Haskell 需要在两个单独的函数调用中等效的类型

考虑Haskell中的以下两个函数（我的实际代码的一个最小示例）：

第一个编译很好，但第二个生成错误：

 Could not deduce (a ~ b)
    from the context (Show a, Show b)
      bound by the type signature for
                 printSequence' :: (Show a, Show b) => a -> b -> IO ()
      at test.hs:8:19-53
      `a' is a rigid type variable bound by
          the type signature for
            printSequence' :: (Show a, Show b) => a -> b -> IO ()
          at test.hs:8:19
      `b' is a rigid type variable bound by
          the type signature for
            printSequence' :: (Show a, Show b) => a -> b -> IO ()
          at test.hs:8:19
    In the first argument of print', namely `y'
    In the second argument of `(>>)', namely `(print' y)'
    In the expression: (print' x) >> (print' y)

---

我理解此错误意味着GHC要求

x

和

y

为等效类型。我不明白的是为什么。像

print“fish”>>print 3.14

这样的语句在解释器中工作得非常好，那么当我两次调用我的

print'

函数时，GHC为什么抱怨

x

和

y

是不同的类型呢？

添加一个明确的类型签名：

printSequence' :: (Show a, Show b) => a -> b -> IO ()
printSequence' x y = print' x >> print' y
    where
    print' :: Show a => a -> IO ()
    print' = putStr . show

或使用：

那么

---

这在逐字复制时有效，但我有一个关于您的第一个解决方案的后续问题。为什么当我在同一行中提供类型签名时失败，比如

where print'=（putStr.show）：：show a=>a->IO（）

？@ApproachingDarknessFish在=符号的右侧注释值，让GHC在左侧推断名称的类型。GHC为没有显式函数参数的模式推断单态类型，并且在右侧声明多态类型不会改变该规则。关于可怕的单态限制：

printSequence' :: (Show a, Show b) => a -> b -> IO ()
printSequence' x y = print' x >> print' y
    where
    print' :: Show a => a -> IO ()
    print' = putStr . show

{-# LANGUAGE NoMonomorphismRestriction #-}

printSequence' :: (Show a, Show b) => a -> b -> IO ()
printSequence' x y = print' x >> print' y
    where
    print' = putStr . show

\> printSequence' 5 "five"
5"five"