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:
Adding just a single tick mark to make a big difference in performance is one of many areas in Haskell that make performance-minded development a pain. With these high-level languages, you really have to understand the implementation of the language to get anything fast written.
