Why I can't make array as class? Vb.net

Question

1.00/5 (1 vote)

See more:

Hi guys
I am trying to make an array as a class. But no luck.
Anybody have idea why this code returns exception and doesn't run?

What I have tried:

VB.NET

Class MainWindow
    Sub New()
        InitializeComponent()
        Dim MyTest(1) As Employee
        MyTest(1).ID = 1
        MyTest(1).Name = "John"
        MsgBox(MyTest(1).ID)
    End Sub
    Public Class Employee
        Public Property ID As Long
        Public Property Name As String
    End Class
End Class

Posted 2-Jul-23 9:49am

Sh.H.

Updated 2-Jul-23 14:19pm

v6

Add a Solution

Comments

Dave Kreskowiak 2-Jul-23 18:26pm

What do you mean by "array as a class"?

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Graeme_Grant · Accepted Answer · 2023-07-02T14:19:00

Solution 2

Arrays in both VB & C# are zero-based, not one-based with .Net Framework and .Net (Core). ref: Arrays - Visual Basic | Microsoft Learn[^].

Also, as pointed out by Ralf, if the class is not static, a Module in VB.Net, then you need to initialise it before using it.

So you need to change to:

VB

Dim MyTest(1) As Employee
MyTest(0) = New Employee() 'initialise before use
MyTest(0).ID = 1
MyTest(0).Name = "John"
MsgBox(MyTest(0).ID)

VB.Net also has some nice syntactical sugar, for readability, when working with setting properties or calling methods (subroutines & functions) on a class using the With[^] syntax:

VB

Dim MyTest(1) As Employee

MyTest(0) = New Employee()
With MyTest(0)
    .ID = 1
    .Name = "John"
End With

MsgBox(MyTest(0).ID)

NOTE: This is only in VB.Net and not supported in other languages.

Posted 2-Jul-23 14:19pm

Graeme_Grant

Updated 2-Jul-23 14:29pm

v3

Comments

Ralf Meier 3-Jul-23 2:35am

where is the difference between your suggestion_1 and my suggestion_2 ?
after dimensioning the Array to 1 it consists of 2 elements (Index 0 and Index 1) - I don't know if that is equal to C# ... but at VB it happens like described ...

Graeme_Grant 3-Jul-23 3:37am

You change from Array to List. Still have an out of bound issue for your second example with array size of 1. Yes, I too tested. ;)

Ralf Meier 3-Jul-23 3:40am

No ... of course List of is also an suggestion from me ... but please read my solution and the added parts carefully ...
There is a difference in dimensioning Arrays ...

Ralf Meier 3-Jul-23 3:42am

my question now : why did my code sample works by me and not by you ? Where is the difference ?

Graeme_Grant 3-Jul-23 3:45am

You did not answer the question but gave a different solution. You then gavea second solution with an error. Then the last solution was a correction after I called out your error. My solution was correct from the first posting. I added a bonus method of working with class properties and methods, as I too was once a VB programmer. I still offer VB code with articles that I write. ;)

Love the down vote Ralf ... very nice of you ;)

Ralf Meier 3-Jul-23 3:54am

9the oppinions are different as I see - I think that either my solution (and the answer) is useful and also all additions to it.
Of course your solution is correct ... but mine isn't incorrect ... and your solution (part 1) is a copy of mine ... and I don't gave you a downvote - I only voted +3 (which was your vote to me ?)

Graeme_Grant 3-Jul-23 3:58am

Part one was not a copy of yours, you used a list, you can see in you post version history. I just fixed the OPs code.

Ralf Meier 3-Jul-23 4:18am

if you speak about version-history you should also look at which time came which suggestions - especially at me ...

Ralf Meier 3-Jul-23 3:56am

and finaly :
I'm very sad that your don't replied to my comments to make it to a real discussion - you only posted comments ... without any corresponding to mine ...

Graeme_Grant 3-Jul-23 3:59am

Every solution has a history, and you can see exactly what was posted and when. So look back and you will see.

Ralf Meier · Accepted Answer · 2023-07-02T12:30:00

Solution 1

A Class is an Object which must be first instanced before using it.
In your code you don`t have such an Instance from Employee - only an Object-Declaration.

