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我在 Haskell 中为编程竞赛实施基于粒子的流体模拟时遇到了一个小问题。我目前有一组粒子，它们在每个模拟步骤中都会被修改。每个粒子都是 2 个向量的元组：位置和速度（我自己的 Vec3D 模块）。在某些时候，我需要从我试图这样做的粒子（有点像解压缩列表）中提取位置： 

let xs = runSTArray $ do
    xs' <- newArray (min, max) (0.0,0.0,0.0) :: ST s (STArray s Int Vec3D)
    forM_ [min..max] $ \j -> do
        (x, v) <- readArray ps' j
        writeArray xs' j x
    return xs'
let displacements = doubleDensityRelaxation xs restDen k kNear t h

whereps'和doubleDensityRelaxationare 类型

type Vec3D = (Double, Double, Double)
ps' :: ST s (STArray s Int (Vec3D, Vec3D))
doubleDensityRelaxation :: Array Int Vec3D -> Double -> Double -> Double -> Double -> Double -> Array Int Vec3D

所以xs应该是 type xs :: Array Int Vec3D。但是，我得到 

Simulator.hs:76:35:
No instance for (MArray (STArray s) (Vec3D, Vec3D) (ST s1))
  arising from a use of `readArray'
Possible fix:
  add an instance declaration for
  (MArray (STArray s) (Vec3D, Vec3D) (ST s1))
In a stmt of a 'do' block: (x, v) <- readArray ps' j
In the expression:
  do { (x, v) <- readArray ps' j;
       writeArray xs' j x }
In the second argument of `($)', namely
  `\ j
     -> do { (x, v) <- readArray ps' j;
             writeArray xs' j x }'

来自我不太了解的编译器，因为它readArray不应该返回整个数组；只有一个(Vec3D, Vec3D)元素。

作为修复，我可以直接 doubleDensityRelaxation拍摄吗？ST s (STArray s Int Vec3D)

如果我改变这样的类型并删除let xs = runSTArray $ do我得到的部分 

Couldn't match expected type `ST s0 (STArray s0 Int Vec3D)'
            with actual type `STArray s Int Vec3D'

但是如果我将它(ST s xs')作为输入而不是仅仅xs'抱怨数据构造函数ST不在范围内。我的进口目前是 

import Data.List
import Data.Array
import Data.Array.ST
import Control.Monad
import Control.Monad.ST
import Vec3D

功能齐全： 

step :: Array Int (Vec3D, Vec3D) -> Vec3D -> Double -> Double -> Double -> Double -> Double -> Array Int (Vec3D, Vec3D)
step ps g restDen k kNear t h = runSTArray $ do
    ps' <- thaw ps :: ST s (STArray s Int Particle)
    --GRAVITY
    forM_ [min..max] $ \i -> do
        (x, v) <- readArray ps' i
        writeArray ps' i (x, addGravity v g t)
    --TODO - VISCOSITY
    --MOVE
    xsOld <- newArray (min, max) (0.0,0.0,0.0) :: ST s(STArray s Int Vec3D)
    forM_ [min..max] $ \i -> do
        (x, v) <- readArray ps' i
        writeArray xsOld i x
        writeArray ps' i (x `add` (v `mulSc` t), v)
    --TODO - SPRINGS
    --DOUBLE DENSITY RELAXATION
    xs' <- newArray (min, max) (0.0,0.0,0.0) :: ST s (STArray s Int Vec3D)
    forM_ [min..max] $ \j -> do
        (x, v) <- readArray ps' j
        writeArray xs' j x
    let displacements = doubleDensityRelaxation (freeze xs') restDen k kNear t h
    ps <- newArray (min, max) ((0.0,0.0,0.0), (0.0,0.0,0.0)) :: ST s (STArray s Int (Vec3D, Vec3D))
    --TODO incomplete
    return ps
    where
        addGravity v g t = v `add` (g `mulSc` t)
        (min, max) = bounds ps
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1 回答 1

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基于评论的社区维基回答：

如果您将freeze调用移出let displacements = ...行xs'' <- freeze xs'，然后使用xs''而不是freeze xs'

这是因为freeze :: (Ix i, MArray a e m, IArray b e) => a i e -> m (b i e). freeze xs'一个IArray b Vec3D => ST s (b Int Vec3D). _ 要获得实际的数组，您需要freeze在runSTArray.

于 2016-03-19T23:20:58.130 回答