我试图在加速之上定义一个类型安全的矩阵计算库,部分用于教育目的,部分是为了看看这是否是一种实用的方法。
但是,当我正确定义矩阵的乘积时,我完全陷入困境 - 即GHC接受/编译我的代码的方式。
我尝试了几次,这是其中的变种:
Linear.hs
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE ScopedTypeVariables #-}
import qualified Data.Array.Accelerate as A
import GHC.TypeLits
import Data.Array.Accelerate ( (:.)(..), Array
, Exp, Shape, FullShape, Slice
, DIM0, DIM1, DIM2, Z(Z)
, IsFloating, IsNum, Elt, Acc
, Any(Any), All(All))
import Data.Proxy
newtype Matrix (rows :: Nat) (cols :: Nat) a = AccMatrix {unMatrix :: Acc (Array DIM2 a)}
(#*#) :: forall k m n a. (KnownNat k, KnownNat m, KnownNat n, IsNum a, Elt a) =>
Matrix k m a -> Matrix m n a -> Matrix k n a
v #*# w = let v' = unMatrix v
w' = unMatrix w
in AccMatrix $ A.generate (A.index2 k' n') undefined
where k' = fromInteger $ natVal (Proxy :: Proxy k)
n' = fromInteger $ natVal (Proxy :: Proxy n)
aux :: Acc (Array (FullShape (Z :. Int) :. Int) e) -> Acc (Array (FullShape (Z :. All) :. Int) e) -> Exp ((Z :. Int) :. Int) -> Exp e
aux v w sh = let (Z:.i:.j) = A.unlift sh
v' = A.slice v (A.lift $ Z:.i:.All)
w' = A.slice w (A.lift $ Z:.All:.j)
in A.the $ A.sum $ A.zipWith (*) v' w'
错误stack build
给我的是
.../src/Linear.hs:196:55:
Couldn't match type ‘A.Plain ((Z :. head0) :. head1)’
with ‘(Z :. Int) :. Int’
The type variables ‘head0’, ‘head1’ are ambiguous
Expected type: Exp (A.Plain ((Z :. head0) :. head1))
Actual type: Exp ((Z :. Int) :. Int)
Relevant bindings include
i :: head0 (bound at src/Linear.hs:196:38)
j :: head1 (bound at src/Linear.hs:196:41)
In the first argument of ‘A.unlift’, namely ‘sh’
In the expression: A.unlift sh
.../src/Linear.hs:197:47:
Couldn't match type ‘FullShape (A.Plain (Z :. head0))’
with ‘Z :. Int’
The type variable ‘head0’ is ambiguous
Expected type: Acc
(Array (FullShape (A.Plain (Z :. head0) :. All)) e)
Actual type: Acc (Array (FullShape (Z :. Int) :. Int) e)
Relevant bindings include
v' :: Acc (Array (A.SliceShape (A.Plain (Z :. head0)) :. Int) e)
(bound at src/Linear.hs:197:34)
i :: head0 (bound at src/Linear.hs:196:38)
In the first argument of ‘A.slice’, namely ‘v’
In the expression: A.slice v (A.lift $ Z :. i :. All)
.../src/Linear.hs:198:39:
Couldn't match type ‘A.SliceShape (A.Plain ((Z :. All) :. head1))’
with ‘A.SliceShape (A.Plain (Z :. head0)) :. Int’
The type variables ‘head0’, ‘head1’ are ambiguous
Expected type: Acc
(Array (A.SliceShape (A.Plain (Z :. head0)) :. Int) e)
Actual type: Acc
(Array (A.SliceShape (A.Plain ((Z :. All) :. head1))) e)
Relevant bindings include
w' :: Acc (Array (A.SliceShape (A.Plain (Z :. head0)) :. Int) e)
(bound at src/Linear.hs:198:34)
v' :: Acc (Array (A.SliceShape (A.Plain (Z :. head0)) :. Int) e)
(bound at src/Linear.hs:197:34)
i :: head0 (bound at src/Linear.hs:196:38)
j :: head1 (bound at src/Linear.hs:196:41)
In the expression: A.slice w (A.lift $ Z :. All :. j)
In an equation for ‘w'’: w' = A.slice w (A.lift $ Z :. All :. j)
.../src/Linear.hs:198:47:
Couldn't match type ‘FullShape (A.Plain ((Z :. All) :. head1))’
with ‘(Z :. Int) :. Int’
The type variable ‘head1’ is ambiguous
Expected type: Acc
(Array (FullShape (A.Plain ((Z :. All) :. head1))) e)
Actual type: Acc (Array (FullShape (Z :. All) :. Int) e)
Relevant bindings include
j :: head1 (bound at src/Linear.hs:196:41)
In the first argument of ‘A.slice’, namely ‘w’
In the expression: A.slice w (A.lift $ Z :. All :. j)
我已经查阅了Accelerate的文档,我也在阅读accelerate-arithmetic,它有类似的目的但不使用TypeLits
来断言数组/向量维度。< / p>
我还尝试制作一个vanilla版本(即没有我自己的矩阵类型),以防我的类型错误,我认为这是对slice
的使用有同样的误解。我只是为了完整性而包含这个,我可以添加错误消息,但我选择省略它们,因为我认为它们与上述问题无关。
(#*#) :: forall a. (IsNum a, Elt a) =>
Acc (Array DIM2 a) -> Acc (Array DIM2 a) -> Maybe (Acc (Array DIM2 a))
v #*# w = let Z:.k :.m = A.unlift $ A.arrayShape $ I.run v
Z:.m':.n = A.unlift $ A.arrayShape $ I.run w
in if m /= m'
then Nothing
else Just $ AccMatrix $ A.generate (A.index2 k n) (aux v w)
where aux :: Acc (Array DIM2 a) -> Acc (Array DIM2 a) -> Exp DIM2 -> Exp a
aux v w sh = let (Z:.i:.j) = A.unlift sh
v' = A.slice v (A.lift $ Z:.i:.All)
w' = A.slice w (A.lift $ Z:.All:.j)
in A.the $ A.sum $ A.zipWith (*) v' w'
答案 0 :(得分:2)
您的代码实际上是正确的。不幸的是,类型检查员还不够聪明,所以你必须帮助它:
let (Z:.i:.j) = A.unlift sh
变为
let (Z:.i:.j) = A.unlift sh :: (Z :. Exp Int) :. Exp Int
这里的关键是A.unlift :: A.Unlift c e => c (A.Plain e) -> e
但A.Plain
是一个关联的类型系列(因此是非单射的),因此如果没有类型签名就无法确定类型e
,{需要{1}}来选择要用于e
的实例。这就是“模糊类型”错误的来源 - 实际上Unlift c e
是不明确的。
您还有一个无关的错误。 e
应该有
aux
或
aux :: (IsNum e, Elt e) => ...
在后一种情况下,aux :: (e ~ a) => ...
是a
类型签名中的(#*#)
,因此它已经具有约束IsNum, Elt