为无限 Stream 派生 Functor

deriving a Functor for an infinite Stream

我正在关注这个 blog 关于 F-代数 它解释说

A terminal coalgebra is usually interpreted in programming as a recipe for generating (possibly infinite) data structures or transition systems.

并说

A canonical example of a coalgebra is based on a functor whose fixed point is an infinite stream of elements of type e. This is the functor:

data StreamF e a = StreamF e a
  deriving Functor

这是它的不动点:

data Stream e = Stream e (Stream e)

我试过代码here

相关部分是

newtype Fix f = Fix (f (Fix f))
unFix :: Fix f -> f (Fix f)
unFix (Fix x) = x

cata :: Functor f => (f a -> a) -> Fix f -> a
cata alg = alg . fmap (cata alg) . unFix

ana :: Functor f => (a -> f a) -> a -> Fix f
ana coalg = Fix . fmap (ana coalg) . coalg

data StreamF e a = StreamF e a
    deriving Functor
data Stream e = Stream e (Stream e)

era :: [Int] -> StreamF Int [Int]
era (p : ns) = StreamF p (filter (notdiv p) ns)
    where notdiv p n = n `mod` p /= 0

primes = ana era [2..]

我遇到了这个错误

main.hs:42:14: error:
• Can’t make a derived instance of ‘Functor (StreamF e)’:
You need DeriveFunctor to derive an instance for this class
• In the data declaration for ‘StreamF’

我哪里错了?

在不使用语言扩展的情况下,

deriving 在 Haskell 中非常有限。由于编译器不能总是计算出 Functor 实例应该是什么,因此 deriving Functor 不是标准的 Haskell.

但是,有一种语言扩展允许这样做,即 -XDeriveFunctor。要启用此扩展程序,请执行以下操作之一:

  • 编译标志-XDeriveFunctor。 (如:编译时运行ghc -XDeriveFunctor Main.hs

  • 在文件顶部写下编译指示 {-# LANGUAGE DeriveFunctor #-}

添加此 pragma 后文件的外观如下:

{-# LANGUAGE DeriveFunctor #-}

newtype Fix f = Fix (f (Fix f))
unFix :: Fix f -> f (Fix f)
unFix (Fix x) = x

cata :: Functor f => (f a -> a) -> Fix f -> a
cata alg = alg . fmap (cata alg) . unFix

ana :: Functor f => (a -> f a) -> a -> Fix f
ana coalg = Fix . fmap (ana coalg) . coalg

data StreamF e a = StreamF e a
    deriving Functor
data Stream e = Stream e (Stream e)

era :: [Int] -> StreamF Int [Int]
era (p : ns) = StreamF p (filter (notdiv p) ns)
    where notdiv p n = n `mod` p /= 0

primes = ana era [2..]

如果您打算使用 GHCi,请在加载文件之前使用 :set -XDeriveFunctor