Pass a function as a first argument in a function call in coffeescript

In the following code

x= (f,n) -> f(n)
...
x( (n) -> n+1 , 5) #parse erro开发者_JS百科r

How can I fix the parse error above ?

Thanks

A pair of parenthesis would fix this problem, just found the answer on IRC.

x( (n) -> n+1  , 5) #parse error
x ((n) -> n+1) , 5 #fixed

I usually do either this:

foo ->
  doStuff('foo')
, 5

or this:

fn = -> doStuff('foo')
foo fn, 5

Wrapping extra parens inside argument lists never sat right with me as it's tough for my brain to parse.

Ali's answer is slightly different to the question he asked. One correct solution is

x = (f,n) -> f(n)

x(( -> n+1), 5)