数据构造函数不在范围内: Monoid :：Bull -> TestBatch

Question

Haskell的书“Haskell编程从第一原则”第1086页有一个例子来说明False不应该用作mempty of Monoid，但是这个页面上的代码没有编译，我也不知道为什么。守则如下：

module BadMonoid where

import Data.Monoid
import Test.QuickCheck
import Test.QuickCheck.Checkers
import Test.QuickCheck.Classes

data Bull =
  Fools
  | Twoo
  deriving (Eq, Show)

instance Arbitrary Bull where
  arbitrary =
    frequency [(1, return Fools)
              ,(1, return Twoo)]

instance Monoid Bull where
  mempty = Fools
  mappend _ _ = Fools

main :: IO ()
main = do
  quickBatch (monoid Twoo)

语法检查器显示此代码的两个错误：

• No instance for (Semigroup Bull)
    arising from the superclasses of an instance declaration
• In the instance declaration for ‘Monoid Bull’

和

• No instance for (EqProp Bull) arising from a use of ‘monoid’
• In the first argument of ‘quickBatch’, namely ‘(monoid Twoo)’
  In a stmt of a 'do' block: quickBatch (monoid Twoo)
  In the expression: do quickBatch (monoid Twoo)

在stack ghci repl中加载此文件将显示：

[1 of 1] Compiling BadMonoid        ( src/Main.hs, interpreted )

src/Main.hs:18:10: error:
    • No instance for (Semigroup Bull)
        arising from the superclasses of an instance declaration
    • In the instance declaration for ‘Monoid Bull’
   |
18 | instance Monoid Bull where
   |          ^^^^^^^^^^^

src/Main.hs:39:15: error:
    • No instance for (EqProp Bull) arising from a use of ‘monoid’
    • In the first argument of ‘quickBatch’, namely ‘(monoid Twoo)’
      In a stmt of a 'do' block: quickBatch (monoid Twoo)
      In the expression: do quickBatch (monoid Twoo)
   |
39 |   quickBatch (monoid Twoo)
   |               ^^^^^^^^^^^

你能帮我修复这段代码吗？根据这本书，实际结果应该是：

Prelude> main
monoid:
  left  identity: *** Failed! Falsifiable (after 1 test):
Twoo
  right identity: *** Failed! Falsifiable (after 2 tests):
Twoo
  associativity:  +++ OK, passed 500 tests.

谢谢!

Answer 1

看来Monoid的定义自从这本书写出来后就发生了变化。As 文件说明

注意：：半群是从Bas-4.11.0.0开始的莫尼特的超类。

实例现在应该如下所示：

instance Monoid Bull where
  mempty = Fools

instance Semigroup Bull where
  _ <> _ = Fools

(mappend现在只需调用<>。)

对于EqProp实例，可以从Eq派生如下：

instance EqProp Bull where
  (=-=) = eq

Answer 2

第二个错误很抱歉：

我失去了一条线

instance EqProp Bull where (=-=) = eq

第一个错误：

我加了

instance Semigroup Bull where _ <> _ = Fools

在Monoid实例化上面，第一个错误消失了。