Question

以下是我的示例代码。据说工程师很特别，如果他的名字是“Hari”（或）id是13.我通过使用View Patterns来做到这一点。我可以在条件运算符上应用视图模式吗？

示例如果他的id是＆lt; 50.

{-# LANGUAGE ViewPatterns #-}

module ABC where
    type Name = String
    type NoOfReportees = Integer
    type EngineerId = Integer
    type ManagerId = Integer
    type DirectorId = Integer

    {- Define an employee type -}
    data Employee = Engineer Name EngineerId
               | Manager Name [EngineerId] ManagerId
               | Director Name [ManagerId] DirectorId
               deriving Show

    getEngineerFirstName :: Employee -> Name
    getEngineerFirstName (Engineer name _) = name

    getEngineerId :: Employee -> EngineerId
    getEngineerId (Engineer _ empId) = empId

    isSpecialEngineer :: Employee -> Bool
    isSpecialEngineer (getEngineerFirstName -> "Hari") = True
    isSpecialEngineer (getEngineerId -> 13) = True
    isSpecialEngineer _ = False

Answer 1

这有效

isSpecialEngineer (getEngineerId -> x) | x < 50 = True

但是，您必须完成getEngineerId函数才能返回每种类型员工的值。请注意，您也可以使用旧式模式在Engineer上进行模式匹配，并且更安全地执行此操作

isSpecialEngineer (Engineer _ x) | x < 50 = True

Haskell：可以在条件运算符上应用视图模式

1 个答案: