Why doesn't Haskell accept arguments after a function composition?

Question

Considering Haskell has currying functions, we can do this:

foo a b = a + b -- equivalent to `foo a = \\b -> a + b`

foo 1 -- ok, returns `\\b -> 1 + b`
fo         


        

Answer 1

bar . foo 1 2 is bar . (foo 1 2) not (bar . foo 1) 2

There's nothing mysterious going on here related to lambdas. Say we expanded the application of foo to 1:

bar . foo 1 2
bar . (\b -> 1 + b) 2

Now, we apply the lambda to the 2

bar . 3

And there is your problem.

Conversely, if we place the parentheses correctly, we evaluate it like this:

(bar . foo 1) 2
(bar . (\b -> 1 + b)) 2
(\x -> bar ((\b -> 1 + b) x)) 2
bar 3