difference between (a -> a) and a -> a

Question

I've noticed that (although I was once told that (a -> a) and a -> a meant the same thing), I get error messages when I use the (a -> a). Should I only use (a -> a) when using brackets amongst the types? (i.e. (5 + 3) instead of 5 + 3)? Just not quite certain of when it's necessary

Answer 1

(a -> a) and a -> a are the same alone,

ff :: (a -> a)   -- this compiles
ff = id

gg :: a -> a
gg = id

h :: a -> a -> Bool
h _ _ = True

i = h ff gg   -- this compiles => ff & gg are of the same type.

but will be different when combined with more types like:

 a -> a  -> b
(a -> a) -> b

This is because -> is right-associative, so a -> a -> b actually means a -> (a -> b) (take an a and return a function), which is different from (a -> a) -> b (take a function and return a b).

This is like (1+2)*3 is different from 1+2*3.