Question

为什么我的merge函数会抱怨它的类型？

我的x不是type 'a seq吗？

type 'a seq = Stop | Cons of 'a * (unit -> 'a seq)

let rec linear start step= (*builds a seq starting with 'start'*)
    Cons (start, fun () -> linear (start+step) step)

let rec take n seq = match seq with (*take first n elem from seq*)
| Stop -> []
| Cons (a, f) -> if n < 1 then [] else a::(take (n-1) (f ()))

let rec merge seq1 seq2 = match seq1, seq2 with 
    | Stop, _ -> seq2
    | _, Stop -> seq1
    | Cons(h1, tf1), _ as x -> 
        Cons(h1, fun () -> merge (x) (tf1 ()))

let l1 = linear 1 1
let l2 = linear 100 100 
let l3 = interleave l1 l2

我希望看到

的正确结果

take 10 l3

int list = [1; 100; 2; 200; 3; 300; 4; 400; 5; 500]

编写我的函数（有效）的另一种方法是

let rec merge seq1 seq2 = match seq1 with
| Stop -> Stop
| Cons (h, tf) -> Cons(h, fun () -> merge seq2 (tf ()))

但是我没理解，为什么第一个merge不起作用。

感谢。

Answer 1

只需写下(_ as x)，因为在这里，您的as x会抓住整个模式。

所以，你看到的是：

| Cons(h1, tf1), (_ as x) -> ...

实际上被解析为

| (Cons(h1, tf1), _) as x -> ...

你实际上可以写：

| Cons(h1, tf1), x -> ...

哪个更好; - ）

甚至

| Cons(h1, tf1), _ -> Cons(h1, fun () -> merge seq2 (tf1 ()))

合并2个懒惰列表类型冲突

1 个答案: