What does $ mean/do in Haskell?

Question

When you are writing slightly more complex functions I notice that $ is used a lot but I don't have a clue what it does?

Answer 1

$ is infix "application". It's defined as

($) :: (a -> b) -> (a -> b) f $ x = f x  -- or  ($) f x = f x -- or ($) = id 

It's useful for avoiding extra parentheses: f (g x) == f $ g x.

A particularly useful location for it is for a "trailing lambda body" like

forM_ [1..10] $ \i -> do   l <- readLine   replicateM_ i $ print l 

compared to

forM_ [1..10] (\i -> do   l <- readLine   replicateM_ i (print l) ) 

Or, trickily, it shows up sectioned sometimes when expressing "apply this argument to whatever function"

applyArg :: a -> (a -> b) -> b applyArg x = ($ x)  >>> map ($ 10) [(+1), (+2), (+3)] [11, 12, 13] 

Answer 2

I like to think of the $ sign as a replacement for parenthesis.

For example, the following expression:

take 1 $ filter even [1..10]  -- = [2] 

What happens if we don't put the $? Then we would get

take 1 filter even [1..10] 

and the compiler would now complain, because it would think we're trying to apply 4 arguments to the take function, with the arguments being 1 :: Int, filter :: (a -> Bool) -> [a] -> [a], even :: Integral a => a -> Bool, [1..10] :: [Int].

This is obviously incorrect. So what can we do instead? Well, we could put parenthesis around our expression:

(take 1) (filter even [1..10])

This would now reduce to:

(take 1) ([2,4,6,8,10])

which then becomes:

take 1 [2,4,6,8,10]

But we don't always want to be writing parenthesis, especially when functions start getting nested in each other. An alternative is to place the $ sign between where the pair of parenthesis would go, which in this case would be:

take 1 $ filter even [1..10]