Prefix form of unary operator in Haskell

问题

In GHCi:

1. Prelude> (+3) 2
   5
2. Prelude> (*3) 2
   6
3. Prelude> (/3) 2
   0.6666666666666666
4. Prelude> (-3) 2
   No instance for (Num (t -> t1))
   arising from the literal 3\' at <interactive>:1:2
Possible fix: add an instance declaration for (Num (t -> t1))
In the expression: 3
In the expression: (- 3) 2
In the definition ofit\': it = (- 3) 2

How can I correct the last one to make it return -1?

回答1:

Haskell's grammar doesn't allow you to use - like that. Use the subtract function instead:

(subtract 3) 2

回答2:

As a footnote to grddev's answer, here's the relevant paragraph from the Haskell 98 Report:

The special form -e denotes prefix negation, the only prefix operator in Haskell, and is syntax for negate (e). The binary - operator does not necessarily refer to the definition of - in the Prelude; it may be rebound by the module system. However, unary - will always refer to the negate function defined in the Prelude. There is no link between the local meaning of the - operator and unary negation.

This is something that frustrated me when I first came across it: I couldn't understand why the operators behaved so differently in this context when :info (+) and :info (-) looked basically identical.

You could use subtract, as grddev suggests, or you could just define a new infix operator:

Prelude> let (#) = (-)
Prelude> (# 3) 2
-1

subtract has the advantage of being familiar to other people who might read your code.

回答3:

You can do

(-) 3 2

but that will give you 1. To have -1, you need to bind the 3 to the second argument of -, which you can do using

flip (-) 3 2

回答4:

If you're intent on keeping your original shape, you can always add the negative:

(+ -3)

It ain't pretty, but it fits your pattern a little bit more.

来源：https://stackoverflow.com/questions/3406320/prefix-form-of-unary-operator-in-haskell