Question

我在Haskell中为编程竞赛实现基于粒子的流体模拟时遇到了一些问题。我目前有一系列粒子在每个模拟步骤中被修改。每个粒子是2个向量的元组：位置和速度（我自己的Vec3D模块）。在某些时候，我需要从粒子中提取位置（有点像解压缩列表），我试图这样做：

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

其中ps'和doubleDensityRelaxation的类型为

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

因此xs应为xs :: Array Int Vec3D类型。但是，我得到了

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 }'

来自编译器我不明白，因为readArray不应该返回整个数组;只有一个(Vec3D, Vec3D)元素。

作为修复，我可以让doubleDensityRelaxation直接接受ST s (STArray s Int Vec3D)吗？

如果我更改类似的类型并删除let xs = runSTArray $ do部分，我会

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

但如果我将(ST s xs')作为输入而不仅仅是xs'，则会抱怨数据构造函数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

完成功能：

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

Answer 1

社区维基回答基于评论：

如果您将freeze来自let displacements = ...行的呼叫移至xs'' <- freeze xs'，然后使用xs''代替freeze xs'

这是因为freeze :: (Ix i, MArray a e m, IArray b e) => a i e -> m (b i e)。因此freeze xs'是IArray b Vec3D => ST s (b Int Vec3D)。要获取实际数组，您需要在freeze内运行runSTArray。

从STArray中提取元素（类似于解压缩）

1 个答案: