One particular example that I recall was that if you calculate something like

    foldl (+) 0 [1..1000000]

(i.e. calculate the sum of the first 1000000 natural numbers)

Then the obvious way to do it is to simply keep a running total, but haskell doesn't do that, it constructs the expression:

    (...((1+2)+3)+4)+5)+...)+1000000)

And tries to evaluate that, which tends to cause stack overflows, and is just generally slow. To fix it you have to use the following expression instead:

    foldl' (+) 0 [1..1000000]

This problem is described in more detail here: https://wiki.haskell.org/Foldr_Foldl_Foldl'

Another discussion today: https://news.ycombinator.com/item?id=11060257

for example: https://news.ycombinator.com/item?id=11060566