错误，以显示实例功能

Question

相关的我有以下的代码，我觉得我失去了一些东西：

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 

工作本身的代码。但奇怪的事情是，虽然功能bruteSolve和bruteSolveWithStartAndEnd具有相同的“结果”类型，bruteSolveWithStartAndEnd的结果不能在GHSi印刷而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 

我在做什么错？我改变了一些东西（不记得是什么），从那以后出现这个错误。从我的理解（Haskell的第一天），打印函数将检查[..]是否派生Show，然后递归地遍历所有内部元素（在这种情况下，列表 - >元组 - >（浮点数&节点））...无论如何，因为“结果”声明是相同的，为什么这种差异？

干杯！

Answer 1

您无法提供足够的参数给bruteSolveWithStartAndEnd。像

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] [] 

应该工作。

Answer 2

(Show ([(Float, [Node])] -> [(Float, [Node])]))没有实例说你正在尝试打印一个函数。这可能不是你想要做的...你忘记了一个参数到bruteSolveWithStartAndEnd？