类型的呼叫

Question

考虑在Haskell以下两个函数（我的实际代码的最小示例）：

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

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

第一编译好，但第二个产生错误：

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) 

我知道这个错误意味着GHC是需要x和y等同类型。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，那么为什么GHC抱怨x和y是不同的类型，当我在两个不同的时间调用我的print'函数？

Answer 1

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NoMonomorphismRestriction：

{-# LANGUAGE NoMonomorphismRestriction #-} 

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

然后，

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