’.


Write a function called ‘penultimate’ that takes a List[A] and returns the penultimate item (i.e. the next to last item) in the list. Hint: Use a tuple.

def penultimate[A](list: List[A]): A =
  list.foldLeft( (list.head, list.tail.head) )((r, c) => (r._2, c) )._1

This one is very much like the function ‘last’, but instead of keeping just the current item it keeps a Pair containing the previous and current items. When foldLeft completes, its result is a Pair containing the next-to-last and last items. The “_1” method returns just the penultimate item.


Write a function called ‘contains’ that takes a List[A] and an item of type A, and returns true if the item is one of the members of the list, and false if it isn’t.

def contains[A](list: List[A], item: A): Boolean =
  list.foldLeft(false)(_ || _==item)

We choose an initial value of false. That is, we’ll assume the item is not in the list until we can prove otherwise. We use each of the two parameters exactly once and in the proper order, so we can use the ‘_’ shorthand in our function literal. That function literal returns the result so far (a Boolean) ORed with a comparison of the current item and the target value. If the target is ever found, the accumulator becomes true and stays true as foldLeft continues.


Write a function called ‘get’ that takes a List[A] and an index Int, and returns the list value at the index position. Throw an exception if the index is out of bounds.

def get[A](list: List[A], idx: Int): A =
  list.tail.foldLeft((list.head,0)) {
    (r,c) => if (r._2 == idx) r else (c,r._2+1)
  } match {
    case (result, index) if (idx == index) => result
    case _ => throw new Exception("Bad index")

This one has two parts. First there’s the foldLeft, and the result is pattern matched. The foldLeft is pretty easy to follow. The accumulator is a Pair containing the current item and the current index. The current item keeps updating and the current index keeps incrementing until the current index equals the passed in idx. Once the correct index is found the same accumulator is returned over and over. This works fine if idx parameter is in bounds. If it’s out of bounds, though, the foldLeft just returns a Pair containing the last item and the last index. That’s where the pattern match comes in. If the Pair contains the right index then we use the result item. Otherwise, we throw an exception.


Write a function called ‘mimicToString’ that mimics List’s own toString method. That is, it should return a String containing a comma-delimited series of string representations of the list contents with “List(” on the left and “)” on the right.

def mimicToString[A](list: List[A]): String = list match {
  case head :: tail => tail.foldLeft("List(" + head)(_ + ", " + _) + ")"
  case Nil => "List()"

This one also uses a pattern match, but this time the match happens first. The pattern match just treats the empty list as a special case. For the general case (a non-empty list) we use, of course, foldLeft. The accumulator starts out as “List(” + the head item. Then each remaining item (notice foldLeft is called on tail) is appended with a leading “, ” and a final “)” is added to the result of foldLeft.


This one’s kind of fun. Make sure to try it before you look at my solution. Write a function called ‘reverse’ that takes a List and returns the same list in reverse order.

def reverse[A](list: List[A]): List[A] =
  list.foldLeft(List[A]())((r,c) => c :: r)

A very simple solution! The initial value of the accumulator is just an empty list. We don’t use Nil, but instead spell out the List type so that Scala will know what type to make ‘r’. As I say, we start with the empty list which is sensible because the reverse of an empty list is an empty list. Then, as we go through the list, we place each item at the front of the accumulator. So the item at the front of list becomes the last item in the accumulator. This goes on until we reach the end of list, and that last member of list goes onto the front of the accumulator. It’s a really neat and tidy solution.


Write a function called ‘unique’ that takes a List and returns the same List, but with duplicated items removed.

def unique[A](list: List[A]): List[A] =
  list.foldLeft(List[A]()) { (r,c) =>
    if (r.contains(c)) r else c :: r

As usual, we start with an empty list. foldLeft looks at each list item and if it’s already contained in the accumulator then then it stays as it is. If it’s not in the accumulator then it’s appended. This code bears a striking similarity to the ‘reverse’ function we wrote earlier except for the “if (r.contains(c)) r” part. Because of this, the foldLeft result is actually the original list with duplicates removed, but in reverse order. To keep the output in the same order as the input, we add the call to reverse. We could also have chained on the foldLeft from the ‘reverse’ function, like so:

def unique[A](list: List[A]): List[A] =
  list.foldLeft(List[A]()) { (r,c) =>
    if (r.contains(c)) r else c :: r
  }.foldLeft(List[A]())((r,c) => c :: r)


Write a function called ‘toSet’ that takes a List and returns a Set containing the unique elements of the list.

def toSet[A](list: List[A]): Set[A] =
  list.foldLeft(Set[A]())( (r,c) => r + c)

Super easy one. You just start out with an empty Set, which would be the right answer for an empty List. Then you just add each list item to the accumulator. Since the accumulator is a Set, it takes care of eliminating duplicates for you.


