Question

我写了这个数据类型：

data Poly = Lit Integer |
            Var |
            Add Poly Poly |
            Mul Poly Poly

我还想为它编写一个函数printPoly，它接受一个Poly表达式并将其转换为字符串。这可以通过为数据类型Poly创建show的自定义实例来实现，如下所示：

instance Show Poly where
   show (Lit x)   = show x
   show (Var)     = "x"
   show (Add x y) = (show x) ++ " + " ++ (show y)
   show (Mul x y) = (show x) ++ "*" ++ (show y)

现在，如果我传递一个像

这样的表达式

main = do
print (Add (Lit 1) $ Add (Var) $ Mul (Var) $ Mul Var Var)

返回1 + x + x*x*x。这就是我想要的。但是，我希望通过函数printPoly完成此操作，如下所示：

printPoly::Poly->String
printPoly (Lit x) = show x
printPoly (Var) = "x"
printPoly (Add x y) = (show x) ++ " + " ++ (show y)
printPoly (Mul x y) = (show x) ++ "*" ++ (show y)

我写这个函数的方式在哪里错了？

Answer 1

要回答你提出的问题，这就是你想要做的。

printPoly :: Poly -> String
printPoly (Lit x) = show x
printPoly (Var) = "x"
printPoly (Add x y) = (printPoly x) ++ " + " ++ (printPoly y)
printPoly (Mul x y) = (printPoly x) ++ "*" ++ (printPoly y)

instance Show Poly where
   show = printPoly

然而，这有一些明显的问题。考虑print (Mul (Lit 2) $ Add Var Var)。这将打印2*x + x，当遵循优先规则时，显然不符合预期。

printPoly :: Poly -> String
printPoly (Lit x) = show x
printPoly (Var) = "x"
printPoly (Add x y) = (b x $ printPoly x) ++ " + " ++ (b y $ printPoly y)
  where b (Add _ _) p = "(" ++ p ++ ")"
        b _         p = p
printPoly (Mul x y) = (b x $ printPoly x) ++ " * " ++ (b y $ printPoly y)
  where b (Mul _ _) p = "(" ++ p ++ ")"
        b _         p = p

这将打印出2 * (x + x)，这是对给定AST的正确解释。此外，您可以使用printf

中的Text.Printf来简化整个过程

printPoly :: Poly -> String
printPoly (Lit x) = show x
printPoly (Var) = "x"
printPoly (Add x y) = printf "%s + %s" (b x $ printPoly x) (b y $ printPoly y)
  where b (Add _ _) = printf "(%s)"
        b _         = id
printPoly (Mul x y) = printf "%s * %s" (b x $ printPoly x) (b y $ printPoly y)
  where b (Mul _ _) = printf "(%s)"
        b _         = id

当然，这仍然会进行大量的链表遍历，将字符串拼凑在一起所以它不会很快，但至少它是正确的;）

如何使函数实现Show的自定义实例的内部。哈斯克尔

1 个答案: