与显示函数实例相关的错误

Question

我有以下代码，我认为我遗漏了一些东西：

bruteSolveWithStartAndEnd :: Node -> Node -> [Node] -> [(Float, [Node])] -> [(Float, [Node])]
bruteSolveWithStartAndEnd st en mp r
           | length mp - 1 == length (snd ( head r)) = sortBy (comparing fst) $ map (\x -> (fst x + (distanceBetweenTwoNodes en $ last $ snd x), snd x ++ [en])) r
           | otherwise = bruteSolveWithStartAndEnd st en mp $ nextStepsForMany r $ filter (/= en) mp

bruteSolve :: [Node] -> [(Float, [Node])] -> [(Float, [Node])]
bruteSolve mp [] = bruteSolve mp $ map (\l -> (0, [l])) mp
bruteSolve mp r
           | length mp == length(snd (head r)) = sortBy (comparing fst) $ r
           | otherwise = bruteSolve mp $ nextStepsForMany r mp

nextSteps :: (Float, [Node]) -> [Node] -> [(Float, [Node])]
nextSteps r n = map (\l -> ((fst r) + (distanceBetweenTwoNodes l $ last $ snd r), snd r ++ [l])) [ nn | nn <- n, nn `notElem` (snd r)]

nextStepsForMany :: [(Float, [Node])] -> [Node] -> [(Float, [Node])]
nextStepsForMany ar n = concat $ map (\l -> nextSteps l n) ar

代码本身有效。但奇怪的是，虽然函数 bruteSolve 和 bruteSolveWithStartAndEnd 具有相同的“结果”类型，但 bruteSolveWithStartAndEnd 的结果无法在 GHSi 中打印，而 bruteSolve 的结果可以：

*> bruteSolve [(Node 0 1), (Node 2 3), (Node 4 5)] []
[(5.656854,[Node {x = 0, y = 1},Node {x = 2, y = 3},Node {x = 4, y = 5}]),(5.656854,[Node {x = 4, y = 5},Node {x = 2, y = 3},Node {x = 0, y = 1}]),(8.485281,[Node {x = 0, y = 1},Node {x = 4, y = 5},Node {x = 2, y = 3}]),(8.485281,[Node {x = 2, y = 3},Node {x = 0, y = 1},Node {x = 4, y = 5}]),(8.485281,[Node {x = 2, y = 3},Node {x = 4, y = 5},Node {x = 0, y = 1}]),(8.485281,[Node {x = 4, y = 5},Node {x = 0, y = 1},Node {x = 2, y = 3}])] 


*> bruteSolveWithStartAndEnd (Node 0 0) (Node 5 5) [Node 0 0, Node 1 1, Node 2 2, Node 3 3, Node 4 4, Node 5 5]

<interactive>:1:1:
    No instance for (Show ([(Float, [Node])] -> [(Float, [Node])]))
      arising from a use of `print'
    Possible fix:
      add an instance declaration for
      (Show ([(Float, [Node])] -> [(Float, [Node])]))
    In a stmt of an interactive GHCi command: print it

我做错了什么？我改变了一些东西（不记得是什么），从那时起这个错误就出现了。根据我的理解（Haskell 的第一天），打印函数将检查 [..] 是否派生 Show，然后递归地检查所有内部元素（在本例中为 list->tuple->(Float & Node)） ...无论如何，既然“结果”声明是相同的，为什么会出现这种差异？

干杯!

Answer 1

(Show ([(Float, [Node])] -> [(Float, [Node])])) 的实例没有表明您正在尝试打印函数。这可能不是您想要做的...您是否忘记了 bruteSolveWithStartAndEnd 的参数？

Answer 2

您没有向 bruteSolveWithStartAndEnd 提供足够的参数。像这样的查询

bruteSolveWithStartAndEnd (Node 0 0) (Node 5 5) [Node 0 0, Node 1 1, Node 2 2, Node 3 3, Node 4 4, Node 5 5] []

应该可以工作。