Function application has the strongest precedence, thus something like

fun1 . fun2 arg

is equivalent to

fun1 . (fun2 arg)

Now, ($) is the same as function application, but has lowest precedence (even lower than (.)).

infixr 0 ($)

($) :: (a -> b) -> a -> b

f $ x = f x

Thus,

fun1 . fun2 $ arg

is equivalent to

(fun1 . fun2) $ arg

== (fun1 . fun2) arg { apply ($) }

On 14 May 2015 at 02:42, Shishir Srivastava <shishir.srivastava@gmail.com> wrote:

Hi,

Could someone please point out what is the difference between using $ and parenthesis in the function composition below. Why does the first expression work whereas the second fails.
---
$$ head.head$[[1,2],[3,4]]
1
$$ head.head([[1,2],[3,4]])

<interactive>:122:12:
Couldn't match expected type ‘a -> [c]’ with actual type ‘[t0]’
Relevant bindings include
---
Thanks,
Shishir

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Regards

Sumit Sahrawat