lambda expressions in vb.net

Question

I have something that is driving me absolutely crazy...

    Public Function GetAccountGroups() As IList(Of AccountGroup)
        Dim raw_account_groups As IList(Of AccountGroup)
        raw_account_groups = _repository.GetAccountGroups().ToList()
        Dim parents = (From ag In raw_account_groups _
                      Where ag.parent_id = 0 _
                      Select ag).ToList()

        parents(0).sub_account_groups = (From sag In raw_account_groups _
                               Where sag.parent_id = 0 _
                                Select sag).ToList()

        Dim sql_func As Func(Of AccountGroup, List(Of AccountGroup)) = Function(p) _
                                                                      (From sag In raw_account_groups _
                                                                       Where sag.parent_id = p.id _
                                                                       Select sag).ToList()

        parents.ForEach(Function(p) p.sub_account_groups = sql_func(p))

        Return parents
    End Function

The line parents.ForEach(Function(p) p.sub_account_groups = sql_func(p)) has this error...

Operator '=' is not defined for types 'System.Collections.Generic.IList(Of st.data.AccountGroup)' and 'System.Collections.Generic.List(Of st.data.AccountGroup)'.

but I really can't see how it is any different from this code from Rob Connery

public IList<Category> GetCategories() {
    IList<Category> rawCategories = _repository.GetCategories().ToList();                 
    var parents = (from c in rawCategories 
        where c.ParentID == 0
        select c).ToList();
     parents.ForEach(p =>
    {
        p.SubCategories = (from subs in rawCategories
        where subs.ParentID == p.ID
        select subs).ToList();
    });

    return parents; 
}

which compiles perfectly... what am I doing incorrectly?

"lambada expressions" - lol :)

Sam Wessel
– Sam Wessel

2008-09-15 23:29:11 +00:00
Commented Sep 15, 2008 at 23:29 — Sam Wessel
– Sam Wessel, Commented Sep 15, 2008 at 23:29
It's the forbidden code!

1800 INFORMATION
– 1800 INFORMATION

2008-09-15 23:34:41 +00:00
Commented Sep 15, 2008 at 23:34 — 1800 INFORMATION
– 1800 INFORMATION, Commented Sep 15, 2008 at 23:34
Forbidden Code? I thought that was Ruby?

GEOCHET
– GEOCHET

2008-09-15 23:37:33 +00:00
Commented Sep 15, 2008 at 23:37 — GEOCHET
– GEOCHET, Commented Sep 15, 2008 at 23:37

chyne · Accepted Answer · 2009-05-21 13:00:06Z

8

Lambda's in VB.Net have to return a value, so your equal sign ('=') is being intepreted as a comparison (so that the lambda returns a boolean), rather than an assignment.

answered May 21, 2009 at 13:00

chyne

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1 Comment

Richard Anthony Hein Over a year ago

chyne is correct. Your code would work fine in C#. I made a similar example to test it. Also you cannot have multiple lines, so you can't even add a Return. So you have to use a ForEach loop.

Konrad Rudolph · Accepted Answer · 2009-05-21 13:11:05Z

8

The accepted answer here is probably wrong, based on your code. chyne has given the correct clue: lambdas in VB always have return values (unlike in C#), statement lambdas are introduced in the next version though.

In the meantime, you simply can't use this code in VB. Use a regular loop instead:

For Each p In parents
    p.sub_account_groups = sql_func(p)
Next

The next version of VB (available as a Beta since yesterday) would allow the following code to be written:

parents.ForEach(Sub (p) p.sub_account_groups = sql_func(p))

edited May 21, 2009 at 13:11

answered May 21, 2009 at 13:06

Konrad Rudolph

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2 Comments

Richard Anthony Hein Over a year ago

You need the EndSub don't you? Or did they remove it in Beta1?

Konrad Rudolph Over a year ago

@Auxon: you don't need it for one-liners as above. At least, you didn't used to. I haven't tested the beta yet: I'm using OS X.

Jeff Moser · Accepted Answer · 2008-09-16 02:09:44Z

I haven't used VB.NET since moving to C# 3.0, but it seems like it could be a type inference issue. The error is a bit odd since List implements IList, so the assignment should work. You can say "p.ID = 123" for the lambda and things seem to work.

For anyone else interested in looking into it, here is code that you can paste into a new VB.NET console project to demonstrate this issue:

Module Module1
    Sub Main()
    End Sub
End Module

Class AccountGroup
    Public parent_id As Integer
    Public id As Integer
    Public sub_account_groups As List(Of AccountGroup)
End Class
Class AccountRepository
    Private _repository As AccountRepository

    Public Function GetAccountGroups() As IList(Of AccountGroup)
        Dim raw_account_groups As IList(Of AccountGroup)
        raw_account_groups = _repository.GetAccountGroups().ToList()
        Dim parents = (From ag In raw_account_groups _
                       Where ag.parent_id = 0 _
                       Select ag).ToList()
        parents(0).sub_account_groups = (From sag In raw_account_groups _
                                         Where sag.parent_id = 0 _
                                         Select sag).ToList()

        Dim sql_func As Func(Of AccountGroup, List(Of AccountGroup)) = Function(p) _
                                                                            (From sag In raw_account_groups _
                                                                             Where sag.parent_id = p.id _
                                                                             Select sag).ToList()



        parents.ForEach(Function(p) p.sub_account_groups = sql_func(p))
        Return parents
    End Function
 End Class

Siddharth Rout · Accepted Answer · 2012-06-09 03:06:26Z

0

Use Sub for assignment operator:

parents.ForEach(Sub(p) p.sub_account_groups = sql_func(p))

edited Jun 9, 2012 at 3:06

Siddharth Rout

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answered May 23, 2010 at 17:58

BSalita

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Comments

Siddharth Rout · Accepted Answer · 2012-06-09 03:06:51Z

0

I guess ag.parent_id = 0 should be Where ag.parent_id == 0?

edited Jun 9, 2012 at 3:06

Siddharth Rout

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answered May 21, 2009 at 11:22

Tore Nestenius

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