Question

我正在尝试为Tower of Hanoi实现一个递归函数。

算法是：

Move n−1 disks from peg AA to peg C using peg B as intermediate storage.

Move the nth disk from peg A to peg B,

Move n−1 disks from peg C to peg BB using peg A as intermediate storage.

Eample：

hanoi 2 "a" "b" "c" =
[("a","c"), ("a","b"), ("c","b")]

这是我的实施

hanoi :: Integer -> Peg -> Peg -> Peg -> [Move] 

hanoi x "a" "b" "c" 
    | x <= 0    = []
    | x == 1    = [("a", "b")]
    | otherwise = (hanoi (x-1) "a" "c" "b") ++ [("a", "b")] ++ (hanoi (x-1) "c" "b" "a")

但是我收到错误there is un-exhausted pattern。我怎么解决它是什么意思？

Answer 1

Haskell函数的参数实际上是提供的值与之匹配的模式。

a是一种无可辩驳的模式，通过将变量a与提供的值（"a"，"b"，"c"或其他内容相匹配，始终可以成功完全是。

"a"也是一种模式，但与匹配的值"a"匹配时仅成功：

~> let f "a" = 1
f :: Num a => [Char] -> a

~> f "a"
1
it :: Num a => a

~> f "c"
*** Exception: <interactive>:3:5-13: Non-exhaustive patterns in function f

因此，如果您希望将它们解释为变量模式，请不要在定义函数时将参数括在引号中。

实施河内塔时Haskell错误

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