问题
> :t (+1)
(+1) :: Num a => a -> a
> :t (-1)
(-1) :: Num a => a
How come the second one is not a function? Do I have to write (+(-1)) or is there a better way?
回答1:
This is because (-1) is interpreted as negative one, however (+1) is interpreted as the curried function (\x->1+x).
In haskell, (a **) is syntactic sugar for (**) a, and (** a) is (\x -> x ** a). However (-) is a special case since it is both a unary operator (negate) and a binary operator (minus). Therefore this syntactic sugar cannot be applied unambiguously here. When you want (\x -> a - x) you can write (-) a, and, as already answered in Currying subtraction, you can use the functions negate and subtract to disambiguate between the unary and binary - functions.
回答2:
Do I have to write (+(-1)) or is there a better way?
I just found a function called subtract, so I can also say subtract 1. I find that quite readable :-)
回答3:
(-1) is negative one, as others have noted. The subtract one function is \x -> x-1, flip (-) 1 or indeed (+ (-1)).
- is treated as a special case in the expression grammar. + is not, presumably because positive literals don't need the leading plus and allowing it would lead to even more confusion.
Edit: I got it wrong the first time. ((-) 1) is the function "subtract from one", or (\x -> 1-x).
来源：https://stackoverflow.com/questions/5210182/the-difference-between-1-and-1