Haskell 实施汉诺塔时出错

Question

我正在尝试为汉诺塔实现递归函数。

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.

示例：

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

这是我的实现

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")

但是我收到一条错误消息 there is un-exhausted pattern。这是什么意思，我该如何解决？

Answer 1

Haskell 函数的参数实际上是提供的值所匹配的模式。

a 是一个无可辩驳的模式，无论是 "a"、"b"、"c"，它总是通过将变量 a 与提供的值匹配来成功或者完全是其他东西。

"a" 也是一种模式，但它只有在与 匹配的 值匹配时才会成功，"a":

~> 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，如果您希望将参数解释为变量模式，则在定义函数时不要将参数括在引号中。

Haskell 实施汉诺塔时出错

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