Haskell thunks - foldl vs foldr

Question

Learning Haskell, I came across the fact that foldl creates thunks and might crash the stack, so it's better to use foldl' from Data.List . Why is it just foldl , and not, for example, foldr ?

Thanks

Answer 1

There is no need for foldr' because you can cause the effect yourself.

Here is why: Consider foldl f 0 [1,2,3] . This expands to f (f (f 0 1) 2) 3 , so by the time you get anything back to work with, thunks for (f 0 1) and (f (f 0 1) 2) have to be created. If you want to avoid this (by evaluating these subexpressions before continuing), you have to instruct foldl to do it for you – that is foldl' .

With foldr , things are different. What you get back from foldr f 0 [1, 2, 3] is f 1 (foldr f 0 [2, 3]) (where the expression in parenthesis is a thunk). If you want to evaluate (parts of) the outer application of f , you can do that now, without a linear number of thunks being created first.

But in general, you are using foldr with lazy functions for f that can already do something (eg produce list constructors) before looking at the second argument.

Using foldr with a strict f (eg (+) ) has the unwanted effect of putting all applications on the stack until the end of the list is reached; clearly not what you want, and not a situation where a however-looking foldr' could help.