Have I corrupted the string in the code example?

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5.00/5 (1 vote)

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Hi All,

First off have I corrupted a string value in this example:

VB

 Dim CommandB As String = ""
'do stuff not related 
' a lot of stuff not related
  Do Until RecString(ICount) = "EOF"
                ICount += 1
                If RecString(ICount) = "EOF" Then GoTo EndofCommandLoop  'Yes that's a Goto No I'm not responcible!
                CommandB(BCount) = RecString(ICount)
                BCount += 1
            Loop
EndofCommandLoop:
            'Take actions

Ah, I spit on the muppet that doodled (not wrote) this mess

VB in it infinite wisdom has decided not to compile this, it returns

Quote:
Error 1 Property 'Chars' is 'ReadOnly'.

with a blue wiggly line, however it compiles this version

VB

 Dim CommandB As String
'do stuff not related 
' a lot of stuff not related
  Do Until RecString(ICount) = "EOF"
                ICount += 1
                If RecString(ICount) = "EOF" Then GoTo EndofCommandLoop
                CommandB(BCount) = RecString(ICount)
                BCount += 1
            Loop
EndofCommandLoop:
            'Take actions

Works with a little green wiggly line under CommandB(BCount) as the following

Quote:
warning BC42104: Variable 'CommandB' is used before it has been assigned a value. A null reference exception could result at runtime.

Why, can anyone help!

Posted 31-Jan-14 0:04am

glennPattonWork3

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4 solutions

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Sergey Vaselenko · Answer 1 · 2014-01-31T00:17:00

Solution 2

Use the StringBuilder class like this:

VB

Dim sb As New System.Text.StringBuilder
'do stuff not related
' a lot of stuff not related
Do Until RecString(ICount) = "EOF"
    ICount += 1
    If RecString(ICount) = "EOF" Then GoTo EndofCommandLoop
    sb.Append(RecString(ICount))
    BCount += 1
Loop
EndofCommandLoop:
'Take actions
Dim CommandB As String = sb.ToString

Posted 31-Jan-14 0:17am

Sergey Vaselenko

Comments

glennPattonWork3 31-Jan-14 6:28am

I wanted to use StringBuilder (I wanted to rewrite it in C# but wasn't allowed) so I might do that... or a mix or your & Pete's answer.

Pete O'Hanlon · Answer 2 · 2014-01-31T00:15:00

Okay, you can fix the second version by doing:

VB

Dim CommandB As String = String.Empty

However, it looks like you are just tring to increment a string there so try changing it to this:

VB

Dim CommandB As StringBuilder = New StringBuilder()
Do Until RectString(ICount) = EofCharacter
  ICount += 1
  If RecString(ICount) = EofCharacter Break
  CommandB.Append(RecString(ICount))

In your example, you are comparing a character to a string "EOF" is a string, so you need to convert the EOF marker into a character.

TnTinMn · Answer 3 · 2014-01-31T13:55:00

This looks like some perverted form of incremental string splitting on a EOF character.
If I am wrong please ignore the rest of this.

VB

'*** assuming these are defined outside of the splitting method
   Const EOF As Char = Chr(26)

   Dim RecString As String = "sequence 1" & EOF & "sequence 2" & EOF & "sequence 3" & EOF
   Dim RecPosition As Int32 = 0 ' This was your ICount variable

'***

   ' the grunt work of find EOF marker and grabbing a substring
   Dim EOFposition As Int32
   Dim CommandB As String

   If RecPosition < RecString.Length Then
      EOFposition = RecString.IndexOf(value:=EOF, startIndex:=RecPosition)
      If EOFposition <> -1 Then
         If RecPosition = EOFposition Then
            CommandB = String.Empty
         Else
            CommandB = RecString.Substring(startIndex:=RecPosition, length:=(EOFposition - RecPosition))
         End If
      Else
         CommandB = RecString.Substring(startIndex:=RecPosition)
      End If
      RecPosition += CommandB.Length + 1 ' start of next sequence
   End If

Dave Kreskowiak · Answer 4 · 2014-01-31T16:28:00

Solution 4

The reason why it says "Chars is ReadOnly" is because, in .NET, once created, a String object is immutable. You cannot change it the contents of the string.

In your case, the StringBuilder suggestions would be the better way to go.

Posted 31-Jan-14 16:28pm

Dave Kreskowiak

Have I corrupted the string in the code example?

4 solutions

Solution 2

Solution 1

Solution 3

Solution 4

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