F＃類重新排序Visual Studio

Question

Im建立一個示例應用程序，其中我的類型層次結構不適用於在Visual Studio中進行類型排序。無論我如何嘗試安排文件（上，下），我都無法定義所有的類。

所以他們的順序是F＃中的項目

type Artist() = 
    let mutable artistId = 0 
    let mutable name = String.Empty 

    member x.ArtistId 
     with get() = artistId 
     and set (value) = artistId <- value 

    member x.Name 
     with get() = name 
     and set (value) = name <- value 

type Genre() = 
    let mutable name = String.Empty 
    let mutable genreId = 0 
    let mutable description = String.Empty 
    let mutable albums = [new Album()] 

    member x.Name 
     with get() = name 
     and set (value) = name <- value 

    member x.GenreId 
     with get() = genreId 
     and set (value) = genreId <- value 

    member x.Description 
     with get() = description 
     and set (value) = description <- value 

    member x.Albums 
     with get() = albums 
     and set (value) = albums <- value 

and Album() = 
    let mutable title = String.Empty 
    let mutable genre = new Genre() 
    let mutable albumId = 0 
    let mutable genreId = 0 
    let mutable artistId = 0 
    let mutable price : decimal = Decimal.Zero 
    let mutable albumArtUrl = String.Empty 
    let mutable artist = new Artist() 

    member x.Title 
     with get() = title 
     and set (value) = title <- value 

    member x.Genre 
     with get() = genre 
     and set (value) = genre <- value 

    member x.AlbumId 
     with get() = albumId 
     and set (value) = albumId <- value 

    member x.GenreId 
     with get() = genreId 
     and set (value) = genreId <- value 

    member x.ArtistId 
     with get() = artistId 
     and set (value) = artistId <- value 

    member x.Price 
     with get() = price 
     and set (value) = price <- value 

    member x.AlbumArtUrl 
     with get() = albumArtUrl 
     and set (value) = albumArtUrl <- value 

    member x.Artist 
     with get() = artist 
     and set (value) = artist <- value 

因此，在上述情況下，我得到的錯誤「專輯」沒有定義。

有沒有辦法解決這個問題？或者我只需要重新考慮我的類型的整個層次結構？

Answer 1

如果您需要定義兩種類型，即相互遞歸（這意味着它們可以互相引用），那麼您需要將它們放在一個文件中並使用type ... and ...語法。

在你的榜樣，這意味着Genre和Album需要被這樣定義：

// Start a definition block using 'type' as normal 
type Genre() = 
    let mutable name = String.Empty 
    let mutable albums = [new Album()] 

    member x.Name 
     with get() = name 
     and set (value) = name <- value 
    member x.Albums 
     with get() = albums 
     and set (value) = albums <- value 

// Continue single type definition block using 'and' 
and Album() = 
    let mutable genre = new Genre() 
    let mutable albumId = 0 
    let mutable artist = new Artist() 

    member x.Genre 
     with get() = genre 
     and set (value) = genre <- value  
    member x.AlbumId 
     with get() = albumId 
     and set (value) = albumId <- value  
    member x.Artist 
     with get() = artist 
     and set (value) = artist <- value 

但是，你的例子是使用在一個非常C＃風格的F＃，因此代碼看上去真的很優雅，它可能不會給你許多功能編程的好處。

如果我想表示您正在使用的結構，那麼我可能不會將該類型的引用添加到Album類型中。當您在Genre中放置相冊列表時，在處理數據結構時（即將其轉換爲其他結構，可能是F＃記錄，可傳遞給數據綁定），您將始終能夠恢復流派， 。 F＃的特點是它可以讓你在幾行上編寫域，但只適用於函數類型。

使用與單一情況下識別聯合，你可以寫：

// Type aliases to make code more readable 
type Name = string 
type AlbumID = int 

// Simple type definitions to represent the domain 
type Artist = Artist of Name 
type Album = Album of AlbumID * Artist 
type Genre = Genre of Name * Album list