Haskell Netwire - 输入错误

Question

我刚刚开始使用netwire,我遇到了很多基础问题.

以下代码对我来说很好:

main :: IO ()
main = testWire clockSession_ (for 3 . yeah)

yeah :: Monad m => Wire s () m a String
yeah = pure "yes"

但这不是:

main :: IO ()
main = testWire clockSession_ forYeah

forYeah :: (Show b, Show e) => Wire s e Identity a b
forYeah = for 3 . yeah

失败并出错:

Could not deduce (b ~ [Char])
from the context (Show b, Show e)
bound by the type signature for
forYeah :: (Show b, Show e) => Wire s e Identity a b
  at /home/fiendfan1/workspace/Haskell/Haskell-OpenGL/src/Main.hs:12:12-54
  `b' is a rigid type variable bound by
      the type signature for
        forYeah :: (Show b, Show e) => Wire s e Identity a b
      at /home/fiendfan1/workspace/Haskell/Haskell-OpenGL/src/Main.hs:12:12
Expected type: Wire s e Identity a b
  Actual type: Wire s () Identity a String
In the second argument of `(.)', namely `yeah'
In the expression: for 3 . yeah
In an equation for `forYeah': forYeah = for 3 . yeah

所以我改成了:

forYeah :: Show e => Wire s e Identity a String

这给了我错误:

Could not deduce (e ~ ())
from the context (Show e)
  bound by the type signature for
             forYeah :: Show e => Wire s e Identity a String
  at /home/fiendfan1/workspace/Haskell/Haskell-OpenGL/src/Main.hs:12:12-49
  `e' is a rigid type variable bound by
      the type signature for
        forYeah :: Show e => Wire s e Identity a String
      at /home/fiendfan1/workspace/Haskell/Haskell-OpenGL/src/Main.hs:12:12
Expected type: Wire s e Identity a String
  Actual type: Wire s () Identity a String
In the second argument of `(.)', namely `yeah'
In the expression: for 3 . yeah
In an equation for `forYeah': forYeah = for 3 . yeah

将其更改为:

forYeah :: Wire s () Identity a String

给出以下错误:

No instance for (HasTime Integer s) arising from a use of `for'
Possible fix: add an instance declaration for (HasTime Integer s)
In the first argument of `(.)', namely `for 3'
In the expression: for 3 . yeah
In an equation for `forYeah': forYeah = for 3 . yeah

有人可以解释为什么会发生这种情况以及如何修复我的第二个代码示例？

Answer 1

编辑：这是此问题的完整、可编译、可运行的解决方案：

module Main (
    main
) where

import Prelude hiding ((.), id)
import qualified Prelude as Prelude
import Control.Wire
import Control.Wire.Interval

main :: IO ()
main = testWire clockSession_ (withoutErrors forYeah)

yeah :: Monad m => Wire s e m a String
yeah = pure "yes"

forYeah :: (Num t, HasTime t s, Monoid e, Monad m) => Wire s e m a String
forYeah = for 3 . yeah

-- This just is an easy way to specify to use () as the type for errors in testWire
withoutErrors :: Wire s () m a b -> Wire s () m a b 
withoutErrors = Prelude.id

这是最初的答案，讨论了为什么我们应该更改 的类型yeah，然后对 的类型进行必要的更改forYeah：

将 的类型更改yeah为Monad m => Wire s e m a String。Monad m => (Wire s e m a)有一个Applicative实例，因此应该存在，而无需指定in类型pure的第二个类型参数是。Wireyeah()

注意：我不使用 netwire，也没有尝试编译它。我只查看了文档中的类型。

编辑：您可能还需要更改forYeah.

Wire还有一个Category实例：

Monad m => Category (Wire s e m)

Category的.运算符具有以下类型：

(.) :: cat b c -> cat a b -> cat a c

所以对于Wires 来说是：

(.) :: Monad m => Wire s e m b c -> Wire s e m a b -> Wire s e m a c

for有以下类型：

for :: (HasTime t s, Monoid e) => t -> Wire s e m a a

所以for 3会有像这样的类型(HasTime Int s, Monoid e) => Wire s e m a a。与 yes 的类型结合Monad m => Wire s e m a String，for 3 . yeah会得到类似的类型

(HasTime Int s, Monoid e, Monad m) => Wire s e m a String

所以我们可能可以将类型更改forYeah为：

forYeah :: (HasTime Int s, Monoid e, Monad m) => Wire s e m a String

编辑：更好的类型forYeah

由于整数（不带小数点）实际上相当于将 fromInteger 应用于数字值作为 Integer , and fromInteger :: (Num a) => Integer -> a，因此文字3实际上具有类型Num t => t。因此，我们可以选择的最佳类型可能是：

forYeah :: (Num t, HasTime t s, Monoid e, Monad m) => Wire s e m a String