[英]Getting the GHC to accept type signature with KnownNat arithmetic
I have been trying to implement the Chinese Remainder Theorem , for the specific case of just two equations, using the Data.Modular package. 我一直在尝试使用Data.Modular包来实现中国剩余定理 ,仅针对两个方程的具体情况。 The idea is that I can specify each equation with only one modular number (
x = a (mod m)
using the number a (mod m)
). 我的想法是,我可以使用数字
a (mod m)
指定每个方程只有一个模数( x = a (mod m)
a (mod m)
)。 Here is my implementation. 这是我的实施。
{-# LANGUAGE DataKinds, ScopedTypeVariables, TypeOperators #-}
import GHC.TypeLits
import Data.Proxy (Proxy (..))
import Data.Modular
crt :: forall k1 k2 i. (KnownNat k1, KnownNat k2, Integral i) => i `Mod` k1 -> i `Mod` k2 -> i `Mod` (k1 * k2)
crt a1 a2 = toMod $ (unMod a1)*n2*(unMod n2') + (unMod a2)*n1*(unMod n1')
where n1 = getModulus a1 :: i
n2 = getModulus a2 :: i
n2' = inv $ (toMod n2 :: i `Mod` k1)
n1' = inv $ (toMod n1 :: i `Mod` k2)
getModulus :: forall n i j. (Integral i, Integral j, KnownNat n) => i `Mod` n -> j
getModulus x = fromInteger $ natVal (Proxy :: Proxy n)
I get the following error: Could not deduce (KnownNat (k1 * k2)) arising from a use of 'toMod'
. 我收到以下错误:
Could not deduce (KnownNat (k1 * k2)) arising from a use of 'toMod'
。 However, shouldn't GHC be able to do the arithmetic for KnownNat (k1 * k2)
? 但是,GHC不能为
KnownNat (k1 * k2)
做算术吗? Also, for some weird reason, it looks like if I had a constructor for the Mod
type instead of the toMod
function, everything would work. 此外,由于一些奇怪的原因,看起来如果我有一个
Mod
类型的toMod
函数而不是toMod
函数,一切都会工作。 I fail to see how that should make a difference either... 我不明白这应该如何有所作为......
I am looking for any fix to help this compile, including whatever extensions. 我正在寻找任何修复来帮助编译,包括任何扩展。 I would, however, like to not have to make my own version of Data.Modular if possible.
但是,如果可能的话,我希望不必制作我自己的Data.Modular版本。 (I think I could make this work inelegantly and clumsily by using the
Mod
constructor directly). (我想我可以通过直接使用
Mod
构造函数来使这项工作变得非常优雅和笨拙)。
The cheap, cheesy way to make this compile is to add FlexibleContexts
, then add KnownNat (k1 * k2)
to the context of crt
. 进行编译的廉价,俗气的方法是添加
FlexibleContexts
,然后将KnownNat (k1 * k2)
到crt
的上下文中。 Once I did this, I could successfully call it in ghci: 一旦我这样做,我可以成功地在ghci中调用它:
> crt (3 :: Mod Integer 5) (5 :: Mod Integer 7)
33
Have fun working out how to express Coprime k1 k2
as a constraint... ;-) 玩得开心如何表达
Coprime k1 k2
作为约束...... ;-)
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