Understanding Folds in Haskell

Question

From what I understand about folds in Haskell, foldl (-) 0 [1..5] gives a result of -15 by calculating 0-1-2-3-4-5, and foldr (-) 0 [1..5] gives a result of -5 by calculating 5-4-3-2-1-0. Why is it then that both foldl (++) "" ["a", "b", "c"] and foldr (++) "" ["a", "b", "c"] give a result of "abc", and the result of foldr is not, instead, "cba"?

Is there something I'm missing in understanding the differences between foldl and foldr?

Answer 1

I think this part from the docs makes it clearer:

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In the case of lists, foldr, when applied to a binary operator, a starting value (typically the right-identity of the operator), and a list, reduces the list using the binary operator, from right to left:

foldr f z [x1, x2, ..., xn] == x1 `f` (x2 `f` ... (xn `f` z)...)

. . .

In the case of lists, foldl, when applied to a binary operator, a starting value (typically the left-identity of the operator), and a list, reduces the list using the binary operator, from left to right:

foldl f z [x1, x2, ..., xn] == (...((z `f` x1) `f` x2) `f`...) `f` xn

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If you look at the example breakdown, the concatenation foldr is equivalent to:

"a" ++ ("b" ++ ("c" ++ ""))

And for foldl, it would be equivalent to:

(("" ++ "a") ++ "b") ++ "c"

For string concatenation, these are the same.

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For subtraction however,

1 - (2 - (3 - 0))

Gives a different result than:

((0 - 1) - 2) - 3

Answer 2

Actually foldr (-) 0 [1..5] equals 3, because it's:

(1 - (2 - (3 - (4 - (5 - 0))))

The answer to this question is in the type of foldr function:

foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b

As we see, (a -> b -> b) function has iterated element as the first argument and accumulator as the second one. That's why with foldr (++) "" ["a", "b", "c"] we have:

("a" ++ ("b" ++ ("c" ++ "")))