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 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