Question

作为一项学习练习，我尝试定义一个新类型作为函数的持有者，可以将Show - 能值转换为Eff，即：

newtype ShEff a = ShEff (forall eff. Show a => a -> Eff eff Unit)

但是，这个：

f :: forall a. ShEff a
f = ShEff logShow

无法使用此错误进行编译：

  Could not match type

    ( console :: CONSOLE
    | t2
    )

  with type

    eff1


while trying to match type Eff
                             ( console :: CONSOLE
                             | t2
                             )
  with type Eff eff1
while checking that expression logShow
  has type t0 -> Eff eff1 Unit
in value declaration f

你能指出我正确的方向吗？

Answer 1

logShow的类型是

forall eff. Show a => a -> Eff (console :: CONSOLE | eff) Unit

因此您无法将其存储在ShEff中，因为它必须适用于每个 eff，而logShow仅适用于包含{{ 1}}效果。

您有两种选择：

将CONSOLE类型参数移到eff之外：
```
ShEff
```
在newtype ShEff eff a = ShEff (a -> Eff eff Unit) f :: forall a eff. Show a => ShEff (console :: CONSOLE | eff) a f = ShEff logShow：
中添加约束
```
ShEff
```

另请注意，我在两种情况下都将newtype ShEff a = ShEff (forall eff. a -> Eff (console :: CONSOLE | eff) Unit) f :: forall a eff. Show a => ShEff a f = ShEff logShow约束移到了Show a之外。

统一记录类型

1 个答案: