不完整的类型签名 [英] Incomplete type signature

问题描述

可以说我们有一个像下面的f这样的函数，它返回一个monad。然而，如果你看到 Int ，假装它是一个非常复杂的类型。

Lets say we've got a function like f below, that returns a monad. However, where you see Int, pretend it's a really complicated type.

f :: (Monad m) => m Int -- Pretend this isn't Int but something complicated
f = return 42

现在可以说我们要强制这个成为 Maybe monad。我们不需要写完整类型的 f 来做到这一点，我们可以做到以下几点：

Now lets say we want to force this into the Maybe monad. We don't need to write the full type of f to do this, we can just do the following:

g :: Maybe a -> Maybe a
g = id

main = print $ (g f)

虚函数 g 强制 f 变成也许 。

我觉得上面比较乱。

I think the above is rather messy. What I'd rather write is this:

main = print $ (f :: Maybe a)

但它会失败并显示以下错误：

But it fails with the following error:

Couldn't match expected type `a' against inferred type `Int'
  `a' is a rigid type variable bound by
      the polymorphic type `forall a. Maybe a' at prog.hs:7:16
  Expected type: Maybe a
  Inferred type: Maybe Int
In the second argument of `($)', namely `(f :: Maybe a)'
In the expression: print $ (f :: Maybe a)

有没有办法做到上面的 g 以不太复杂的方式做，而不需要创建一个新的函数？我不想写 f :: Maybe Int ，因为如果返回类型改变，它就成为一个维护问题。 GHC扩展可以在答案中找到答案。

Is there a way to do what g above does in a less messy way that doesn't involve creating a new function? I don't want to write f :: Maybe Int, as it becomes a maintenance problem if the return type changes. GHC extensions are okay in answers.

推荐答案

使用 asTypeOf 。它将第一个参数与第二个参数的类型统一起来，返回第一个参数。它只是 const 的类型限制版本，但对于这种情况很有用。

Use asTypeOf. It returns the first argument while unifying its type with that of the second. It's just a type-restricted version of const, but useful for situations like this.

main = print $ f `asTypeOf` (undefined :: Maybe a)

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