Instead of that you could use the following code :

VB.NET

Sub New()
    InitializeComponent()
    Dim MyTest As New List(Of Employee)
    Dim x As New Employee
    x.ID = 1
    x.Name = "John"
    MyTest.Add(x)
    MsgBox(MyTest(0).ID)
End Sub

Public Class Employee
    Public Property ID As Long
    Public Property Name As String
End Class

or like that :

VB

Sub New()
    InitializeComponent()
    Dim MyTest(1) As Employee
    MyTest(1) = New Employee
    MyTest(1).ID = 1
    MyTest(1).Name = "John"
    MsgBox(MyTest(1).ID)
End Sub

completed after discussion with Grame_Grant :

VB

Sub New()
    InitializeComponent()
    Dim MyTest(1) As Employee
    MyTest(0) = New Employee
    MyTest(0).ID = 1
    MyTest(0).Name = "John"

    MyTest(1) = New Employee
    MyTest(1).ID = 2
    MyTest(1).Name = "Jack"

    MsgBox(MyTest(0).ID.ToString + " " + MyTest(0).Name + vbCrLf + MyTest(1).ID.ToString + " " + MyTest(1).Name)
End Sub

and if I add the following I will get the decribed error / exception :

VB

MyTest(2) = New Employee
MyTest(2).ID = 3
MyTest(2).Name = "Jim"

Posted 2-Jul-23 12:30pm

Ralf Meier

Updated 2-Jul-23 21:41pm

v6

Comments

Graeme_Grant 2-Jul-23 20:14pm

"A Class is an Object which must be first instanced before using it." only true if the class is not static. If a method of a class is public and static, then the class also does not need to have an instance generated to call that method.

Also, your code example will fail. Read my first sentence in my solution, I think that some missed that detail.

Ralf Meier 3-Jul-23 2:29am

OK ... you are right ... but if you make a Class Static you create an Instance of it ...
Where do you think thta my code-sample will fail ? I have tested both versions ...

Graeme_Grant 3-Jul-23 3:08am

A collection size of (1) = 1 element. Array/collection is base 0. So accessing element(1) = 2nd element, not the first, which is element(0), is out of bounds. So if you run your code you will see the error message:
System.NullReferenceException: 'Object reference not set to an instance of an object.'

Ralf Meier 3-Jul-23 3:25am

No ... as I allready wrote - I tested it and it works ...
See also my comment to your Solution ...

Ralf Meier 3-Jul-23 3:34am

I made also an addition to my Solution as an add to our discussion ...

Graeme_Grant 3-Jul-23 3:38am

Dim MyTest(1) As Employee

is an array size of 1, not 2 items. VB6 was base 1, not base 0 by default.

Ralf Meier 3-Jul-23 4:08am

So please answer :
why does my code-part 3 work ? I get the execption only with code-part 4 ...
If I create an Array with the Dimension 1 my Array has 2 Elements (Index 0 and Index 1) ...

Richard MacCutchan 3-Jul-23 4:06am

Your second and third sample code blocks are both incorrect. You declare an array of one element and then try to access the second one. While the code may not fail immediately you are overwriting unallocated memory.
Dim x(1) actually does allocate two elements. I should know that VB has its own set of standards.

Ralf Meier 3-Jul-23 4:10am

as I wrote to Grame_Grant - part 3 doesn't fail ... and the dimnsioned Array (1) has 2 Elements ...

Richard MacCutchan 3-Jul-23 5:10am

I apologise, my VB skills need sharpening.

Ralf Meier 3-Jul-23 5:15am

you are welcome - and thanks a lot ...

Graeme_Grant 3-Jul-23 4:16am

Yeah, totally. Declaring an array size of one is only one element, not two.

Ralf Meier 3-Jul-23 4:25am

this discussion goes to nowhere - inside my Visual-Studio an Array(1) has 2 Elements ... and my code-part 3 doesn't fail.
That has nothing in common with unallocated memory as mentioned by Richard - it simply works and as I remember that was every time like this.
If you create an Array (10) it is the same - it has 11 Elements and I can access the Index 10.
If wwe speak about Collections or Lists there is a Difference ...
So ... to finalize it : if no one can explain me why my code works (and my development system) I'm out of this discussion.
And ... also thanks for downvoting (to 1 !!!) my Solution ... very very nice ...

Richard MacCutchan 3-Jul-23 5:13am

I have upvoted to compensate.

Ralf Meier 3-Jul-23 5:03am

I must add a last comment :
In your Solution you added a Link for explaining Arrays. When you take a look at that Description there you will see that my Explaining of Array-Dimensioning was complete right. An Array (4) consists of 5 Elements (0 .. 4).
This only for your Info ...

Ralf Meier 3-Jul-23 5:04am

I must add a last comment :
Inside the Solution of Grame_Grant was added a Link for explaining Arrays. When you take a look at that Description there you will see that my Explaining of Array-Dimensioning was complete right. An Array (4) consists of 5 Elements (0 .. 4).
This only for your Info ...