(^)上的类型推断问题

因此,我正在尝试为Prelude编写自己的替换,我已经(^)实现了:

{-# LANGUAGE RebindableSyntax #-}

class Semigroup s where
    infixl 7 *
    (*) :: s -> s -> s

class (Semigroup m) => Monoid m where
    one :: m

class (Ring a) => Numeric a where
    fromIntegral :: (Integral i) => i -> a
    fromFloating :: (Floating f) => f -> a

class (EuclideanDomain i, Numeric i, Enum i, Ord i) => Integral i where
    toInteger :: i -> Integer
    quot :: i -> i -> i
    quot a b = let (q,r) = (quotRem a b) in q
    rem :: i -> i -> i
    rem a b = let (q,r) = (quotRem a b) in r
    quotRem :: i -> i -> (i, i)
    quotRem a b = let q = quot a b; r = rem a b in (q, r)

-- . . .

infixr 8 ^
(^) :: (Monoid m, Integral i) => m -> i -> m
(^) x i
    | i == 0 = one
    | True   = let (d, m) = (divMod i 2)
                   rec = (x*x) ^ d in
               if m == one then x*rec else rec

(注意,这里使用的积分是我定义的积分,而不是前奏曲中的积分,尽管它很相似。此外, one

数字类型是幺半群,所以我可以尝试这样做,比如说2^3,但是类型检查器会给我:

*AlgebraicPrelude> 2^3

<interactive>:16:1: error:
    * Could not deduce (Integral i0) arising from a use of `^'
      from the context: Numeric m
        bound by the inferred type of it :: Numeric m => m
        at <interactive>:16:1-3
      The type variable `i0' is ambiguous
      These potential instances exist:
        instance Integral Integer -- Defined at Numbers.hs:190:10
        instance Integral Int -- Defined at Numbers.hs:207:10
    * In the expression: 2 ^ 3
      In an equation for `it': it = 2 ^ 3

<interactive>:16:3: error:
    * Could not deduce (Numeric i0) arising from the literal `3'
      from the context: Numeric m
        bound by the inferred type of it :: Numeric m => m
        at <interactive>:16:1-3
      The type variable `i0' is ambiguous
      These potential instances exist:
        instance Numeric Integer -- Defined at Numbers.hs:294:10
        instance Numeric Complex -- Defined at Numbers.hs:110:10
        instance Numeric Rational -- Defined at Numbers.hs:306:10
        ...plus four others
        (use -fprint-potential-instances to see them all)
    * In the second argument of `(^)', namely `3'
      In the expression: 2 ^ 3
      In an equation for `it': it = 2 ^ 3

我知道这是因为Int和Integer都是整数类型,但为什么在正常的前奏中我可以很好地做到这一点呢

Prelude> :t (2^)
(2^) :: (Num a, Integral b) => b -> a
Prelude> :t 3
3 :: Num p => p
Prelude> 2^3
8

即使部分应用于我的签名看起来完全相同?

*AlgebraicPrelude> :t (2^)
(2^) :: (Numeric m, Integral i) => i -> m
*AlgebraicPrelude> :t 3
3 :: Numeric a => a

我怎样才能使2^3实际上起作用,从而得到8呢?