1

我正在尝试将以下 C# 转换为 F#：

    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;
         }
       }
     }

这是我到目前为止所拥有的：

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]

我上面的类型似乎有两个问题我不知道如何解决。第一个是矩阵。[(x, y)] 应该具有类型 `a[] 但具有类型 double[,]。第二个是类型定义必须在成员和接口定义之前有 let/do 绑定。问题是我试图在 do 块中填充索引属性，这意味着我必须先创建它。

提前致谢，

鲍勃

4

2 回答 2

6

关于您的第一个问题，您想使用matrix.[x,y]而不是matrix.[(x,y)]- 您的矩阵由两个整数索引，而不是由整数元组索引（尽管这些在概念上相似）。

这是大致相当于您的 C# 的内容：

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

这假设您的矩阵实际上不能被重新分配（例如，在您发布的 C# 中，您可以创建您的matrix字段readonly- 除非您隐藏了其他代码）。因此，行数和列数可以在构造函数中计算一次，因为矩阵的条目可能会改变，但它的大小不会改变。

但是，如果您想要更直接地翻译代码，可以为新构建的实例命名（this在这种情况下）：

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
于 2011-03-05T21:44:56.693 回答
2 
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
于 2011-03-05T21:45:37.397 回答