haskell定义新类型

Question

这可能是一个愚蠢的问题,但在我发布之前,我花了四个小时来指出问题所在.

data Film = Film {title :: String
                ,name :: String
                ,year :: Int}
    deriving (Show)
testDatabase :: [Film]
 testDatabase = [ ("Blade Runner", "Ridley Scott",1982)]
 --(i) Add new film to the database
addFilm :: String -> String -> Int -> [Film] -> [Film]
 addFilm title director year film = film + Film title director year 

 --(ii) Give all film in the database
getFilm :: [Film]
getFilm =  testDatabase

我想要做的是定义一个新的类型名称Film,其中包含:电影片名,电影导演和制作年份.
Testdatabase用于存储数据.
addFilm是一个向数据库添加更多电影的功能.
getfilm用于打印出电影列表.

这就是错误的样子.

coursework.hs:21:18:
Couldn't match expected type `Film'
            with actual type `([Char], [Char], Integer)'
In the expression: ("Blade Runner", "Ridley Scott", 1982)
In the expression: [("Blade Runner", "Ridley Scott", 1982)]
In an equation for `testDatabase':
    testDatabase = [("Blade Runner", "Ridley Scott", 1982)]

coursework.hs:24:43:
Couldn't match expected type `[Film]' with actual type `Film'
In the return type of a call of `Film'
In the second argument of `(+)', namely `Film title director year'
In the expression: film + Film title director year
Failed, modules loaded: none.

谢谢!!

Answer 1

你的类型

data Film = Film {title :: String
                ,name :: String
                ,year :: Int}

等价于元组类型(String,String,Int),但与它不同.

("Blade Runner", "Ridley Scott",1982) :: (String,String,Int)

但你想要

Film {title = "Blade Runner", name="Ridley Scott",year=1982} :: Film

你写了

addFilm :: String -> String -> Int -> [Film] -> [Film]
 addFilm title director year film = film + Film title director year

但+只适用于数字,而不是列表,我们在列表中使用新的东西:,所以'!':"Hello"给出"!Hello",所以你需要

addFilm :: String -> String -> Int -> [Film] -> [Film]
addFilm title director year films  
    = Film {title=title,name=director,year=year}:films

(你应该将函数体与它的类型声明排成一行,但只要缩进就可以开始一个新的部分.)