> :t (+1)
(+1) :: Num a => a -> a
> :t (-1)
(-1) :: Num a => a
How come the sec开发者_开发技巧ond one is not a function? Do I have to write (+(-1))
or is there a better way?
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.
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 :-)
(-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)
.
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