Question

我有一个程序给出了一组食谱，所需的输出将告诉我需要制作多少中间产品。

data Recipe = Recipe { name :: String
                     , ingredients :: [(Recipe, Int)]
                     } deriving (Eq)

instance Show Recipe where
        show Recipe{name=n} = show n

foldRecipes :: (Integral a) => [(Recipe, a)] -> (Recipe, a) -> [(Recipe, a)] 
foldRecipes acc r@(Recipe{ingredients=i}, n) = 
         acc ++ requirements [(recipe, fromIntegral count * n) | (recipe, count) <- i]

requirements :: (Integral a) => [(Recipe, a)] -> [(Recipe, a)] 
requirements m = 
        foldl foldRecipes m m

main = 
        let dough = Recipe{ name = "dough", ingredients = [(flour, 200), (water, 200)] }
            flour = Recipe{ name = "flour", ingredients = [] }
            water = Recipe{ name = "water", ingredients = [] }

        in putStrLn $ show $ requirements [(dough, 2)]

输出：[("dough",2),("flour",400),("water",400)

在我做这个的过程中，我遇到了以下版本不起作用的事实，有人可以解释原因吗？如果我使用明确的Int代替Integral a作为类型签名，那么它确实有效。

foldRecipes :: (Integral a) => [(Recipe, a)] -> (Recipe, a) -> [(Recipe, a)]
foldRecipes acc r@(Recipe{ingredients=i}, _) =
         acc ++ requirements i

输出：

test_unused.hs:10:30:
    Could not deduce (a ~ Int)
    from the context (Integral a)
      bound by the type signature for
                 foldRecipes :: Integral a =>
                                [(Recipe, a)] -> (Recipe, a) -> [(Recipe, a)]
      at test_unused.hs:8:16-76
      `a' is a rigid type variable bound by
          the type signature for
            foldRecipes :: Integral a =>
                           [(Recipe, a)] -> (Recipe, a) -> [(Recipe, a)]
          at test_unused.hs:8:16
    Expected type: [(Recipe, a)]
      Actual type: [(Recipe, Int)]
    In the first argument of `requirements', namely `i'
    In the second argument of `(++)', namely `requirements i'
    In the expression: acc ++ requirements i

Answer 1

foldRecipes :: (Integral a) => [(Recipe, a)] -> (Recipe, a) -> [(Recipe, a)]
foldRecipes acc r@(Recipe{ingredients=i}, _) =
     acc ++ requirements i

上述方法不起作用，因为ingredients始终包含类型[(Recipe, Int)]的列表，而您的函数接受使用泛型整数类型Integral a的列表。 a和Int是不同的类型，您无法使用++连接不同类型的列表。

您的其他版本的功能有效，因为您使用fromIntegral将Int转换为更通用的类型。

无法推断未使用参数的类型 - 如果使用参数则消失

1 个答案: