为什么以下尝试在列表理解中进行模式匹配不起作用?
示例:同时替换术语数据类型中的原子。
数据类型:
data Term a
= Atom a
| Compound (Term a) (Term a)
deriving Show
用原子来表示原子的替换算法(如果有匹配则选择第一个匹配替换,然后忽略其余匹配):
subs :: [(Term a, Term a)] -> Term a -> Term a
subs subList term = case term of
atom@(Atom x) -> let substitutions =
[ s | s@(Atom x, _) <- subList ]
in if null substitutions
then atom
else snd . head $ substitutions
(Compound t1 t2) -> Compound (subs subList t1) (subs subList t2)
一些测试数据:
subList = [((Atom 'a'), Compound (Atom 'b') (Atom 'c'))]
term1 = Atom 'a'
term2 = Atom 'x'
运行示例将导致:
>: subs subList term1
Compound (Atom 'b') (Atom 'c')
这是理想的行为,并且
>: subs subList term2
Compound (Atom 'b') (Atom 'c')
不是。
粗暴的显式匹配工作原理:
subs'' :: [(Term Char, Term Char)] -> Term Char -> Term Char
subs'' subList term = case term of
atom@(Atom _) -> let substitutions =
[ s | s@(Atom 'a', _) <- subList ]
in if null substitutions
then atom
else snd . head $ substitutions
(Compound t1 t2) -> Compound (subs subList t1) (subs subList t2)
subs''' subList term = case term of
atom@(Atom _) -> let substitutions =
[ s | s@(Atom 'x', _) <- subList ]
in if null substitutions
then atom
else snd . head $ substitutions
(Compound t1 t2) -> Compound (subs subList t1) (subs subList t2)
将测试数据输入以下内容:
>: subs'' subList term1
或者 >: subs'' subList term2
Compound (Atom 'b') (Atom 'c')
>: subs''' subList term1
或者 >: subs''' subList term2
Atom 'x'
我想念什么?
Haskell具有线性模式,这意味着模式中不得有重复的变量。同样,内部表达式中的模式变量会遮盖外部变量,而不是建立相同变量的相等性。
您正在尝试执行以下操作:
charEq :: Char -> Char -> Bool
charEq c c = True
charEq _ _ = False
但是由于重复的变量,这是一个错误。如果我们将第二个c
移到内部表达式,它会编译,但仍无法按预期工作:
charEq :: Char -> Char -> Bool
charEq c d = case d of
c -> True
_ -> False
这里的innerc
只是一个新的变量,它遮蔽了外部c
,因此charEq
总是返回True
。
如果我们想检查是否相等,则必须==
显式使用:
subs :: [(Term a, Term a)] -> Term a -> Term a
subs subList term = case term of
atom@(Atom x) -> let substitutions =
[ s | s@(Atom x', _) <- subList, x == x' ]
in if null substitutions
then atom
else snd . head $ substitutions
(Compound t1 t2) -> Compound (subs subList t1) (subs subList t2)
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