Question

This has been driving me crazy for a few days. Why doesn't the following work?

    Dim arr(3, 3) As Integer

    For y As Integer = 0 To arr.GetLength(0) - 1
        For x As Integer = 0 To arr.GetLength(y) - 1
            arr(y, x) = y + x
        Next
    Next

Also, what if the array looked like this instead?

{ {1, 2, 3},
  {4, 5, 6, 7, 8, 9, 9, 9},
  {5, 4, 3, 2}
}
Was it helpful?

Solution

Ok, so what you really need is a "jagged array". This will allow you to have an "array that contains other arrays of varying lengths".

  Dim arr As Integer()() = {New Integer() {1, 2, 3}, New Integer() {4, 5, 6, 7, 8, 9, 9, 9}, New Integer() {5, 4, 3, 2}}

  For x = 0 To arr.GetUpperBound(0)
      Console.WriteLine("Row " & x & " has " & arr(x).GetUpperBound(0) & " columns")
      For y = 0 To arr(x).GetUpperBound(0)
          Console.WriteLine("(" & x & "," & y & ") = " & arr(x)(y))
      Next
   Next

Output:

Row 0 has 2 columns
(0,0) = 1
(0,1) = 2
(0,2) = 3
Row 1 has 7 columns
(1,0) = 4
(1,1) = 5
(1,2) = 6
(1,3) = 7
(1,4) = 8
(1,5) = 9
(1,6) = 9
(1,7) = 9
Row 2 has 3 columns
(2,0) = 5
(2,1) = 4
(2,2) = 3
(2,3) = 2

OTHER TIPS

Because there is no '2' or '3' dimension. Should be .GetLength(1) instead of .GetLength(y)

Also: in VB.Net array declarations work a little differently. The subscript you specify in the declaration is the last index, not the number of items created like with C# or C++. But the array is still 0-indexed like C# or C++, instead of 1-indexed like VB6. That means that if you move to VB.Net from a different language your array instincts are probably wrong, no matter which language it is. In VB.Net, Dim arr(3,3) As Integer actually creates a 4x4 array.

arr.GetLength(y)

should be

arr.GetLength(1)

Well what if I had an array that looked like this

{ {1, 2, 3},
  {4, 5, 6, 7, 8, 9, 9, 9},
  {5, 4, 3, 2}
}

How would GetLength(1) still know the length of each row?


Basically what I want is.... a way to find the number of elements in any given row.

Dim arr(3, 3) As Integer
Dim y As Integer
Dim x As Integer

For x = 0 To arr.Rank - 1
    For y = 0 To arr.GetLength(x) - 2
        arr(x, y) = x + y
    Next
Next

The above code worked for me.

Edit, the code feels dirty though. I'm wondering what it is you are trying to accomplish?

This code en C# is to get all the combinations of items in a jagged array:

    static void Main(string[] args)
    {
        bool exit = false;
        int[] indices = new int[3] { 0, 0, 0 };
        string[][] vectores = new string[3][];

        vectores[0] = new string[] { "A", "B", "C" };
        vectores[1] = new string[] { "A", "B" };
        vectores[2] = new string[] { "B", "D", "E", "F" };

        string[] item;
        int[] tamaños = new int[3]{vectores[0].GetUpperBound(0), 
            vectores[1].GetUpperBound(0), 
            vectores[2].GetUpperBound(0)};

        while (!exit)
        {
            item = new string[]{ vectores[0][indices[0]],
                    vectores[1][indices[1]],
                    vectores[2][indices[2]]};

            Console.WriteLine("[{0},{1},{2}]={3}{4}{5}", indices[0], indices[1], indices[2], item[0], item[1], item[2]);
            GetVector(tamaños, ref indices, ref exit);
        }
        Console.ReadKey();
    }

    public static void GetVector(int[] tamaños, ref int[] indices, ref bool exit)
    {
        for (int i = tamaños.GetUpperBound(0); i >= 0; i--)
        {
            if (tamaños[i] > indices[i])
            {
                indices[i]++;
                break;
            }
            else
            {
                //ULTIMO ITEM EN EL ARRAY, VALIDAR LAS OTRAS DIMENSIONES SI YA ESTA EN EL ULTIMO ITEM
                if (!ValidateIndexes(tamaños, indices))
                    indices[i] = 0;
                else
                {
                    exit = true;
                    break;
                }
            }
        }
    }

    public static bool ValidateIndexes(int[] tamaños, int[] indices)
    {
        for (int i = 0; i < tamaños.Length; i++)
        {
            if (tamaños[i] != indices[i])
                return false;
        }
        return true;
    }

The output looks like [0,0,0]=AAB [0,0,1]=AAD [0,0,2]=AAE [0,0,3]=AAF [0,1,0]=ABB [0,1,1]=ABD [0,1,2]=ABE [0,1,3]=ABF [1,0,0]=BAB [1,0,1]=BAD [1,0,2]=BAE [1,0,3]=BAF [1,1,0]=BBB [1,1,1]=BBD [1,1,2]=BBE [1,1,3]=BBF [2,0,0]=CAB [2,0,1]=CAD [2,0,2]=CAE [2,0,3]=CAF [2,1,0]=CBB [2,1,1]=CBD [2,1,2]=CBE [2,1,3]=CBF

Your declaration: DIM arr(3,3) As Integer allready specifies that there are 3 elements in any given row (or 4, I'm not so sure about VB)

You could try:

Dim arr(3) as Integer()

You should then be able to do:

arr(n).Length

To find the length of row n.

I'm a bit rusty on VB6 and never learned VB.NET, but this should give you a 'jagged' array. Check out the msdn documentation on multidimensioned arrays.

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