Haskell dot operator

问题

I try to develop a simple average function in Haskell. This seems to work:

lst = [1, 3]

x = fromIntegral (sum lst)
y = fromIntegral(length lst)

z = x / y

But why doesn't the following version work?

lst = [1, 3]

x = fromIntegral.sum lst
y = fromIntegral.length lst

z = x / y

回答1:

You're getting tripped up by haskell's precedence rules for operators, which are confusing.

When you write

x = fromIntegral.sum lst

Haskell sees that as the same as:

x = fromIntegral.(sum lst)

What you meant to write was:

x = (fromIntegral.sum) lst

回答2:

. (composition) has a lower precedence than function application, so

fromIntegral.sum lst

is interpreted as

fromIntegral . (sum lst)

which is wrong since sum lst is not a function.

回答3:

I just wanted to add "$ to the rescue!":

x = fromIntegral $ sum lst
y = fromIntegral $ length lst

It has the lowest precedence and it's there to avoid too many parenthesis levels. Note that unlike (.), it doesn't do function composition, it evaluates the argument to the right and pass it to the function on the left. The type says it all:

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

来源：https://stackoverflow.com/questions/2834626/haskell-dot-operator