Haskell 使用GHCi中的记录通配符扩展时出错

我创建了一个函数，该函数使用RecordWildCards语法对Haskell记录类型进行模式匹配：

布拉格马

我已将pragmas放在文件的顶部。我还尝试添加：set-xrecord通配符

类型定义

作用

错误

当我在CompanyR上使用此函数来匹配使用as模式的PersonR时，我得到：

作用

错误

---

在这里的第一个案例中，您做得很好，尽管我在您有++的地方修复了一个+：

但此处firstName等不是CompanyR中的记录，因此CompanyR{..}不将其纳入范围：

greet2 CompanyR { .. } = "hello " ++ firstName ++ " " ++ lastName ++ "who works as a " ++ duty ++ " " ++ clientRName + " "

您必须像上面第一个greet2案例中那样：

greet2 CompanyR {person = PersonR { .. }, .. } = "hello " ++ firstName ++ " " ++ lastName ++ "who works as a " ++ duty ++ " " ++ clientRName ++ " "

我正在努力c@CompanyR{..}=hello++firstName$person c+++lastName$person c使用记录通配符将person从CompayR类型中删除。

greet2 :: ClientR -> String
greet2 IndividualR { person = PersonR { .. } } = "hi" ++ firstName ++ " " ++ lastName + " "
greet2 CompanyR { .. } = "hello " ++ firstName ++ " " ++ lastName ++ "who works as a " ++ duty ++ " " ++ clientRName + " "
greet2 GovOrgR {} = "Welcome"

    • Couldn't match expected type ‘[Char]’
                  with actual type ‘PersonR -> String’
    • Probable cause: ‘lastName’ is applied to too few arguments
      In the first argument of ‘(++)’, namely ‘lastName’
      In the second argument of ‘(++)’, namely
        ‘lastName ++ "who works as a " ++ duty ++ " " ++ clientRName + " "’
      In the second argument of ‘(++)’, namely
        ‘" "
         ++
           lastName ++ "who works as a " ++ duty ++ " " ++ clientRName + " "’
Failed, modules loaded: none.

greet2 c@(CompanyR { .. }) = "hello " ++ (firstName $ person c) ++ " " ++ (lastName $ person c) 

Couldn't match expected type ‘ClientR -> PersonR’
                  with actual type ‘PersonR’
    • The function ‘person’ is applied to one argument,
      but its type ‘PersonR’ has none
      In the second argument of ‘($)’, namely ‘person c’
      In the first argument of ‘(++)’, namely ‘(firstName $ person c)’

    • Couldn't match expected type ‘ClientR -> PersonR’
                  with actual type ‘PersonR’
    • The function ‘person’ is applied to one argument,
      but its type ‘PersonR’ has none
      In the second argument of ‘($)’, namely ‘person c’
      In the second argument of ‘(++)’, namely ‘(lastName $ person c)’

greet2 :: ClientR -> String
greet2 IndividualR { person = PersonR { .. } } = "hi" ++ firstName ++ " " ++ lastName ++ " "

greet2 CompanyR { .. } = "hello " ++ firstName ++ " " ++ lastName ++ "who works as a " ++ duty ++ " " ++ clientRName + " "

greet2 CompanyR {person = PersonR { .. }, .. } = "hello " ++ firstName ++ " " ++ lastName ++ "who works as a " ++ duty ++ " " ++ clientRName ++ " "