Question

是否可以从其中一个子功能中的功能定义引用类型？（有关更多详细信息，请参见下面代码中的注释）

module TypesAndSubFunctions where

class DingBat k where

foo :: DingBat k => k -> String
foo kv = "foo"
  where
    -- how can I define bar so that the type 'k2' is the same 
    -- as k from foo? Or is this not possible?
    --
    -- The following causes an error, because k2 isn't the same type
    -- as foo's k. But I can't seem to find a way to reference foo's k
    -- without specifying it as an arg (as I do in bar2). Is this just
    -- the way haskell is?
    --
    -- I would think there should be a simple way to reference k from
    -- foo with the type definition of bar, but I can't see how.
    -- bar :: DingBat k2 => k2 -> String
    -- bar kv2 = ding2 kv kv2

    -- To restate the problem, I would like the same functionality for
    -- bar as bar2, below, but without having to specify the first arg
    bar2 :: DingBat k => k -> k -> String
    bar2 kv kv2 = ding2 kv kv2

ding2 :: DingBat k => k -> k -> String
ding2 = undefined

Answer 1

这应该可以完成工作。要重用k，您需要打开一个非常通用的扩展名。

{-# LANGUAGE ScopedTypeVariables #-}

foo :: forall k . DingBat k => k -> String
foo kv = "foo"
  where
  bar :: k -> String
  bar kv2 = ding2 kv kv2

如何从子函数中的函数引用类型

1 个答案: