Question

我正在尝试做作业但是我遇到了一个我不理解的语法错误。

data CTree a b =
        CNode (CTree a b) b (CTree a b) | -- left, value, right
        CLeaf a                           -- value
    deriving (Show, Ord)

-- a)
--

instance (Ord a, Ord b) => Ord (CTree a b) where

    --
    -- Compares two CNode objects
    --
    (CNode left1 v1 right1) `compare` (CNode left2 v2 right2) =

        -- If the value of the first node is greater than
        -- the value of the second node, the result is GT.
        if (v1 > v2) then GT

        -- If the value of the first node is smaller than
        -- the value of the second node, the result is LT
        else if (v2 < v1) then LT

        -- We must compare the child nodes if the values of
        -- boths nodes equal.
        else if (v1 == v2) then

            if (left1 == left2) then
                (right1 `compare` right2)
            else
                (left1 `compare` left2)

main = do
    print "foo"

错误是（在main = do）行

解析错误（可能是错误的缩进或括号不匹配）

我知道比较函数有一些重载缺失，但语法错误不应该源于这个事实。

Answer 1

在您的代码中，if (v1 == v2)有一个then分支，但缺少else分支。您应该删除if，因为您之前检查过v1 < v2和v1 > v2，这是没有用的。

Answer 2

你应该缩进：

if (left1 == left2) then
    (right1 `compare` right2)
else
    (left1 `compare` left2)

再一层，并为相应的else提供if (v1 == v2) then。

然后，您应该从Ord移除deriving并添加Eq。由于您的b类型为Ord，因此您只需将程序重写为：

data CTree a b =
        CNode (CTree a b) b (CTree a b) |
        CLeaf a
    deriving (Show, Eq)

instance (Ord a, Ord b) => Ord (CTree a b) where
    (CNode _ v1 _) `compare` (CNode _ v2 _) = v1 `compare` v2

main = do
    print "foo"

Live demo

我注意到我已经改变了CTree instance (Ord a, Ord b) => Ord (CTree a b) where (CNode left1 v1 right1) `compare` (CNode left2 v2 right2) = if (v1 == v2) then if (left1 == left2) then (right1 `compare` right2) else (left1 `compare` left2) else v1 `compare` v2函数的含义，因为你的函数永远不会返回compare，这实际上是预期的值。

但如果你想保持这种行为，你可以这样做：

{{1}}

EQ

嵌套ifs的Haskell语法错误

2 个答案: