[英]Haskell pattern matching error with data constructor
我在Haskell中有以下代码:
module testData where
import SImpL
changeName :: String -> String -> ProgT -> ProgT
...
changeName x y (Seq []) = Seq[]
changeName x y (Seq (oneStatement:moreStatements)) = Seq (changeName x y oneStatement : changeName x y (Seq moreStatements))
ProgT的定义在模块SImpL中定义:
data StmtT = Assign NameT AExprT |
If BExprT StmtT StmtT |
While BExprT StmtT |
Seq [StmtT] -- If the list is empty, this is a statement that does nothing.
deriving (Show,Eq)
type ProgT = StmtT
简单地说,Seq [StmtT]是在构造函数中定义的Assign,If或While语句的列表。 函数changeName检查ALL语句中是否存在等于x的变量,并将其替换为y。 运行代码时,出现以下错误:
Assignment3.hs:12:89:错误:•无法将类型'StmtT'与'[ProgT]'匹配预期类型:[ProgT]实际类型:ProgT•在'(:)'的第二个参数中,即'changeName xy(Seq moreStatements)'在'Seq'的第一个参数中,即'(changeName xy oneStatement:changeName xy(Seq moreStatements))'在表达式中:Seq(changeName xy oneStatement:changeName xy(Seq moreStatements))
根据错误消息,问题在最后一行:
changeName xy(Seq(oneStatement:moreStatements))= Seq(changeName xy oneStatement: changeName xy(Seq moreStatements) )
我可以理解为什么会引发错误,但是我必须遍历每个语句来更改语句中的每个变量。 抱歉,这很简单,但是我不知道我可以通过Seq [StmtT]类型进行递归而不会出现错误。
注意:我不认为其他数据类型是什么(即BExprT)无关紧要,以防万一,这里有更多的模块:
module SImpL where
data AExprT = ALit ValT -- a literal value (an Int)
| AName NameT -- a variable name (a String)
| Add AExprT AExprT -- one arithmetic expression added to another
| Sub AExprT AExprT -- one arithmetic expression subtracted from another
| Mult AExprT AExprT -- one arithmetic expression multiplied by another
deriving (Show,Eq)
data BExprT = BLit Bool -- a literal value (True or False)
| Eq AExprT AExprT -- an equality test between two arithmetic expressions
| Less AExprT AExprT -- a "less than" test between two arithmetic expressions
| Greater AExprT AExprT -- a "greater than" test between two arithmetic expressions
| Not BExprT -- the negation of a boolean expression
| And BExprT BExprT -- the "and" of two boolean expressions
| Or BExprT BExprT -- the "or" of two boolean expressions
deriving (Show,Eq)
type ValT = Integer
type NameT = String
data StmtT = Assign NameT AExprT |
If BExprT StmtT StmtT |
While BExprT StmtT |
Seq [StmtT] -- If the list is empty, this is a statement that does nothing.
deriving (Show,Eq)
If BExprT StmtT StmtT |
While BExprT StmtT |
Seq [StmtT] -- If the list is empty, this is a statement that does nothing.
deriving (Show,Eq)
type ProgT = StmtT
type StateT = [(NameT, ValT)]
编辑: @Ben使用地图有助于解决该错误(因为该函数现在不返回列表)。
changeName x y (Seq (oneStatement:moreStatements)) = Seq(changeName x y oneStatement:changeName x y Seq(moreStatements))
应该:
changeName x y (Seq (oneStatement:moreStatements)) = Seq (changeName x y oneStatement : changeName x y (Seq moreStatements))
具体来说,问题在于您拥有changeName xy Seq(moreStatements)
的最后一部分被视为将changeName
应用于4个参数:
x
y
Seq
(moreStatements)
Seq(moreStatements)
并不意味着“将Seq
应用于moreStatements
,就像在使用类似于C的语法的语言中那样,而只是术语Seq
(moreStatements)
彼此相邻,完全等同于您是否编写了Seq moreStatements
因为Seq
是该子表达式的“头”,所以它可以做您想要的事情,但是它跟随在changeName xy
,因此将其应用于Seq
,然后应用于moreStatements
。这不适合changeName
的类型。
(请注意,这几乎完全是错误消息所说的内容)
您的第二个错误是changeNames
类型为String -> String -> ProgT -> ProgT
(记住ProgT = StmtT
)。 但是,您使用的结果changeName
上的右侧:
,仿佛它返回的东西的清单 。
可能您需要改为在列表上changeName
名称,而不是将其应用于头部和尾部? 实际上,您甚至不需要在列表上进行模式匹配,您可以:
changeName x y (Seq statements) = Seq (map (changeName x y) statements)
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