Haskell 编译错误 - 保护类型不匹配
Haskell Compilation Error - Guard Type mismatching
我一直停留在这个特定阶段，一直在解决我遇到的问题，我对 haskell 的体验仍处于初学者水平。
在尝试创建一个函数来创建一个节点（由一个字符串 "Key" 和一个值组成）插入到二叉搜索树中时，我尝试了一种使用保护括号的方法。我似乎误解了如何在这里做某事，我收到的错误对我来说没有意义。
module Ex03 where
import Data.List ((\))
-- Datatypes -------------------------------------------------------------------
-- do not change anything in this section !
-- Binary Tree
data BT a b
= Leaf
| Branch (BT a b) a b (BT a b)
deriving (Eq, Show)
-- association list
type Assoc a b = [(a,b)]
-- lookup binary (search) tree
lkpBST :: Ord a1 => BT a1 a -> a1 -> Maybe a
lkpBST Leaf _ = Nothing
lkpBST (Branch left k d right) k'
| k < k' = lkpBST left k'
| k > k' = lkpBST right k'
| otherwise = Just d
-- Coding Part 1 (13 Marks)
-- insert into binary (search) tree
insBST :: Ord a => a -> b -> BT a b -> BT a b
insBST labelX valueX Leaf = Branch Leaf labelX valueX Leaf
{-
insBST a 1 Branch Leaf b 1 Leaf
@?= Branch Branch Leaf a 1 Leaf b 2 Leaf
-}
insBST labelX valueX (Branch left labelRoot valueRoot right)
| valueX < valueRoot = (Branch (insBST labelX valueX left) labelRoot valueRoot right)
| valueX > valueRoot = (Branch left labelRoot valueRoot (insBST labelX valueX right))
| otherwise = Branch left labelX valueX right
-- Coding Part 2 (6 Marks)
-- convert an association list to a binary search tree
assoc2bst :: Ord a => Assoc a b -> BT a b
assoc2bst _ = error "assoc2bst not yet implemented"
-- Coding Part 3 (6 Marks)
-- convert a binary search tree into an (ordered) association list
bst2assoc :: Ord c => BT c e -> Assoc c e
bst2assoc _ = error "bst2assoc not yet
我收到的错误如下
Hours of hacking await!
If I break, you can:
1. Restart: M-x haskell-process-restart
2. Configure logging: C-h v haskell-process-log (useful for debugging)
3. General config: M-x customize-mode
4. Hide these tips: C-h v haskell-process-show-debug-tips
src/Ex03.hs:35:14: Could not deduce (Ord b) arising from a use of ‘<’ …
from the context (Ord a)
bound by the type signature for
insBST :: Ord a => a -> b -> BT a b -> BT a b
at /Users/eoindowling/college/3year2/michaelmas/CS3016-_Introduction_to_Functional_Programming/CSU34016-1920/Exercise03/src/Ex03.hs:28:11-45
Possible fix:
add (Ord b) to the context of
the type signature for
insBST :: Ord a => a -> b -> BT a b -> BT a b
In the expression: valueX < valueRoot
In a stmt of a pattern guard for
an equation for ‘insBST’:
valueX < valueRoot
In an equation for ‘insBST’:
insBST labelX valueX (Branch left labelRoot valueRoot right)
| valueX < valueRoot
= (Branch (insBST labelX valueX left) labelRoot valueRoot right)
| valueX > valueRoot
= (Branch left labelRoot valueRoot (insBST labelX valueX right))
| otherwise = Branch left labelX valueX right
Compilation failed.
我这里的 guard 语句有什么问题？我见过类似的方法，但我不明白我做错了什么。
您正在比较类型 b 的 valueX 和 valueRoot。错误消息说这样做需要一个约束 Ord b，但你只有一个约束 Ord a。
也许您是想比较 labelX 和 labelRoot（它们的类型为 a）？
我一直停留在这个特定阶段，一直在解决我遇到的问题，我对 haskell 的体验仍处于初学者水平。
在尝试创建一个函数来创建一个节点（由一个字符串 "Key" 和一个值组成）插入到二叉搜索树中时，我尝试了一种使用保护括号的方法。我似乎误解了如何在这里做某事，我收到的错误对我来说没有意义。
module Ex03 where
import Data.List ((\))
-- Datatypes -------------------------------------------------------------------
-- do not change anything in this section !
-- Binary Tree
data BT a b
= Leaf
| Branch (BT a b) a b (BT a b)
deriving (Eq, Show)
-- association list
type Assoc a b = [(a,b)]
-- lookup binary (search) tree
lkpBST :: Ord a1 => BT a1 a -> a1 -> Maybe a
lkpBST Leaf _ = Nothing
lkpBST (Branch left k d right) k'
| k < k' = lkpBST left k'
| k > k' = lkpBST right k'
| otherwise = Just d
-- Coding Part 1 (13 Marks)
-- insert into binary (search) tree
insBST :: Ord a => a -> b -> BT a b -> BT a b
insBST labelX valueX Leaf = Branch Leaf labelX valueX Leaf
{-
insBST a 1 Branch Leaf b 1 Leaf
@?= Branch Branch Leaf a 1 Leaf b 2 Leaf
-}
insBST labelX valueX (Branch left labelRoot valueRoot right)
| valueX < valueRoot = (Branch (insBST labelX valueX left) labelRoot valueRoot right)
| valueX > valueRoot = (Branch left labelRoot valueRoot (insBST labelX valueX right))
| otherwise = Branch left labelX valueX right
-- Coding Part 2 (6 Marks)
-- convert an association list to a binary search tree
assoc2bst :: Ord a => Assoc a b -> BT a b
assoc2bst _ = error "assoc2bst not yet implemented"
-- Coding Part 3 (6 Marks)
-- convert a binary search tree into an (ordered) association list
bst2assoc :: Ord c => BT c e -> Assoc c e
bst2assoc _ = error "bst2assoc not yet
我收到的错误如下
Hours of hacking await!
If I break, you can:
1. Restart: M-x haskell-process-restart
2. Configure logging: C-h v haskell-process-log (useful for debugging)
3. General config: M-x customize-mode
4. Hide these tips: C-h v haskell-process-show-debug-tips
src/Ex03.hs:35:14: Could not deduce (Ord b) arising from a use of ‘<’ …
from the context (Ord a)
bound by the type signature for
insBST :: Ord a => a -> b -> BT a b -> BT a b
at /Users/eoindowling/college/3year2/michaelmas/CS3016-_Introduction_to_Functional_Programming/CSU34016-1920/Exercise03/src/Ex03.hs:28:11-45
Possible fix:
add (Ord b) to the context of
the type signature for
insBST :: Ord a => a -> b -> BT a b -> BT a b
In the expression: valueX < valueRoot
In a stmt of a pattern guard for
an equation for ‘insBST’:
valueX < valueRoot
In an equation for ‘insBST’:
insBST labelX valueX (Branch left labelRoot valueRoot right)
| valueX < valueRoot
= (Branch (insBST labelX valueX left) labelRoot valueRoot right)
| valueX > valueRoot
= (Branch left labelRoot valueRoot (insBST labelX valueX right))
| otherwise = Branch left labelX valueX right
Compilation failed.
我这里的 guard 语句有什么问题？我见过类似的方法，但我不明白我做错了什么。
您正在比较类型 b 的 valueX 和 valueRoot。错误消息说这样做需要一个约束 Ord b，但你只有一个约束 Ord a。
也许您是想比较 labelX 和 labelRoot（它们的类型为 a）？