Question

我今天在Haskell遇到了令人沮丧的某事。

以下是发生的事情：

我在ghci中写了一个函数并给它一个类型签名
ghci抱怨类型
我删除了类型签名
ghci接受了这个功能
我检查了推断类型
推断类型与我尝试提供的类型完全相同
我非常苦恼
我发现我可以在任何let-expression
咬牙切齿;决定咨询SO

尝试使用类型签名定义函数：

Prelude Control.Monad> let myFilterM f m = do {x <- m; guard (f x); return x} :: (MonadPlus m) => (b -> Bool) -> m b -> m b

<interactive>:1:20:
    Inferred type is less polymorphic than expected
      Quantified type variable `b' is mentioned in the environment:
        m :: (b -> Bool) -> m b -> m b (bound at <interactive>:1:16)
        f :: (m b -> m b) -> Bool (bound at <interactive>:1:14)
      Quantified type variable `m' is mentioned in the environment:
        m :: (b -> Bool) -> m b -> m b (bound at <interactive>:1:16)
        f :: (m b -> m b) -> Bool (bound at <interactive>:1:14)
    In the expression:
          do { x <- m;
               guard (f x);
               return x } ::
            (MonadPlus m) => (b -> Bool) -> m b -> m b
    In the definition of `myFilterM':
        myFilterM f m
                    = do { x <- m;
                           guard (f x);
                           return x } ::
                        (MonadPlus m) => (b -> Bool) -> m b -> m b

定义没有类型签名的函数，检查推断类型：

Prelude Control.Monad> let myFilterM f m = do {x <- m; guard (f x); return x}
Prelude Control.Monad> :t myFilterM 
myFilterM :: (MonadPlus m) => (b -> Bool) -> m b -> m b

使用该功能非常好 - 它工作正常：

Prelude Control.Monad> myFilterM (>3) (Just 4)
Just 4
Prelude Control.Monad> myFilterM (>3) (Just 3)
Nothing

我最好猜测发生了什么：

奖励积分：
标准的Haskell发行版中有一个函数可以做到这一点吗？我很惊讶filterM做了一些非常不同的事情。

Answer 1

问题是类型运算符（::）的优先级。你试图描述myFilterM的类型，但你实际上在做的是：

ghci> let myFilterM f m = (\
        do {x <- m; guard (f x); return x} \
        :: \
        (MonadPlus m) => (b -> Bool) -> m b -> m b)\
      )

（插入反斜杠仅为了可读性，不是合法的ghci语法）

你看到了这个问题吗？对于像

这样的简单问题我也遇到了同样的问题

ghci> let f x = x + 1 :: (Int -> Int)
<interactive>:1:15:
    No instance for (Num (Int -> Int))
      arising from the literal `1'
    Possible fix: add an instance declaration for (Num (Int -> Int))
    In the second argument of `(+)', namely `1'
    In the expression: x + 1 :: Int -> Int
    In an equation for `f': f x = x + 1 :: Int -> Int

解决方案是将类型签名附加到适当的元素：

ghci> let f :: Int -> Int ; f x = x + 1
ghci> let myFilterM :: (MonadPlus m) => (b -> Bool) -> m b -> m b; myFilterM f m = do {x <- m; guard (f x); return x}

对于奖励积分，您需要mfilter（hoogle is your friend）。

Answer 2

这可能只是类型注释语法和绑定优先级的问题。如果你把你的例子写成，

let myFilterM :: (MonadPlus m) => (b -> Bool) -> m b -> m b; myFilterM f m = do {x <- m; guard (f x); return x}

然后GHCi会给你一个高五并送你前往。

Answer 3

我不知道你使用什么样的编译器，但在我的平台上（GHC 7.0.3），我得到一个简单的类型不匹配：

$ ghci
GHCi, version 7.0.3: http://www.haskell.org/ghc/  :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Loading package ffi-1.0 ... linking ... done.
Prelude> :m +Control.Monad
Prelude Control.Monad> let myFilterM f m = do {x <- m; guard (f x); return x} :: (MonadPlus m) => (b -> Bool) -> m b -> m b

<interactive>:1:30:
    Could not deduce (t1 ~ ((b1 -> Bool) -> m1 b1 -> m1 b1))
    from the context (MonadPlus m)
      bound by the inferred type of
               myFilterM :: MonadPlus m => t -> t1 -> (b -> Bool) -> m b -> m b
      at <interactive>:1:5-100
    or from (MonadPlus m1)
      bound by an expression type signature:
                 MonadPlus m1 => (b1 -> Bool) -> m1 b1 -> m1 b1
      at <interactive>:1:21-100
      `t1' is a rigid type variable bound by
           the inferred type of
           myFilterM :: MonadPlus m => t -> t1 -> (b -> Bool) -> m b -> m b
           at <interactive>:1:5
    In a stmt of a 'do' expression: x <- m
    In the expression:
        do { x <- m;
             guard (f x);
             return x } ::
          MonadPlus m => (b -> Bool) -> m b -> m b
    In an equation for `myFilterM':
        myFilterM f m
          = do { x <- m;
                 guard (f x);
                 return x } ::
              MonadPlus m => (b -> Bool) -> m b -> m b

<interactive>:1:40:
    Could not deduce (t ~ ((m1 b1 -> m1 b1) -> Bool))
    from the context (MonadPlus m)
      bound by the inferred type of
               myFilterM :: MonadPlus m => t -> t1 -> (b -> Bool) -> m b -> m b
      at <interactive>:1:5-100
    or from (MonadPlus m1)
      bound by an expression type signature:
                 MonadPlus m1 => (b1 -> Bool) -> m1 b1 -> m1 b1
      at <interactive>:1:21-100
      `t' is a rigid type variable bound by
          the inferred type of
          myFilterM :: MonadPlus m => t -> t1 -> (b -> Bool) -> m b -> m b
          at <interactive>:1:5
    The function `f' is applied to one argument,
    but its type `t' has none
    In the first argument of `guard', namely `(f x)'
    In a stmt of a 'do' expression: guard (f x)
Prelude Control.Monad>

我想问题在于::没有达到论证的事实。这个小变化（注意单独的类型声明）

let myFilterM f m = do {x <- m; guard (f x); return x}; myFilterM :: (MonadPlus m) => (b -> Bool) -> m b -> m b

运行没有问题。它可能与GHC 7中的新型检查器有关。

Haskell let-expression中的奇怪类型错误 - 问题是什么？

3 个答案: