Overriding interface members

[Sebastian]

Hi,

I'm confronted with a problem that seems not to be solvable. In
general: How can I override an interface member of my base class and
call the overridden method from my derived class?

This is my class:
class RemoteXmlDataSo urce : System.Web.UI.W ebControls.XmlD ataSource

And I want to override
DataSourceView System.Web.UI.I DataSource.GetV iew(string viewName)
but I want to make a call to the overridden method right after my
stuff.

Declaring the member with override, I get the message
The modifier 'override' is not valid for this item

When trying to access the "overridden " member with
(IDataSource)ba se).GetView(vie wName), I get the message
Use of keyword 'base' is not valid in this context

Ok I understand, my interface member completely replaces the member
from the base class. Is there any way to do this even though?
I tried something like
((IDataSource)( XmlDataSource)t his).GetView(vi ewName)
but that just calls my new member :)

That makes deriving much harder than just overriding a virtual method
and calling base.X()... But I hope I just did not fully understand ;)

Kind regards, Sebastian

[Ignacio Machin \( .NET/ C# MVP \)]

Re: Overriding interface members

Hi,

"Sebastian" <sebpaul79.work @googlemail.com wrote in message
news:1182413815 .648232.224930@ u2g2000hsc.goog legroups.com...

Hi,
>
I'm confronted with a problem that seems not to be solvable. In
general: How can I override an interface member of my base class and
call the overridden method from my derived class?
>
This is my class:
class RemoteXmlDataSo urce : System.Web.UI.W ebControls.XmlD ataSource
>
And I want to override
DataSourceView System.Web.UI.I DataSource.GetV iew(string viewName)
but I want to make a call to the overridden method right after my
stuff.
>
Declaring the member with override, I get the message
The modifier 'override' is not valid for this item

Did you declare the base as public virtual?

See if this code helps you:

class Program
{
static void Main(string[] args)
{
B b = new C();
b.M();

II i = b;
i.M();
Console.Read();
}
}

class B : II
{
public virtual void M() { Console.WriteLi ne("B.M"); }
}
class C : B
{
public override void M()
{
Console.WriteLi ne("C.M");
}
}

public interface II
{
void M();
}

[Ben Voigt [C++ MVP]]

Re: Overriding interface members


"Ignacio Machin ( .NET/ C# MVP )" <machin TA laceupsolutions .comwrote in
message news:e2BHU4AtHH A.4424@TK2MSFTN GP04.phx.gbl...

Hi,
>
"Sebastian" <sebpaul79.work @googlemail.com wrote in message
news:1182413815 .648232.224930@ u2g2000hsc.goog legroups.com...

>Hi,
>>
>I'm confronted with a problem that seems not to be solvable. In
>general: How can I override an interface member of my base class and
>call the overridden method from my derived class?
>>
>This is my class:
>class RemoteXmlDataSo urce : System.Web.UI.W ebControls.XmlD ataSource
>>
>And I want to override
>DataSourceVi ew System.Web.UI.I DataSource.GetV iew(string viewName)
>but I want to make a call to the overridden method right after my
>stuff.
>>
>Declaring the member with override, I get the message
>The modifier 'override' is not valid for this item

>
Did you declare the base as public virtual?
>
See if this code helps you:
>
class Program
{
static void Main(string[] args)
{
B b = new C();
b.M();
>
II i = b;
i.M();
Console.Read();
}
}
>
class B : II
{
public virtual void M() { Console.WriteLi ne("B.M"); }

I assume that the base class method implements an interface, but doesn't use
the virtual keyword (that might make it sealed).

}
class C : B
{
public override void M()
{
Console.WriteLi ne("C.M");
}
}
>
public interface II
{
void M();
}
>

[Sebastian]

Re: Overriding interface members

It turns out that declaring explicit interface implementations as
virtual is not possible - only implicit interface implementation
support that. Seems to be a limitation of the C# language. Or is there
a way I do not see?

