在OCaml中键入错误

Question

我正在尝试构建列表最小值出现的索引列表.

let rec max_index l =
let rec helper inList min builtList index = 
match inList with
| [] -> builtList
| x :: xs ->
    if (x < min) then
        helper xs x index :: builtList index + 1 //line 63
    else
        helper xs min builtList index + 1
in helper l 100000 [] 0;;

第63行给出了以下错误.

Error: This expression has type 'a list -> 'a list
       but an expression was expected of type 'a
       The type variable 'a occurs inside 'a list -> 'a list

表达式是'a？' 我不确定为什么这么说.我猜这与它有关index::builtList

Answer 1

        helper xs x index :: builtList index + 1 //line 63
    else
        helper xs x index min index + 1

您遇到的问题是,您尝试将第63 行(min)上的非列表传递给辅助函数,同时尝试将int list第63行传递给第63行的相同参数.请尝试min使用[min]或替换min::[].

编辑:

在更新之后,问题在于函数调用是左关联的并且优先于二元运算符(参见此处),因此helper xs x index将在之前执行index :: builtList并且同样helper xs x index :: builtList将在之前执行index + 1.要获得正确的评估顺序,您需要在其他函数调用周围添加括号,::并+加上它们的参数,如下所示:

        helper xs x (index :: builtList) (index + 1) //line 63
    else
        helper xs x index min (index + 1)