How does foldr work?

Question

Can anybody explain how does foldr work?

Take these examples:

Prelude> foldr (-) 54 [10, 11]
53
Prelude> foldr (\x y -> (x+y)/2) 54 [12, 4, 10, 6]
12.0

I am confused about these executions. Any suggestions?

Answer 1

Ok, lets look at the arguments:

• a function (that takes a list element and a value (a possible partial result) of the same kind of the value it returns);
• a specification of the initial result for the empty list special case
• a list;

return value:

• some final result

It first applies the function to the last element in the list and the empty list result. It then reapplies the function with this result and the previous element, and so forth until it takes some current result and the first element of the list to return the final result.

Fold "folds" a list around an initial result using a function that takes an element and some previous folding result. It repeats this for each element. So, foldr does this starting at the end off the list, or the right side of it.

folr f emptyresult [1,2,3,4] turns into f(1, f(2, f(3, f(4, emptyresult) ) ) ) . Now just follow parenthesis in evaluation and that's it.

One important thing to notice is that the supplied function f must handle its own return value as its second argument which implies both must have the same type.

Source: my post where I look at it from an imperative uncurried javascript perspective if you think it might help.

Answer 2

It helps to understand the distinction between foldr and foldl. Why is foldr called "fold right"?

Initially I thought it was because it consumed elements from right to left. Yet both foldr and foldl consume the list from left to right.

• foldl evaluates from left to right (left-associative)
• foldr evaluates from right to left (right-associative)

We can make this distinction clear with an example that uses an operator for which associativity matters. We could use a human example, such as the operator, "eats":

foodChain = (human : (shark : (fish : (algae : []))))

foldl step [] foodChain
  where step eater food = eater `eats` food  -- note that "eater" is the accumulator and "food" is the element

foldl `eats` [] (human : (shark : (fish : (algae : []))))
  == foldl eats (human `eats` shark)                              (fish : (algae : []))
  == foldl eats ((human `eats` shark) `eats` fish)                (algae : [])
  == foldl eats (((human `eats` shark) `eats` fish) `eats` algae) []
  ==            (((human `eats` shark) `eats` fish) `eats` algae)

The semantics of this foldl is: A human eats some shark, and then the same human who has eaten shark then eats some fish, etc. The eater is the accumulator.

Contrast this with:

foldr step [] foodChain
    where step food eater = eater `eats` food.   -- note that "eater" is the element and "food" is the accumulator

foldr `eats` [] (human : (shark : (fish : (algae : []))))
  == foldr eats (human `eats` shark)                              (fish : (algae : []))))
  == foldr eats (human `eats` (shark `eats` (fish))               (algae : [])
  == foldr eats (human `eats` (shark `eats` (fish `eats` algae))) []
  ==            (human `eats` (shark `eats` (fish `eats` algae) 

The semantics of this foldr is: A human eats a shark which has already eaten a fish, which has already eaten some algae. The food is the accumulator.

Both foldl and foldr "peel off" eaters from left to right, so that's not the reason we refer to foldl as "left fold". Instead, the order of evaluation matters.

Answer 3

foldr means fold from the right, so foldr (-) 0 [1, 2, 3] produces (1 - (2 - (3 - 0))). In comparison foldl produces (((0 - 1) - 2) - 3).

When the operators are not commutative foldl and foldr will get different results.

In your case, the first example expands to (10 - (11 - 54)) which gives 53.