foldl / foldr query

Question

I'm a beginner at Haskell, and even after reading several explanations of foldr/foldl, I can't understand why I'm getting different results below. What is the explanation?

Prelude> foldl (\_ -> (+1)) 0 [1,2,3]
4
Prelude> foldr (\_ -> (+1)) 0 [1,2,3]
3

Thanks!

Answer 1

That's because the order of the arguments is flipped in foldl. Compare their type signatures:

foldl :: (a -> b -> a) -> a -> [b] -> a
foldr :: (a -> b -> b) -> b -> [a] -> b

So you see, in your code using foldl, you repeatly increment the accumulator, ignoring the list. But in the code with foldr, you don't even touch the accumulator, but just increment the element of the list. As the last element is 3, the result is 3 + 1 = 4.

You could see your misstake more easy, if you'd use a list of characters aka string instead:

ghci> foldr (\_ -> (+1)) 0 ['a','b','c']
3
ghci> foldl (\_ -> (+1)) 0 ['a','b','c']

:1:20:
    No instance for (Num Char)
      arising from the literal `0'
    Possible fix: add an instance declaration for (Num Char)
    In the second argument of `foldl', namely `0'
    In the expression: foldl (\ _ -> (+ 1)) 0 ['a', 'b', 'c']
    In an equation for `it':
        it = foldl (\ _ -> (+ 1)) 0 ['a', 'b', 'c']
ghci>

Answer 2

In the foldl case, the lambda is being passed the accumulator as the first argument, and the list element as the second. In the foldr case, the lambda is being passed the list element as the first argument, and the accumulator as the second.

Your lambda ignores the first argument, and adds 1 to the second, so in the foldl case you are adding 1 to the last element, and in the foldr case, you are counting the number of elements in the list.