Write a function called ‘double’ that takes a List and a new List in which each item appears twice in a row. For example double(List(1, 2, 3)) should return List(1, 1, 2, 2, 3, 3).

def double[A](list: List[A]): List[A] =
  list.foldLeft(List[A]())((r,c) => c :: c :: r).reverse

Again, pretty easy. Are you starting to see a pattern. When you use foldLeft to transform one list into another, you usually end up with the reverse of what you really want.

Alternately, you could have used the foldRight method instead. This does the same thing as foldLeft, except it accumulates its result from back to front instead of front to back. I can’t recommend using it, though, due to problems I point out in my other post on foldLeft and foldRight. But here’s what it would look like:

def double[A](list: List[A]): List[A] =
  list.foldRight(List[A]())((c,r) => c :: c :: r)


This one takes some thinking. Write a function called ‘insertionSort’ that uses foldLeft to sort the input List using the insertion sort algorithm. Try it on your own before you look at the solution.

Need a hint? Use List’s ‘span’ method.

Did you find a solution? Here’s mine:

def insertionSort[A <% Ordered[A]](list: List[A]): List[A] =
  list.foldLeft(List[A]()) { (r,c) =>
    val (front, back) = r.span(_ < c)
    front ::: c :: back

First, the type parameter ensures that we have elements that can be arranged in order. We start, predictably, with an empty list as our initial accumulator. Then, for each item we assume the accumulator is in order (which it always will be), and use span to split it into two sub-lists: all already-sorted items less than the current item, and all already-sorted items greater than or equal to the current item. We put the current item in between these two and the accumulator remains sorted. This is, of course, not the fastest way to sort a list. But it’s a neat foldLeft trick.


Speaking of sorting, you can implement part of quicksort with foldLeft, the pivot. Write a function called ‘pivot’ that takes a List, and returns a Tuple3 containing: (1) a list of all elements less than the original list’s first element, (2) the first element, and (3) a List of all elements greater than or equal to the first element.

def pivot[A <% Ordered[A]](list: List[A]): (List[A],A,List[A]) =
  list.tail.foldLeft[(List[A],A,List[A])]( (Nil, list.head, Nil) ) {
    (result, item) =>
    val (r1, pivot, r2) = result
    if (item < pivot) (item :: r1, pivot, r2) else (r1, pivot, item :: r2)

We’re using the first element, head, as the pivot value, so we skip the head and call foldLeft on list.tail. We initialize the accumulator to a Tuple3 containing the head element with an empty list on either side. Then for each item in the list we just pick which of the two lists to add to based on a comparison with the pivot value.

If you take the additional step of turning this into a recursive call, you can implement a quicksort algorithm. It probably won’t be a very efficient one because it will involve a lot of building and rebuilding lists. Give it a try if you like, and then look at my solution:

def quicksort[A <% Ordered[A]](list: List[A]): List[A] = list match {
  case head :: _ :: _ =>
    list.foldLeft[(List[A],List[A],List[A])]( (Nil, Nil, Nil) ) {
      (result, item) =>
      val (r1, r2, r3) = result
      if      (item < head) (item :: r1, r2, r3)
      else if (item > head) (r1, r2, item :: r3)
      else                  (r1, item :: r2, r3)
    } match {
      case (list1, list2, list3) =>
        quicksort(list1) ::: list2  ::: quicksort(list3)
  case _ => list

Basically, for all lists that have more than 1 element the function chooses the head element as the pivot value, uses foldLeft to divide the list into three (less than, equal to, and greater than the pivot), recursively sorts the less-than and greater-than lists, and knits the three together.


Ok, we got a little into the weeds with that last one. Here’s a simpler one. Write a function called ‘encode’ that takes a List and returns a list of Pairs containing the original values and the number of times they are repeated. So passing List(1, 2, 2, 2, 2, 2, 3, 2, 2) to encode will return List((1, 1), (2, 5), (3, 1), (2, 2)).

def encode[A](list: List[A]): List[(A,Int)] =
list.foldLeft(List[(A,Int)]()){ (r,c) =>
    r match {
      case (value, count) :: tail =>
        if (value == c) (c, count+1) :: tail
        else            (c, 1) :: r
      case Nil =>
        (c, 1) :: r


You knew this was coming. Write a function called ‘decode’ that does the opposite of encode. Calling ‘decode(encode(list))’ should return the original list.

def decode[A](list: List[(A,Int)]): List[A] =
list.foldLeft(List[A]()){ (r,c) =>
    var result = r
    for (_ <- 1 to c._2) result = c._1 :: result

Encode and decode could both have been written by using foldRight and dropping the call to reverse.


