用Data.Binary对列表进行延迟解码 [英] Lazy decoding of a list with Data.Binary

问题描述

我使用此代码懒惰地编码列表（摘自此 SO问题）：

I am lazily encoding lists using this code (taken from this SO question):

import Data.Binary

newtype Stream a = Stream { unstream :: [a] }

instance Binary a => Binary (Stream a) where

    put (Stream [])     = putWord8 0
    put (Stream (x:xs)) = putWord8 1 >> put x >> put (Stream xs)

问题是解码实现不是懒惰的：

The problem is that the decoding implementation is not lazy:

    get = do
        t <- getWord8
        case t of
            0 -> return (Stream [])
            1 -> do x         <- get
                    Stream xs <- get
                    return (Stream (x:xs))

这在我看来应该是懒惰的，但是如果我们运行这个测试代码：

This looks to me like it should be lazy, but if we run this test code:

head $ unstream (decode $ encode $ Stream [1..10000000::Integer] :: Stream Integer)

<内存使用情况爆炸。出于某种原因，它想先解码整个列表，然后再让我看看第一个元素。

memory usage explodes. For some reason it wants to decode the whole list before letting me look at the first element.

为什么这不是懒惰，我怎么能使它懒惰？获取 monad是一个严格的状态monad（p>

Why is this not lazy, and how I can make it lazy?

推荐答案

在 binary-0.5.0.2到0.5.1.1 ;这是一个懒惰的状态monad之前和 binary-0.6。* 它已经成为继续monad，我没有分析这种变化的严格影响）：

It is not lazy because the Get monad is a strict state monad (in binary-0.5.0.2 to 0.5.1.1; it was a lazy state monad before, and in binary-0.6.* it has become a continuation monad, I haven't analysed the strictness implications of that change):

-- | The parse state
data S = S {-# UNPACK #-} !B.ByteString  -- current chunk
           L.ByteString                  -- the rest of the input
           {-# UNPACK #-} !Int64         -- bytes read

-- | The Get monad is just a State monad carrying around the input ByteString
-- We treat it as a strict state monad. 
newtype Get a = Get { unGet :: S -> (# a, S #) }

-- Definition directly from Control.Monad.State.Strict
instance Monad Get where
    return a  = Get $ \s -> (# a, s #)
    {-# INLINE return #-}

    m >>= k   = Get $ \s -> case unGet m s of
                             (# a, s' #) -> unGet (k a) s'
    {-# INLINE (>>=) #-}

因此最终递归

thus the final recursive

get >>= \x ->
get >>= \(Stream xs) ->
return (Stream (x:xs))

强制整个 Stream 在它可以被返回之前被读取。

forces the entire Stream to be read before it can be returned.

我不认为可以懒惰地解码 Stream 在获取 monad（所以不适用于 Binary 实例）。但你可以使用 runGetState 编写一个懒惰的解码函数：

I don't think it's possible to lazily decode a Stream in the Get monad (so a fortiori not with the Binary instance). But you can write a lazy decoding function using runGetState:

-- | Run the Get monad applies a 'get'-based parser on the input
-- ByteString. Additional to the result of get it returns the number of
-- consumed bytes and the rest of the input.
runGetState :: Get a -> L.ByteString -> Int64 -> (a, L.ByteString, Int64)
runGetState m str off =
    case unGet m (mkState str off) of
      (# a, ~(S s ss newOff) #) -> (a, s `join` ss, newOff)

首先写一个 Get 解析器返回一个也许一个，

First write a Get parser that returns a Maybe a,

getMaybe :: Binary a => Get (Maybe a)
getMaybe = do
    t <- getWord8
    case t of
      0 -> return Nothing
      _ -> fmap Just get

然后使用它来创建类型的函数（ByteString， Int64） - >可能（a，（ByteString，Int64））：

step :: Binary a => (ByteString,Int64) -> Maybe (a,(ByteString,Int64))
step (xs,offset) = case runGetState getMaybe xs offset of
                     (Just v, ys, newOffset) -> Just (v,(ys,newOffset))
                     _                       -> Nothing

然后您可以使用 Data.List.unfoldr


and then you can use Data.List.unfoldr to lazily decode a list,

lazyDecodeList :: Binary a => ByteString -> [a]
lazyDecodeList xs = unfoldr step (xs,0)

a Stream

lazyDecodeStream :: Binary a => ByteString -> Stream a
lazyDecodeStream = Stream . lazyDecodeList

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