Look at this snippet:

interface IFace {
string iFaceMethod();
}

public class BaseClass : IFace {
// works (implicit interface implementation)
public virtual string iFaceMethod() { return "BaseClass public
virtual string iFaceMethod()"; }

// works (explicit interface implementation - obove method won't be
anymore an implicit interface implementation)
string IFace.iFaceMeth od() { return "BaseClass string
IFace.iFaceMeth od()"; }

// The modifier 'public' is not valid for this item
// The modifier 'virtual' is not valid for this item
//public virtual string IFace.iFaceMeth od() { }
}

public class DerivedClass : BaseClass, IFace {

// works
public override string iFaceMethod() {
// Use of keyword 'base' is not valid in this context
//return ((IFace)base).i FaceMethod();

// works
//return base.iFaceMetho d(); // calls BaseClass public virtual
string iFaceMethod()

return "DerivedCla ss public override string iFaceMethod()";
}

// works, but would definitely hide base.iFaceMetho d
string IFace.iFaceMeth od() {
// Use of keyword 'base' is not valid in this context
//return ((IFace)base).i FaceMethod();

// works
//return base.iFaceMetho d(); // calls BaseClass public virtual
string iFaceMethod()

// works, but leads to infinite recursion
//return ((IFace)this).i FaceMethod();

return "DerivedCla ss string IFace.iFaceMeth od()";
}

// The modifier 'public' is not valid for this item
// The modifier 'override' is not valid for this item
//public override string IFace.iFaceMeth od() { }
}

Kind regards, Sebastian

[Ben Voigt [C++ MVP]]

Re: Overriding interface members


"Sebastian" <sebpaul79.work @googlemail.com wrote in message
news:1182939783 .654061.181680@ c77g2000hse.goo glegroups.com.. .

It turns out that declaring explicit interface implementations as
virtual is not possible - only implicit interface implementation
support that. Seems to be a limitation of the C# language. Or is there
a way I do not see?

No, you're right. explicit interface implementations are always private
(therefore can't be virtual), and implicit implementations are always
public. There's no way to get anything in-between.

>
Look at this snippet:
>
interface IFace {
string iFaceMethod();
}
>
public class BaseClass : IFace {
// works (implicit interface implementation)
public virtual string iFaceMethod() { return "BaseClass public
virtual string iFaceMethod()"; }
>
// works (explicit interface implementation - obove method won't be
anymore an implicit interface implementation)
string IFace.iFaceMeth od() { return "BaseClass string
IFace.iFaceMeth od()"; }
>
// The modifier 'public' is not valid for this item
// The modifier 'virtual' is not valid for this item
//public virtual string IFace.iFaceMeth od() { }
}
>
public class DerivedClass : BaseClass, IFace {
>
// works
public override string iFaceMethod() {
// Use of keyword 'base' is not valid in this context
//return ((IFace)base).i FaceMethod();
>
// works
//return base.iFaceMetho d(); // calls BaseClass public virtual
string iFaceMethod()
>
return "DerivedCla ss public override string iFaceMethod()";
}
>
// works, but would definitely hide base.iFaceMetho d
string IFace.iFaceMeth od() {
// Use of keyword 'base' is not valid in this context
//return ((IFace)base).i FaceMethod();
>
// works
//return base.iFaceMetho d(); // calls BaseClass public virtual
string iFaceMethod()
>
// works, but leads to infinite recursion
//return ((IFace)this).i FaceMethod();
>
return "DerivedCla ss string IFace.iFaceMeth od()";
}
>
// The modifier 'public' is not valid for this item
// The modifier 'override' is not valid for this item
//public override string IFace.iFaceMeth od() { }
}
>
Kind regards, Sebastian
>

[Sebastian]

Re: Overriding interface members

No, you're right. explicit interface implementations are always private

(therefore can't be virtual), and implicit implementations are always
public. There's no way to get anything in-between.