One last example. Write a function called ‘group’ that takes a List and an Int size that groups elements into sublists of the specified sizes. So calling “group( List(1, 2, 3, 4, 5, 6, 7), 3)” should return List(List(1, 2, 3), List(4, 5, 6), List(7)). Don’t forget to make sure list items are in the right order. Try it yourself before you look at the solution below.

def group[A](list: List[A], size: Int): List[List[A]] =
  list.foldLeft( (List[List[A]](),0) ) { (r,c) => r match {
    case (head :: tail, num) =>
      if (num < size)  ( (c :: head) :: tail , num + 1 )
      else             ( List(c) :: head :: tail , 1 )
    case (Nil, num) => (List(List(c)), 1)
  }._1.foldLeft(List[List[A]]())( (r,c) => c.reverse :: r)

This code uses the first foldLeft to group the items in a way that’s convenient to list operations, and that last foldLeft to fix the order, which would otherwise be wrong in both the outer and inner lists.

The End!

That’s all for now. If you know of any neat foldLeft tricks, please do leave a comment. I’d be interested to hear about it.

One of my favorite functional programming tricks is folding. The fold left and fold right functions can do a lot of complicated things with a small amount of code. Today, I’d like to (1) introduce folding, (2) make note of some surprising, nay, shocking fold behavior, (3) review the folding code used in Scala’s List class, and (4) make some presumptuous suggestions on how to improve List.

Update: I’ve created a new post in which I list lots and lots of foldLeft examples in case you’d like to learn more about what folding can accomplish.

Know When to Hold ‘Em, Know When to Fold ‘Em

In case you’re not familiar with folding, I’ll describe it as briefly as I can.

Here’s the signature of the foldLeft function from List[A], a list of items of type A:

def foldLeft[B](z: B)(f: (B, A) => B): B

Firstly, foldLeft is a curried function (So is foldRight). If you don’t know about currying, that’s ok; this function just takes its two parameters (z and f) in two sets of parentheses instead of one. Currying isn’t the important part anyway.

The first parameter, z, is of type B, which is to say it can be different from the list contents type. The second parameter, f, is a function that takes a B and an A (a list item) as parameters, and it returns a value of type B. So the purpose of function f is to take a value of type B, use a list item to modify that value and return it.

The foldLeft function goes through the whole List, from head to tail, and passes each value to f. For the first list item, that first parameter, z, is used as the first parameter to f. For the second list item, the result of the first call to f is used as the B type parameter.

For example, say we had a list of Ints 1, 2, and 3. We could call foldLeft(“X”)((b,a) => b + a). For the first item, 1, the function we define would add string “X” to Int 1, returning string “X1″. For the second list item, 2, the function would add string “X1″ to Int 2, returning “X12″. And for the final list item, 3, the function would add “X12″ to 3 and return “X123″.

Here are a few more examples.

list.foldLeft(0)((b,a) => b+a)
list.foldLeft(1)((b,a) => b*a)
list.foldLeft(List[Int]())((b,a) => a :: b)

The first line is super simple. It’s almost like the example I described above, but the z value is the number 0 instead of string “X”. This fold combines the elements of the list by addition instead of concatenation. So the fold returns the sum of all Ints in the list. Line 2 combines them through multiplication. Do you see why the z value is 1 in this case?

Line 3 is a little more complex. Can you guess what it does? It starts out with an empty list of Ints and adds each item to the accumulator (We call the b parameter of our function the accumulator because it accumulates data from each of our list items). Because it starts with the head and adds to the beginning of the accumulator list until it gets to the last item of the original list, it returns the original list in reverse order.

The foldRight function works in much the same way as foldLeft. Can you guess the difference? You got it. It starts at the end of the list and works its way up to the head.

Folds can be used for MUCH more than I’ve shown here. With folds, you can solve lots of different problems with a standard construct. You should read up on them if you’re just starting out in functional programming.

All That Glitters Is Not Fold

Now for the moment you’ve been waiting for. Fold’s dirty little secret! The below is taken from a scala interpreter session.

scala> var shortList = 1 to 10 toList
shortList: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> var longList = 1 to 325000 toList
longList: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, ...

scala> shortList.foldLeft("")((x,y) => "X")
res1: java.lang.String = X

scala> shortList.foldRight("")((x,y) => "X")
res2: java.lang.String = X

scala> longList.foldLeft("")((x,y) => "X")
res3: java.lang.String = X

scala> longList.foldRight("")((x,y) => "X")
        at scala.List.foldRight(List.scala:1079)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRight(List.scala:1081)
        at scala.List.foldRig...

We created two lists: shortList with 10 items, and longList with 325,000 items. Then we perform a trivial foldLeft and foldRight on shortList. It’s trivial because the passed-in function always returns the string “X”; it doesn’t even use the list data.

Then we do a foldLeft on longList. This goes off without a hitch. Finally we try to do a foldRight, the same foldRight that succeeded on the shorter list, and it fails! The foldLeft worked. Why didn’t the foldRight work? It’s a perfectly reasonable call against a perfectly reasonable List. Something funny is going on here.

