1
爲什麼我的merge函數抱怨它的類型？合併2懶惰列表類型衝突
是不是我x一個type 'a seq？
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
我想看到正確的結果爲
take 10 l3
INT列表= [1; 100; 2; 200; 3; 300; 4; 400; 5; 500]
另一種方式來寫我的功能（工作）將
let rec merge seq1 seq2 = match seq1 with
| Stop -> Stop
| Cons (h, tf) -> Cons(h, fun() -> merge seq2 (tf()))
，但我不明白，爲什麼第一次merge不起作用。
感謝。