Why are implicit implementations always public and explicit interface
implementations are always private?
Is there any logical explanation for that? For me, it sounds just
stupid. When implementing two methods from different interfaces that
have the same name, the methods can never be overridden because they
have to be explicit.

Kind regards, Sebastian

[Jon Skeet [C# MVP]]

Re: Overriding interface members

On Jul 4, 3:19 pm, Sebastian <sebpaul79.w... @googlemail.com wrote:

No, you're right. explicit interface implementations are always private
(therefore can't be virtual), and implicit implementations are always
public. There's no way to get anything in-between.

>
Why are implicit implementations always public and explicit interface
implementations are always private?
Is there any logical explanation for that? For me, it sounds just
stupid. When implementing two methods from different interfaces that
have the same name, the methods can never be overridden because they
have to be explicit.

I think I've got a solution, actually. You just need to reimplement
the interface. You don't specify that you're overriding it, just that
you're implementing it. Here's some code to demonstrate what I mean -
if it doesn't do what you want it to, it would be good to know exactly
what's still missing. Note that you have to redeclare that you're
implementing IDisposable in the derived class.

using System;

public class Base : IDisposable
{
void IDisposable.Dis pose()
{
Console.WriteLi ne ("Base");
}
}

public class Derived : Base, IDisposable
{
void IDisposable.Dis pose()
{
Console.WriteLi ne ("Derived");
}
}

public class Test
{
static void Main()
{
IDisposable a = new Base();
IDisposable b = new Derived();

a.Dispose();
b.Dispose();
}
}

Jon

[Marc Gravell]

Re: Overriding interface members

This is a feature of the C# compiler (I believe you have other options
in VB for instance). You can, however, simply write a protected inner
method for each, and forward from the explicit interface call - i.e.

void ISomeInterface. SomeMethod() {SomeInterfaceS omeMethod();}
void IOtherInterface .SomeMethod() {OtherInterface SomeMethod();}
protected virtual void SomeInterfaceSo meMethod() {...}
protected virtual void OtherInterfaceS omeMethod() {...}

Marc

[Sebastian]

Re: Overriding interface members

On Jul 4, 4:33 pm, "Jon Skeet [C# MVP]" <s...@pobox.com wrote:

I think I've got a solution, actually. You just need to reimplement
the interface. You don't specify that you're overriding it, just that
you're implementing it.

In that case, you cannot call the overridden method, as it would be
possible in virtual methods (base keyword). I tried to show this in my
last code example - maybe it's hard legible because of the line
wraps...

On Jul 4, 4:37 pm, "Marc Gravell" <marc.grav...@g mail.comwrote:

This is a feature of the C# compiler

LOL In my oppinion it's a defect :D

simply write a protected inner
method for each, and forward from the explicit interface call

Yes this would work. But this is not possible with existing (i.e.
built in .NET) classes that do not implement that way...

Sebastian

[Jon Skeet [C# MVP]]

Re: Overriding interface members

On Jul 4, 3:56 pm, Sebastian <sebpaul79.w... @googlemail.com wrote:

I think I've got a solution, actually. You just need to reimplement
the interface. You don't specify that you're overriding it, just that
you're implementing it.

>
In that case, you cannot call the overridden method, as it would be
possible in virtual methods (base keyword). I tried to show this in my
last code example - maybe it's hard legible because of the line
wraps...

That's true. It can be done with reflection, but I agree that's pretty
grim:

void IDisposable.Dis pose()
{
Console.WriteLi ne ("Derived");
InterfaceMappin g mapping =
typeof(Base).Ge tInterfaceMap(t ypeof(IDisposab le));
MethodInfo method = mapping.TargetM ethods[0];
method.Invoke(t his, null);
}

(I only used 0 blindly because IDisposable has only one member.)

Jon

[Sebastian]

Re: Overriding interface members

That's true. It can be done with reflection, but I agree that's pretty

grim

Many thanks for your workaround!

Nevertheless, I can't imagine that this was the intention by the C#
language developers...

Kind regards, Sebastian