Re: Multiline textbox scrollbars cannot be moved - C# Discussion Boards

9-Jan-15 13:00

Hi,

I am outputting log style info into a multiline textbox. I have the vertical and horizontal scrollbar property enabled. At runtime when data is added to the multiline textbox, I can see the scrollbars but they cannot be moved.

Do I need to do something more to make them actually work?

Thanks,
Rob

Eddy Vluggen9-Jan-15 22:04

9-Jan-15 22:04

The scrollbars should work once there is more text in the textbox than it can show, when the application is running and the textbox'Enabled property is set to "true".

A disadvantage will be the speed of adding stuff to the textbox; consider using a listbox once you encounter that problem.

Bastard Programmer from Hell Suspicious | :suss:

If you can't read my code, try converting it here[^]

robwm112-Jan-15 7:39

12-Jan-15 7:39

I found that the textbox Enabled property was set to false. I set it to true and now the scrollbars work properly. I just assumed that it was enabled by default. I thought that was true for all controls but you have changed my mind!

Thank you!

Eddy Vluggen12-Jan-15 7:56

12-Jan-15 7:56

robwm1 wrote:
I just assumed that it was enabled by default

It is

Bastard Programmer from Hell Suspicious | :suss:

If you can't read my code, try converting it here[^]

robwm112-Jan-15 7:59

12-Jan-15 7:59

I must have changed it at some point. This is my first C# project and I have done a lot of experimentation. Just need to remember to undo any changes I make that don't give the results I am looking for so everything remains in its default state.

I am my own job security... Smile | :)

Eddy Vluggen12-Jan-15 10:14

12-Jan-15 10:14

Are you using a code-repository like SourceSafe where you can easily undo all changes?

I usually copy the complete project to a RAM-disk for experiments Smile | :)

Bastard Programmer from Hell Suspicious | :suss:

If you can't read my code, try converting it here[^]

robwm112-Jan-15 10:28

12-Jan-15 10:28

My code is stored within Team Foundation Server so I could roll back if needed. That's a great idea with the ram drive although using disk storage, if you have the space available, might work as well. Either way, I have been making changes as I progress to learn the language and IDE better without making copies. Probably not the best idea but I don't plan to code this way in the long term.

3 Dimensional Array C#

Member 113642179-Jan-15 10:35

9-Jan-15 10:35

I have a problem with my script:

Public partial class Form_Process : Form
{
   Static Object[ , , ] normW = null;
   //3 dimensional variable

   Private void process(){
       for(int i=0; i<3; i++){
          /*bla..bla..bla..
             //..............................
             //my process is here
             //..............................
         */
          //The result of the process
          //is in variable Normali[ , ].
          //Normali is a 2 dimensional array variable.

          //than..
          normalW[i] = Normali;  //this is where an error take place
       }
   }
}

Error: Wrong number of indices inside []; expected 3

I hope anyone can help me for my error script above.
I'm sorry for my bad english. Thank you.

modified 9-Jan-15 16:46pm.

PIEBALDconsult9-Jan-15 10:51

PIEBALDconsult

9-Jan-15 10:51

There are a number of problems, including that you never instantiate the array.

First off, why an array? Rather than a Tuple or a List?

Are you actually trying to make an array of three two-dimensional arrays?

Try
Object[,][] normalW = new Object[,][ 3 ] ;

(Untested)

(That ain't working.) Sigh | :sigh:

Member 1136421710-Jan-15 7:08

10-Jan-15 7:08

Yes, i wanna try to use 3 dimensional array, just for learning.
I also used a List. Thank's a lot.

Eddy Vluggen9-Jan-15 12:37

9-Jan-15 12:37

Is normW the same as normalW? If yes, then you declared it as a 3 dimensional variable.

If you solve it, you'll get the same error on Normali if that is indeed a 2 dimensional variable.