The error message says there was a stack overflow, and the stack trace shows a long list of calls at List.scala:1081. If you’ve read my post about tail-recursion, then you probably suspect that some recursive code is to blame.

Let’s look into List.scala, maybe the single most important Scala source file.

Fool’s Fold

Without further ado, here’s the code for foldLeft and foldRight from List.scala:

override def foldLeft[B](z: B)(f: (B, A) => B): B = {
  var acc = z
  var these = this
  while (!these.isEmpty) {
    acc = f(acc, these.head)
    these = these.tail

override def foldRight[B](z: B)(f: (A, B) => B): B = this match {
  case Nil => z
  case x :: xs => f(x, xs.foldRight(z)(f))

Wow! Those two definitions are very different!

The foldLeft function is the one that worked for short and long lists. You can see why? It isn’t head-recursive. In fact, it isn’t recursive at all. It is implemented as a while loop. On each iteration, the next list item is passed to the function f and the accumulator (called acc) is updated. When there are no more list items, the accumulator is returned. No recursion means no stack overflows.

The foldRight function is implemented in a totally different way. If the list is empty, the z parameter is returned. Otherwise, a recursive call is made on the tail (the whole list minus the first item) of this list, and that result is passed to the function f. Study the foldRight definition. Do you understand how it works? It’s an elegant recursive solution, and the code really is quite pretty, but it’s not tail recursive so it fails for large lists.

Why didn’t Mr Odersky just write foldRight using a while loop, too? Then this problem wouldn’t exist, right? The reason is that Scala’s List is a implemented as a singly-linked list. Each list element has access to the next item in the list, but not to the previous item. You can only traverse a list in one direction! This works fine for foldLeft, which goes from head to tail, but foldRight has to start at the end of the list and work its way forward to the head. If foldRight uses recursion, it must recurse all the way to the end and then use the results of those recursive calls as the accumulator passed into function f.

See? The results of the recursive call must be used for further calculation, so the recursive call can’t be the last thing that happens, so it can’t be written as a tail-recursive function. If you don’t know what I’m talking about, read my introduction to tail-recursion.

Out With The Fold, In With The New

So is that it for foldRight? Is it hopeless? I say no!

There is a way to get the same result as foldRight, but using foldLeft. Can you guess what it is? Here’s how:

list.foldRight("X")((a,b) => a + b)
list.reverse.foldLeft("X")((b,a) => a + b)

These two lines are equivalent! They give the same result no matter what’s in list. Since foldRight processes list elements from last to first, that’s the same as processing the reversed list from first to last.

Here are three possible implementations of foldRight that could replace the current one.

def foldRight[B](z: B)(f: (A, B) => B): B = 
  reverse.foldLeft(z)((b,a) => f(a,b))

def foldRight[B](z: B)(f: (A, B) => B): B =
  if (length > 50) reverse.foldLeft(z)((b,a) => f(a,b))
  else             originalFoldRight(z)(f)

def foldRight[B](z: B)(f: (A, B) => B): B =
  try {
  } catch {
    case e1: StackOverflowError => reverse.foldLeft(z)((b,a) => f(a,b))

The first one simply replaces the original recursive logic with the equivalent call to reverse and foldLeft. Why wasn’t foldRight implemented this way to begin with? It may be, in part, that the authors thought the extra overhead of reversing the list was unwarranted. To me, it doesn’t seem that bad. The original foldRight and foldLeft functions are O(n), meaning they run in an amount of time roughly proportional to the number of items in the list. If you look at the source for the reverse function, you’ll see it’s also O(n). So running reverse followed by foldLeft is O(n).

The second implementation is a compromise. It uses the original recursive version of foldRight (referred to as originalFoldRight in the above code) only when the list is shorter than 50 elements. The reverse.foldLeft is used for lists of 50 elements or longer. 50 is just an arbitrary number, just a guess at a sensible limit on the number of recursive calls to allow.

The third implementation tries the original foldRight logic first and if the call stack overflows then it uses reverse.foldLeft. This solution is, of course, completely ridiculous, but even this would be better than a foldRight which sometimes crashes your program.

That’s All, Folds!

As I pointed out before, the reverse.foldLeft implementation of foldRight is O(n), same as the original recursive version. The original foldRight may work just fine when your Scala application is young and working with small data sets. Over time more customers are added, more products are created, more orders are placed, and then one day, *POOF*, a runtime error! It’s a ticking time-bomb.

As you may well guess, I would like to see the reverse.foldLeft logic used instead of the recursive version. That would prevent the stack overflow errors. But I would settle for just deprecating foldRight. It would be better to eliminate foldRight and force the coder to work around it than to leave it in its current state. In fact, I don’t think any head-recursive functions belong in the List class.

Do any readers have any insight into why foldRight is coded the way it is?