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Folds for Imperative Programmers

Posted on January 5, 2007, in Programming

Coming from an imperative language such as Java/C/C# and moving to a functional language, you will almost certainly encounter ‘folding functions’. There are usually two folding functions (sometimes there are minor variants) – one folds to the right and the other folds to the left. In Haskell, these functions are named foldr and foldl.

This first encounter with folding functions can often be quite daunting, but I propose an alternative approach. In fact, these folding functions are almost certainly quite familiar to the imperative programmer – just without consciousness. If I had a dollar for every time I saw something like this (I’d buy every CS student a book on the lambda calculus :)):

...
B accum = b;
for(final A a : as) {
  accum = f.f(a, accum);
}
return accum;

Look familiar? Not so daunting? Great, because it’s a fold – specificially, a fold to the left. It is a fold to the left because the sequence (list, array, whatever) is iterated from the left/start to the right/end. Conversely, a fold right iterates from the right/end of the sequence to the left/start.

Let’s take a closer look at the ‘fold left’ above. The first line initialises an accumulator of type B to the given argument. Then the sequence (of A) is iterated and each element is passed to some function and its result assigned to the accumulator. Finally, the accumulated value is returned.

Have I just trivialised a seemingly complicated concept? Not at all! I have undermined its apparent complexity perhaps. On the contrary, it is the imperative language version that has complicated a trivial concept. Let’s take a look at a complete and compilable example:

class Foldr {
  static interface F<A, B> {
    B f(A a, B b);
  }

  static <A, B> B foldr(final F<A, B> f, final B b, final A[] as) {
    B accum = b;

    for(int i = as.length - 1; i >= 0; i--) {
      accum = f.f(as[i], accum);
    }

    return accum;
  }
}

class Foldl {
  static interface F<A, B> {
    A f(A a, B b);
  }

  static <A, B> A foldl(final F<A, B> f, final A a, final B[] bs) {
    A accum = a;

    for(final B b : bs) {
      accum = f.f(accum, b);
    }

    return accum;
  }
}

public class Folds {
  public static void main(final String[] args) {
    final Integer[] i = new Integer[]{7,8,9,42,11,13,45,54,45,46,64,74};

    final Integer right = Foldr.foldr(new Foldr.F<Integer, Integer>() {
      public Integer f(final Integer a, final Integer b) {
        return a - b;
      }
    }, 79, i);

    System.out.println(right);

    final Integer left = Foldl.foldl(new Foldl.F<Integer, Integer>() {
      public Integer f(final Integer a, final Integer b) {
        return a - b;
      }
    }, 97, i);

    System.out.println(left);
  }
}

Phew! What an effort – pass me a beer!

We notice in the main method that the types of both A and B are the same (Integer). This is purely consequential and these types may, and often do, differ. Also notice that our function in each case is not commutative. That is, f(a, b) is not necessarily equivalent to f(b, a). We know this because a - b is not necessarily equivalent to b - a. If the functions were commutative, then the result of a fold right would be equivalent to the result of a fold left. If you execute the code above, you will not get the same output for each fold. Try changing the implementation of each function to be commutative (like +) and observe the equivalent results.

Let’s take a look at the Haskell equivalent at an interpreter:

Prelude> let i = [7,8,9,42,11,13,45,54,45,46,64,74]
Prelude> foldr (-) 97 i
41
Prelude> foldl (-) 79 i
-321

Same outputs as the imperative code? Good, so that’s folds out of the way – easy peasey :)

By the way, yes the Haskell equivalent is supposed to be 100 times shorter – that is the nature of a relatively expressive programming language.