sml - 在 SML 获取问题上展平数据类型列表

Question

我有一个关于我的作业的问题，问题是
有使用数据类型:

datatype 'a llist = LList of 'a llist list| Elem of 'a;

嵌套列表由多态类型的元素或嵌套列表的列表组成。这
以下是一些示例:

Elem(1);
LList [];
LList([Elem(1), LList([Elem(2), LList([Elem 1, Elem(3)]), Elem(4)])]);

Write a function flatten that takes a nested list as input and returnsa flat list of allelements in the nested list. Note that the elements in the resulting list are in the same orderas in the nested list.

- flatten;
val flatten = fn : 'a llist -> 'a list

Examples:

- flatten(Elem(3));
val it = [3] : int list
- flatten(LList([]));
val it = [] : ?.X1 list
- flatten(LList([Elem(1),LList([Elem(2),LList([]),Elem(3)]),Elem(4)]
));
val it = [1,2,3,4] : int list

但我的代码是

fun flatten Elem x = [x] | LList x = (List.concat (map (fn a => flatten(a)) x));

有问题

- fun flatten Elem x = [x] | LList x = (List.concat (map (fn a => flatten(a)) x));
stdIn:13.1-17.71 Error: clauses do not all have same function name

stdIn:13.1-17.71 Error: clauses do not all have same number of patterns

stdIn:17.4-17.8 Error: data constructor Elem used without argument in pattern

stdIn:13.1-17.71 Error: types of rules do not agree [tycon mismatch]
  earlier rule(s): 'Z * 'Y -> 'Y list
  this rule: 'X list -> 'W list
  in rule:
    x => List.concat ((map (fn a => flatten a)) x)

stdIn:13.1-17.71 Error: right-hand-side of clause does not agree with function result type [tycon mismatch]
  expression:  'Z -> 'Z list
  result type:  'Y list
  in declaration:
    flatten =
      (fn arg =>
            (fn arg =>
                  (case (arg,arg)
                  of (_,x) => x :: nil
                   | x => List.concat ((map <exp>) x))))

我不知道我的代码有什么问题。

Answer 1

您的代码中存在一些语法问题。
否则逻辑没问题。
这是一个更正的版本:

fun flatten (Elem x) = [x]
  | flatten (LList x) = (List.concat (map (fn a => flatten(a)) x));