How to declare a fixed-length string in VB.NET? - vb.net

Question

How do i Declare a string like this:
Dim strBuff As String * 256
in VB.NET?

Answer 1

Use the VBFixedString attribute. See the MSDN info here
<VBFixedString(256)>Dim strBuff As String

Answer 2

It depends on what you intend to use the string for. If you are using it for file input and output, you might want to use a byte array to avoid encoding problems. In vb.net, A 256-character string may be more than 256 bytes.
Dim strBuff(256) as byte
You can use encoding to transfer from bytes to a string
Dim s As String
Dim b(256) As Byte
Dim enc As New System.Text.UTF8Encoding
...
s = enc.GetString(b)
You can assign 256 single-byte characters to a string if you need to use it to receive data, but the parameter passing may be different in vb.net than vb6.
s = New String(" ", 256)
Also, you can use vbFixedString. I'm not sure exactly what this does, however, because when you assign a string of different length to a variable declared this way, it becomes the new length.
<VBFixedString(6)> Public s As String
s = "1234567890" ' len(s) is now 10

Answer 3

To write this VB 6 code:
Dim strBuff As String * 256
In VB.Net you can use something like:
Dim strBuff(256) As Char

Answer 4

Use stringbuilder
'Declaration
Dim S As New System.Text.StringBuilder(256, 256)
'Adding text
S.append("abc")
'Reading text
S.tostring

Answer 5

Try this:
Dim strbuf As New String("A", 80)
Creates a 80 character string filled with "AAA...."'s
Here I read a 80 character string from a binary file:
FileGet(1,strbuf)
reads 80 characters into strbuf...

Answer 6

You can use Microsoft.VisualBasic.Compatibility:
Imports Microsoft.VisualBasic.Compatibility
Dim strBuff As New VB6.FixedLengthString(256)
But it's marked as obsolete and specifically not supported for 64-bit processes, so write your own that replicates the functionality, which is to truncate on setting long values and padding right with spaces for short values. It also sets an "uninitialised" value, like above, to nulls.
Sample code from LinqPad (which I can't get to allow Imports Microsoft.VisualBasic.Compatibility I think because it is marked obsolete, but I have no proof of that):
Imports Microsoft.VisualBasic.Compatibility
Dim U As New VB6.FixedLengthString(5)
Dim S As New VB6.FixedLengthString(5, "Test")
Dim L As New VB6.FixedLengthString(5, "Testing")
Dim p0 As Func(Of String, String) = Function(st) """" & st.Replace(ChrW(0), "\0") & """"
p0(U.Value).Dump()
p0(S.Value).Dump()
p0(L.Value).Dump()
U.Value = "Test"
p0(U.Value).Dump()
U.Value = "Testing"
p0(U.Value).Dump()
which has this output:
"\0\0\0\0\0"
"Test "
"Testi"
"Test "
"Testi"

Answer 7

This object can be defined as a structure with one constructor and two properties.
Public Structure FixedLengthString
Dim mValue As String
Dim mSize As Short
Public Sub New(Size As Integer)
mSize = Size
mValue = New String(" ", mSize)
End Sub
Public Property Value As String
Get
Value = mValue
End Get
Set(value As String)
If value.Length < mSize Then
mValue = value & New String(" ", mSize - value.Length)
Else
mValue = value.Substring(0, mSize)
End If
End Set
End Property
End Structure
https://jdiazo.wordpress.com/2012/01/12/getting-rid-of-vb6-compatibility-references/

Answer 8

Have you tried
Dim strBuff as String
Also see Working with Strings in .NET using VB.NET
This tutorial explains how to
represent strings in .NET using VB.NET
and how to work with them with the
help of .NET class library classes.

Answer 9

Dim a as string
a = ...
If a.length > theLength then
a = Mid(a, 1, theLength)
End If

Answer 10

This hasn't been fully tested, but here's a class to solve this problem:
''' <summary>
''' Represents a <see cref="String" /> with a minimum
''' and maximum length.
''' </summary>
Public Class BoundedString
Private mstrValue As String
''' <summary>
''' The contents of this <see cref="BoundedString" />
''' </summary>
Public Property Value() As String
Get
Return mstrValue
End Get
Set(value As String)
If value.Length < MinLength Then
Throw New ArgumentException(String.Format("Provided string {0} of length {1} contains less " &
"characters than the minimum allowed length {2}.",
value, value.Length, MinLength))
End If
If value.Length > MaxLength Then
Throw New ArgumentException(String.Format("Provided string {0} of length {1} contains more " &
"characters than the maximum allowed length {2}.",
value, value.Length, MaxLength))
End If
If Not AllowNull AndAlso value Is Nothing Then
Throw New ArgumentNullException(String.Format("Provided string {0} is null, and null values " &
"are not allowed.", value))
End If
mstrValue = value
End Set
End Property
Private mintMinLength As Integer
''' <summary>
''' The minimum number of characters in this <see cref="BoundedString" />.
''' </summary>
Public Property MinLength() As Integer
Get
Return mintMinLength
End Get
Private Set(value As Integer)
mintMinLength = value
End Set
End Property
Private mintMaxLength As Integer
''' <summary>
''' The maximum number of characters in this <see cref="BoundedString" />.
''' </summary>
Public Property MaxLength As Integer
Get
Return mintMaxLength
End Get
Private Set(value As Integer)
mintMaxLength = value
End Set
End Property
Private mblnAllowNull As Boolean
''' <summary>
''' Whether or not this <see cref="BoundedString" /> can represent a null value.
''' </summary>
Public Property AllowNull As Boolean
Get
Return mblnAllowNull
End Get
Private Set(value As Boolean)
mblnAllowNull = value
End Set
End Property
Public Sub New(ByVal strValue As String,
ByVal intMaxLength As Integer)
MinLength = 0
MaxLength = intMaxLength
AllowNull = False
Value = strValue
End Sub
Public Sub New(ByVal strValue As String,
ByVal intMinLength As Integer,
ByVal intMaxLength As Integer)
MinLength = intMinLength
MaxLength = intMaxLength
AllowNull = False
Value = strValue
End Sub
Public Sub New(ByVal strValue As String,
ByVal intMinLength As Integer,
ByVal intMaxLength As Integer,
ByVal blnAllowNull As Boolean)
MinLength = intMinLength
MaxLength = intMaxLength
AllowNull = blnAllowNull
Value = strValue
End Sub
End Class