小编nul*_*nge的帖子

说我有一个功能:

f :: Int -> (Rational, Integer)
f b = ((toRational b)+1,(toInteger b)+1)

我想像这样抽象出(+1):

f :: Int -> (Rational, Integer)
f b = (h (toRational b)
      ,h (toInteger b))
    where h = (+1)

这显然不会起作用,但如果我指定类型签名,它将起作用:

f :: Int -> (Rational, Integer)
f b = (h (toRational b)
      ,h (toInteger b))
    where h :: Num a => a -> a
          h = (+1)

假设我现在想通过传递h作为参数来进一步抽象函数:

f :: Num a => Int -> (a -> a) -> (Rational, Integer)
f b g = (h …

说我有以下(错误的)代码.

data A a b where
  APure ::  (A a b)
  AApply :: A (A b c) c

test :: (A a b) -> a -> b
test (APure) a = a
test AApply a = undefined

然后GHC会给我这个错误:

Couldn't match type `b' with `A b1 b'
  `b' is a rigid type variable bound by
      the type signature for test :: A a b -> a -> b
Inaccessible code in
  a pattern with constructor
    AApply :: forall c b. A …

说我有以下GADT:

data Stage a b where
    Comb :: Stage a b -> Stage b c -> Stage a c
    FMap :: (a -> b) -> Stage a b

我现在想要一个像这样工作的函数:

run (a `Comb` b) = (a,b)
run (FMap f)     = (FMap f,FMap id)

我该如何构建这样的函数？

我尝试了不同的绑定类型的方法,但没有成功.是否存在我缺少的扩展,可以实现更广泛的类型绑定？

这是错误消息:

Couldn't match type `t' with `(Stage t1 b, Stage b t2)'
  `t' is a rigid type variable bound by
      the inferred type of run :: Stage t1 t2 -> t at <interactive>:11:5
In the expression: …