Question

嘿大家我正在尝试编写此代码，而我遇到的数据类型有问题

data Collection = Set [Int] deriving (Show)

remove :: Int -> Collection -> Collection
remove _ (Set []) = (Set [])
remove numberToRemove (Set (x:xs))
    |x == numberToRemove = (Set xs)
    |otherwise = Set ([x]:remove numberToRemove xs)

我收到此错误，这是类型的问题：

 Couldn't match expected type `Int' with actual type `[t0]'
In the first argument of `(:)', namely `[x]'
In the first argument of `Set', namely
  `([x] : remove numberToRemove xs)'
In the expression: Set ([x] : remove numberToRemove xs)
Failed, modules loaded: none.

感谢任何帮助感谢

Answer 1

第一个问题，在表达式中：

Set ([x] : remove numberToRemove xs)

列表的头部（在：之前）必须是Int，而不是[Int]，替换为：

Set (x : remove numberToRemove xs)

然后，第二个问题。在同一表达式中，子表达式：

remove numberToRemove xs

是一个Collecion，但是在运算符之后必须有一个[Int]：所以，一个可能的解决方案：

data Collection = Set [Int] deriving (Show)

col_to_list :: Collection -> [Int]
col_to_list (Set xs) = xs

remove :: Int -> Collection -> Collection
remove _ (Set []) = (Set [])
remove numberToRemove (Set (x:xs))
    |x == numberToRemove = (Set xs)
    |otherwise = Set (x : col_to_list (remove numberToRemove (Set xs)))

另一个haskell类型的困境

1 个答案: