Question

let rec move_robot (pos: int) (dir: string) (num_moves: int) : int =

    let new_forward_position = pos + num_moves in
    if (new_forward_position > 99) then failwith "cannot move beyond 99 steps"
    else new_forward_position

    let new_backward_position = pos - num_moves in
    if (new_backward_position pos < 0) then failwith "cannot move less than 0 steps"
    else new_backward_position

    begin match dir with 
    | "forward" -> new_forward position
    | "backward" -> new_backward_position
    end

我继续为let new_backward_position行获取“意外令牌”。我的错误是什么？

Answer 1

这是一个编译代码：

let rec move_robot pos dir num_moves =
    let new_forward_position = pos + num_moves in
    if new_forward_position > 99 then failwith "cannot move beyond 99 steps";

    let new_backward_position = pos - num_moves in
    if new_backward_position < 0 then failwith "cannot move less than 0 steps";

    begin match dir with
    | "forward" -> new_forward_position
    | "backward" -> new_backward_position
    end

我修改了几件事：

重要提示：if foo then bar else qux是OCaml中的表达式，其值为bar或qux。因此bar和qux需要具有相同的类型。
new_backward_position代替new_backward_position pos
您不需要类型注释：OCaml具有类型推断
围绕if子句
new_forward position

此外，根据您的代码逻辑，let _ = move_robot 0 "forward" 5失败。不应该返回5而不是吗？我建议你为pos定义一个和类型，然后先对它进行模式匹配。

Answer 2

如果您认为不会发生故障，则您的代码具有此基本结构：

let f () =
    let p = 3 in p
    let q = 5 in q
    ...

目前还不清楚你要做什么，但这不是很好的OCaml（正如编译器告诉你的那样）。也许你想要的更像是这样的东西：

let f () =
    let p = 3 in
    let q = 5 in
    match ...

如果是这样，您需要在if之前移动in：

let f () =
    let p = if badp then failwith "" else 3 in
    let q = if badq then failwith "" else 5 in
    match ...

或者这可能更符合您的要求：

let f () =
    let p = 3 in
    let () = if badp p then failwith "" in
    let q = 5 in
    let () = if badq q then failwith "" in
    match ...

（我希望这很有帮助。）

为什么这个OCaml代码收到“意外令牌”？

2 个答案: