这可能是一个愚蠢的问题,但在发布此问题之前,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
是用于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!
答案 0 :(得分:6)
您的类型
data Film = Film {title :: String
,name :: String
,year :: Int}
等同于元组类型(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+
仅适用于数字而非列表,我们会使用:
将新内容放入列表中,因此'!':"Hello"
会提供"!Hello"
,因此您需要
addFilm :: String -> String -> Int -> [Film] -> [Film]
addFilm title director year films
= Film {title=title,name=director,year=year}:films
(您应该将函数体与其类型声明对齐,但只要它缩进就可以开始一个新的行。)
答案 1 :(得分:3)
您需要使用电影构造函数:[Film "Blade Runner" "Ridley Scott" 1982]
。另外,我认为您需要:
而不是+
。 :
的参数也需要与当前的参数进行交换。