不在范围内：数据构造函数“Song”- Haskell

Question

type Song = (String, String, Int) --(title, artist, sales)

database :: [Song]
database = [("Amon Amarth","Ravens flight", 1), 
  ("Amon Amarth","Shield wall", 11),
  ("Amon Amarth","The way of vikings", 105),
  ("Elijah Nang","Journey to the west", 1000),
  ("Elijah Nang","Tea house", 7),
  ("Pink Floyd","Wish you were here", 123),
  ("Amon Amarth","Raise your horns", 9001),
  ("NLE Choppa","Walk 'em down'", 69420),
  ("Elijah Nang","Kumite", 1337),
  ("NLE Choppa","Shotta flow 6", 511),
  ("Pink Floyd","Comfortably numb", 9),
  ("Pink Floyd","Shotta flow 6", 711), -- changed to match the name of an nle choppa song as requested
  ("Johannes Chrysostomus Wolfgangus Theophilus Mozart","Requiem", 10203948),
  ("Elijah Nang","Kenjutsu water style", 1),
  ("NLE Choppa","Shotta flow 5", 1),
  ("Pink Floyd","High hopes", 1),
  ("Amon Amarth","Deceiver of the gods", 1),
  ("Johannes Chrysostomus Wolfgangus Theophilus Mozart","Turkish march", 1),
  ("Chance The Rapper","Cocoa butter kisses", 1),
  ("Chance The Rapper","Favourite song", 1),
  ("Chance The Rapper","Hot shower", 1),
  ("Chance The Rapper","High hopes", 1)] 


getTrackSale :: Int -> String -> String -> String --(index, artist, track, sales)
getTrackSale index artist track
  | ((getArtist(database!!index) == artist) && (getTrack(database!!index) == track)) = getTrackSale(database!!index)
  | otherwise = getTrackSale(index + 1 artist track)


task2 = getTrackSale(0 "Chance The Rapper" "Hot Shower")

getArtist :: Song -> String
getArtist (Song y _ _) = y

getTrack :: Song -> String
getTrack (Song _ z _) = z

getSale :: Song -> Int
getSale (Song _ _ x) = x

我不明白这是什么意思或如何解决它，我之前已经为三个“get”函数编写了相同的函数并且它们没有问题，但我之前确实使用了“type”声明，所以我想就是这样。我有使用相同类型声明的示例代码，它工作正常，所以我在这里有点迷路。

Answer 1

你写

type Song = (String, String, Int)

这意味着 Song 是 (String, String, Int) 的 类型同义词 。也就是说，Song 实际上只是一个有序三元组，使用有序三元组构造函数 (,,) 构造。但是后来你写

getArtist :: Song -> String
getArtist (Song y _ _) = y

getTrack :: Song -> String
getTrack (Song _ z _) = z

getSale :: Song -> Int
getSale (Song _ _ x) = x

（不存在的）Song 数据构造函数上的模式匹配。

你可以通过写

来坚持类型同义词

getArtist :: Song -> String
getArtist (y, _, _) = y

等等。或者你可以让 Song 成为它自己的数据类型：

data Song = Song
  { getArtist :: String
  , getTrack :: String
  , getSale :: Int }

在这种情况下，您必须修改 database 定义以使用 Song 构造函数。使用自定义数据类型声明更加惯用且更加自文档化，它使 Haskell 的类型系统可以做更多的工作来帮助您查找代码中的错误。作为一般规则，我建议 Haskell 初学者完全避免类型同义词，更有经验的 Haskell 程序员在某些特殊情况下很少使用它们。特别是，类型同义词在与 TypeFamilies 扩展结合使用时非常有用，在与 ConstraintKinds 和 RankNTypes 扩展结合使用时也有些用处，但除此之外往往更令人困惑而不是有用，在我的经验。

不在范围内：数据构造函数“Song”- Haskell

Not in scope: data constructor `Song' - Haskell

database

constructor

haskell

types

scope