Haskell中的$ mean / do是什么？

Question

当你编写稍微复杂的函数时，我注意到$被大量使用，但我不知道它的作用是什么？

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 (gx) == f $ gx .

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]