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
Let's assume that you've define the following in GHCi:
λ> let foo a b = a + b
λ> let bar x = x * x
Based on some of your follow-up comments, it seems that you believe
bar . foo 1 2
to be equivalent to
(bar . foo 1) 2
However, remember that function application (space) has higher precedence than the composition operator (.
); therefore
bar . foo 1 2
is really equivalent to
bar . ((foo 1) 2)
Now, let's look at the types:
.
has type (b -> c) -> (a -> b) -> a -> c
; its two arguments are functions (that can be composed).bar
has type Num a => a -> a
, and is therefore compatible with the type (b -> c
) of the first argument of .
.foo 1 2
has type Num a => a
; it's a (polymorphic) numeric constant, not a function, and is therefore not compatible with the type (a -> b
) of the second argument of .
.That's why you're getting a type error in bar . foo 1 2
. What you can do, though, is
bar $ foo 1 2
because the $
operator has type (a -> b) -> a -> b
. See Haskell: difference between . (dot) and $ (dollar sign)