在Haskell中,是否存在无穷大:: Num a => a？

Question

我正在尝试实现一个数据结构,如果我使用无穷大进行数字比较,那就简单了.请注意,这不是maxBound/minBound,因为值可以<= maxbound,但所有值都是<infinity.

没希望？

Answer 1

那怎么样!事实证明,如果你只是键入1/0它返回Infinity!在ghci:

Prelude> 1/0
Infinity
Prelude> :t 1/0
1/0 :: (Fractional t) => t
Prelude> let inf=1/0
Prelude> filter (>=inf) [1..]

然后它当然会永远运行,永远不会找到比无穷大更大的数字.(但请参阅以下关于实际行为的ephemient的评论[1..])

Answer 2

infinity = read "Infinity"

Answer 3

也许你想要一个Maybe类型？

data Infinite a = Infinite | Only a

然后使用您需要的数字规则为Num a => Infinite a写一个Num实例.

Answer 4

尝试这样的事情.但是,要获取Num操作(如+或-),您需要Num为Infinitable a类型定义实例.就像我Ord上课一样.

data Infinitable a = Regular a | NegativeInfinity | PositiveInfinity deriving (Eq, Show)

instance Ord a => Ord (Infinitable a) where
    compare NegativeInfinity NegativeInfinity = EQ
    compare PositiveInfinity PositiveInfinity = EQ
    compare NegativeInfinity _ = LT
    compare PositiveInfinity _ = GT
    compare _ PositiveInfinity = LT
    compare _ NegativeInfinity = GT
    compare (Regular x) (Regular y) = compare x y    

main =
    let five = Regular 5
        pinf = PositiveInfinity::Infinitable Integer
        ninf = NegativeInfinity::Infinitable Integer
        results = [(pinf > five), (ninf < pinf), (five > ninf)]
    in
        do putStrLn (show results)

Answer 5

?: let infinity = (read "Infinity")::Double
?: infinity > 1e100
True
?: -infinity < -1e100
True