在類型使用F＃索引屬性

我想下面的C＃轉換爲F＃：在類型使用F＃索引屬性

public class Matrix 
    { 
     double[,] matrix; 

public int Cols 
     { 
      get 
      { 
       return this.matrix.GetUpperBound(1) + 1; 
      } 
     } 

public int Rows 
     { 
      get 
      { 
       return this.matrix.GetUpperBound(0) + 1; 
      } 
     } 

     public Matrix(double[,] sourceMatrix) 
     { 
     this.matrix = new double[sourceMatrix.GetUpperBound(0) + 1, sourceMatrix.GetUpperBound(1) + 1]; 
     for (int r = 0; r < this.Rows; r++) 
     { 
      for (int c = 0; c < this.Cols; c++) 
      { 
       this[r, c] = sourceMatrix[r, c]; 
      } 
     } 
     } 

     public double this[int row, int col] 
     { 
     get 
     { 
      return this.matrix[row, col]; 
     } 
     set 
     { 
      this.matrix[row, col] = value; 
     } 
     } 
    }

這是我到目前爲止有：以上

type Matrix(sourceMatrix:double[,]) = 
let mutable (matrix:double[,]) = Array2D.create (sourceMatrix.GetUpperBound(0) + 1) (sourceMatrix.GetUpperBound(1) + 1) 0.0 
member this.Item 
    with get(x, y) = matrix.[(x, y)] 
    and set(x, y) value = matrix.[(x, y)] <- value 
do 
    for i = 0 to matrix.[i].Length - 1 do 
    for j = (i + 1) to matrix.[j].Length - 1 do 
     this.[i].[j] = matrix.[i].[j]

我的類型似乎有兩個問題我不知道如何解決。第一個是矩陣。[（x，y）]預計有類型`a []但是類型爲double [，]。第二種是類型定義在成員和接口定義之前必須有let/do綁定。這個問題是我試圖填充do塊中的索引屬性，這意味着我必須先創建它。

由於提前，

鮑勃

來源

2011-03-05 Beaker

關於你的第一個問題，你想用matrix.[x,y]而不是matrix.[(x,y)] - 你的矩陣是由兩個整數索引，而不是整數的元組（雖然這些概念類似）。

這裏的東西大致相當於你的C＃：

type Matrix(sourceMatrix:double[,]) = 
    let rows = sourceMatrix.GetUpperBound(0) + 1 
    let cols = sourceMatrix.GetUpperBound(1) + 1 
    let matrix = Array2D.zeroCreate<double> rows cols 
    do 
    for i in 0 .. rows - 1 do 
    for j in 0 .. cols - 1 do 
     matrix.[i,j] <- sourceMatrix.[i,j] 
    member this.Rows = rows 
    member this.Cols = cols 
    member this.Item 
    with get(x, y) = matrix.[x, y] 
    and set(x, y) value = matrix.[x, y] <- value

這是假設你的矩陣實際上不能重新分配（例如，在C＃中你張貼，你可以做出你matrix場readonly - 除非還有其他隱藏的代碼）。因此，行和列的數量可以在構造函數中計算一次，因爲矩陣的條目可能會更改，但其大小不會更改。

但是，如果你希望你的代碼更直譯，你可以給你新建成的情況下（在這種情況下this）名稱：

type Matrix(sourceMatrix:double[,]) as this = 
    let mutable matrix = Array2D.zeroCreate<double> (sourceMatrix.GetUpperBound(0) + 1) (sourceMatrix.GetUpperBound(1) + 1) 
    do 
    for i in 0 .. this.Rows - 1 do 
    for j in 0 .. this.Cols - 1 do 
     this.[i,j] <- sourceMatrix.[i,j] 
    member this.Rows = matrix.GetUpperBound(0) + 1 
    member this.Cols = matrix.GetUpperBound(1) + 1 
    member this.Item 
    with get(x, y) = matrix.[x, y] 
    and set(x, y) value = matrix.[x, y] <- value

來源

2011-03-05 21:44:56 kvb

對不起，我只是編輯它，粘貼在錯誤的構造函數中。 – Beaker 2011-03-05 21:48:20

優秀的答案，以及只讀的這是你的一個正確的假設。原來的C＃不是我的，我忽略了這一點。對C＃和F＃都很好。 :) – Beaker 2011-03-05 23:35:46

如果有人感興趣，我添加了這個問題的後續問題：http://stackoverflow.com/questions/5212570/adding-overloaded-constructors-to-implicit-f-type – Beaker 2011-03-06 19:07:13

type Matrix(sourceMatrix:double[,]) = 
    let matrix = Array2D.copy sourceMatrix 
    member this.Item 
     with get(x, y) = matrix.[x, y] 
     and set(x, y) value = matrix.[x, y] <- value

來源

2011-03-05 21:45:37 desco

在類型使用F＃索引屬性

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