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Reference to a variablelike this: Public Class MyClass Private m_Widget As MyWidget End Class I realise that if you do something like: m_Widget = New MyWidget The m_Widget holds a reference to the new MyWidget object. If you then do something like: Dim m_Reference As MyWidget = m_Widget Then m_Reference now holds a reference to the same MyWidget object. However, what I want is the equivilent of the C++ ** mechanism whereby m_Reference is actually a reference to the m_Width reference if you see what I mean :-) So that if the m_Width variable is changed to another object, then the m_Reference also changes. In C++ (which I'm real rusty on), I would have done something like: Widget *m_Widget, **m_Reference; m_Widget = New Widget; m_Reference = &m_Widget; Sorry if that's a real mixture - ages since I wrote a line of C++! And I know I'm mixing pointers and references. Thanks, Rob.
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"Rob Nicholson" <informed@community.nospam> schrieb If you need a reference to a reference, put the 2nd reference into a class > How do you handle references to a variable in VB.NET? Consider > something like this: > > Public Class MyClass > > Private m_Widget As MyWidget > > End Class > > I realise that if you do something like: > > m_Widget = New MyWidget > > The m_Widget holds a reference to the new MyWidget object. If you > then do something like: > > Dim m_Reference As MyWidget = m_Widget > > Then m_Reference now holds a reference to the same MyWidget object. > > However, what I want is the equivilent of the C++ ** mechanism > whereby m_Reference is actually a reference to the m_Width reference > if you see what I mean :-) So that if the m_Width variable is > changed to another object, then the m_Reference also changes. > > In C++ (which I'm real rusty on), I would have done something like: > > Widget *m_Widget, **m_Reference; > m_Widget = New Widget; > m_Reference = &m_Widget; > > Sorry if that's a real mixture - ages since I wrote a line of C++! > And I know I'm mixing pointers and references. and store a reference to an instance of that class. Armin > If you need a reference to a reference, put the 2nd reference into a class Hmm, I thought of that using ByRef in a class method and then storing that > and store a reference to an instance of that class. in a class variable: Public Class RefClass Private m_Reference as MyWidget Sub StoreReference(ByRef Widget As MyWidget) m_Reference = Widget End Sub But whilst this does indeed pass in a reference, what gets store in m_Reference is the reference to the Widget object, not the reference to the variable referencing the Widget object. Cheers, Rob.
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"Rob Nicholson" <informed@community.nospam> schrieb Correct.> > > If you need a reference to a reference, put the 2nd reference into > > a class and store a reference to an instance of that class. > > Hmm, I thought of that using ByRef in a class method and then > storing that in a class variable: > > Public Class RefClass > > Private m_Reference as MyWidget > > Sub StoreReference(ByRef Widget As MyWidget) > m_Reference = Widget > End Sub > > But whilst this does indeed pass in a reference, what gets store in > m_Reference is the reference to the Widget object, not the reference > to the variable referencing the Widget object. (there's nothing I can add) Armin > Correct. Hmm, bit of an omission that. I can code around it by in effect creating a > (there's nothing I can add) little class that contains the pointer and do it that way. Or was that what you were suggesting in the first place? Thanks, Rob. "Rob Nicholson" <informed@community.nospam> schrieb Yes, that's what I suggested. But, I've never needed it.> > Correct. > > (there's nothing I can add) > > Hmm, bit of an omission that. I can code around it by in effect > creating a little class that contains the pointer and do it that > way. Or was that what you were suggesting in the first place? Armin Hi Rob,
In addition to Armin's comment, I hope the following information is also helpful for you. Based on my knowledge, there is a way which is called "RefByRef", just like: shared sub Main dim myObject as new SomeClass Test(myObject) End sub Public sub Test(ByRef v1 as SomeClass) //now v1 has a reference to MyObject(reference) and not to the actual Object End sub I hope the above information is helpful for you. Thanks and have a nice day! Best Regards, Terry Fei [MSFT] Microsoft Community Support Get Secure! www.microsoft.com/security -------------------- Show quoteHide quote >From: "Rob Nicholson" <informed@community.nospam> >Subject: Reference to a variable >Date: Thu, 16 Feb 2006 18:09:55 -0000 >Lines: 39 >X-Priority: 3 >X-MSMail-Priority: Normal >X-Newsreader: Microsoft Outlook Express 6.00.3790.1830 >X-RFC2646: Format=Flowed; Original >X-MimeOLE: Produced By Microsoft MimeOLE V6.00.3790.1830 >Message-ID: <ueYxbQyMGHA.3***@TK2MSFTNGP14.phx.gbl> >Newsgroups: microsoft.public.dotnet.languages.vb >NNTP-Posting-Host: host217-45-2-100.in-addr.btopenworld.com 217.45.2.100 >Path: TK2MSFTNGXA01.phx.gbl!TK2MSFTNGP08.phx.gbl!TK2MSFTNGP14.phx.gbl >Xref: TK2MSFTNGXA01.phx.gbl microsoft.public.dotnet.languages.vb:318033 >X-Tomcat-NG: microsoft.public.dotnet.languages.vb > >How do you handle references to a variable in VB.NET? Consider something >like this: > >Public Class MyClass > > Private m_Widget As MyWidget > >End Class > >I realise that if you do something like: > >m_Widget = New MyWidget > >The m_Widget holds a reference to the new MyWidget object. If you then do >something like: > >Dim m_Reference As MyWidget = m_Widget > >Then m_Reference now holds a reference to the same MyWidget object. > >However, what I want is the equivilent of the C++ ** mechanism whereby >m_Reference is actually a reference to the m_Width reference if you see what >I mean :-) So that if the m_Width variable is changed to another object, >then the m_Reference also changes. > >In C++ (which I'm real rusty on), I would have done something like: > > Widget *m_Widget, **m_Reference; > m_Widget = New Widget; > m_Reference = &m_Widget; > >Sorry if that's a real mixture - ages since I wrote a line of C++! And I >know I'm mixing pointers and references. > >Thanks, Rob. > > > > > |
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