Bastard Programmer from Hell Suspicious | :suss:

If you can't read my code, try converting it here[^]

Member 1136421710-Jan-15 6:51

10-Jan-15 6:51

Yes i'm sorry. Thank you. NormW is same as NormalW. I try to save the Normali[i, j] (2 dimensional) in NormW[z, i, j] (3 dimensional)

BillWoodruff9-Jan-15 14:05

BillWoodruff

9-Jan-15 14:05

Sample 3-dimensional Array initialization:

int[,,] d3ary = new int[3, 4, 5];

for (int i = 0; i < d3ary.GetUpperBound(0); i++)
{
    for (int j = 0; j < d3ary.GetUpperBound(1); j++)
    {
        for (int k = 0; k < d3ary.GetUpperBound(2); k++)
        {
            d3ary[i, j, k] = // add a value
        }
    }
}

«A man will be imprisoned in a room with a door that's unlocked and opens inwards ... as long as it does not occur to him to pull rather than push» Wittgenstein

Member 1136421710-Jan-15 6:41

10-Jan-15 6:41

Yes, you're right. I used the code like yours. But i used d3ary.GetLength(0). Is that same with d3ary.GetUpperBound(0)?

BillWoodruff10-Jan-15 8:19

BillWoodruff

10-Jan-15 8:19

In C#, for all practical purposes, it is legitimate to consider the value returned by 'GetUpperBound to be the same as the value returned by GetLength. Normally in C#: Array.GetLowerBound(#n) will return zero.

It is possible, though unusual, to have Arrays in C# whose lower bounds are not zero; for example: you might inter-operate with Excel and get an Array whose indexes are one-based. Or you can create such strange arrays by using the Array.CreateInstance(Type, Int32[], Int32[]) overload of Array.CreateInstance.

Here's a one-based array of string of length #5:

Array oneBasedArray =  Array.CreateInstance(typeof(string), new int[] { 5 }, new int[] { 1 });

// force an error of type: An unhandled exception of type 'System.IndexOutOfRangeException' occurred in mscorlib.dll

// oneBasedArray.SetValue("wtf",0)

// this is okay ...
oneBasedArray.SetValue("hello", 1);

And:

> ? oneBasedArray.GetUpperBound(0)

// examine in Command Window:
5
> ? oneBasedArray.GetLength(0)
5
> ? oneBasedArray.GetLowerBound(0)
1

I suggest you engage with non-zero based Arrays as you need to in the course of inter-op with COM, or whatever.

«A man will be imprisoned in a room with a door that's unlocked and opens inwards ... as long as it does not occur to him to pull rather than push» Wittgenstein

Member 1136421711-Jan-15 16:10

11-Jan-15 16:10

Ok, I got the point. So, GetUpperBound and GetLowerBound are more flexible than using GetLength, aren't they?
They're used as we need.
Thank you BillWoodruff.

Richard Deeming12-Jan-15 2:03

Richard Deeming

12-Jan-15 2:03

If you're going to use GetUpperBound, then your loop condition needs to be <= rather than <.
GetUpperBound returns the last index for the dimension, not the length.

int[,,] d3ary = new int[3, 4, 5];

Console.WriteLine("{0}, {1}, {2}", d3ary.GetUpperBound(0), d3ary.GetUpperBound(1), d3ary.GetUpperBound(2));
// Prints: 2, 3, 4

Console.WriteLine("{0}, {1}, {2}", d3ary.GetLength(0), d3ary.GetLength(1), d3ary.GetLength(2));
// Prints: 3, 4, 5

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

OriginalGriff9-Jan-15 22:45

OriginalGriff

9-Jan-15 22:45

When you declare a three dimensional array, you are saying that each row, column and zaxis-whatever-it-is-called is the same size:

object[,,] ar3d = new object[2,3,4];

But that isn't necessarily the case with the code you show - we have no guarantee that each of the 2D arrays loaded into Normali is the same each time. And if we aren't sure, nor is the compiler!
So when you try to assign a new 2D array to a row, it complains because a 3D array is not made up of arrays of references to arrays of references to objects - it is in fact a flat linear space of objects which is interpreted as a 3D array.

Probably, what you are trying to do is construct a jagged array:

object[][,] normW = new object[3][,];
for (int i = 0; i < normW.Length; i++)
    {
    object[,] normali = MyProcess(i);
    normW[i] = normali;
    }

Where it doesn't matter what the dimensions of the array you generate in your processing are.

Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...

Member 1136421710-Jan-15 6:34

10-Jan-15 6:34

Ok, Thank you for your help. I got it.