PR Stanley
Hi folks
let f = sum.map (^2).filter even f [1..4] 20
So far so good!
sum.map (^2).filter even [1..4] returns an error.
How is it that the composition above as part of a function equation doesn't return any errors while on its own it requires parentheses? I can understand the need for the parentheses - the composition operator takes two arguments/operand,s each of which is a mapping which must correspond to specific domain and range requirements i.e. (b->c -> (a->b) -> (a->c) Why aren't the parentheses required in the function definition?
The answer is precedence. The (.) operator is right-associative and has lower precedence than function application, meaning that: a . b . c . d . e == a . (b . (c . (d . e))) so the second form is equivalent to: sum . ((map (^2)) . (filter even [1..4])) Typing :t (filter even [1..4]) should give you your explanation. If you want to avoid parentheses, you can write that expression as: sum . map (^2) . filter even $ [1..4] which really means (sum . map (^2) . filter even) [1..4] which is the same as "f [1..4]" above. Hope that helps, G