I can't do this
interface InterfaceA
{
void MethodA();
}
class ClassA : InterfaceA
{
virtual void InterfaceA.MethodA()
// Error: The modifier 'virtual' is not valid for this item
{
}
}
Where the following works
class ClassA : InterfaceA
{
public virtual void MethodA()
{
}
}
Why? How to circumvent this?
I think this is because when a member is explicitly implemented, it cannot be accessed through a class instance, but only through an instance of the interface.
So making something 'virtual' really does not make sense in this case, since virtual means that you intend to override it in an inherited class. Implementing an interface explicitly and making it virtual would be contradictory. Which also may be why the compiler disallows this.
To work around it I think csharptest.net's or Philip's answer sounds like it would do the trick
It is part of the C# language specification:
It is a compile-time error for an explicit interface member implementation to include access modifiers, and it is a compile-time error to include the modifiers abstract, virtual, override, or static.
According to C# language spec:
It is a compile-time error for an
explicit interface member
implementation to include access
modifiers, and it is a compile-time
error to include the modifiers
abstract, virtual, override, or
static.
You can "get around" it only by calling your virtual method from the explicit interface implementation.
You have to do something like this:
interface InterfaceA
{
void MethodA();
}
class ClassA : InterfaceA
{
void InterfaceA.MethodA()
{ MethodB(); }
protected virtual void MethodB()
{
}
}
Often this is a superior approach anyway as the internal method may change signature without changing the interface. Take a more real-word example:
interface IOrderDetails
{
void PlaceOrder();
}
//Sometime later you extend with:
interface IOrderDetails2 : IOrderDetails
{
void PlaceOrder(IUser user);
}
//Implementation
class Order : IOrderDetails, IOrderDetails2
{
static readonly IUser AnonUser;
void IOrderDetails.PlaceOrder()
{ OnPlaceOrder(AnonUser); }
void IOrderDetails2.PlaceOrder(IUser user)
{ OnPlaceOrder(user); }
protected virtual void OnPlaceOrder(IUser user)
{
}
}
Here you can see as the IOrderDetails2 was added we can safely refactor the existing virtual method (thus generating compile-time errors for derivations). Additionally this often allows you to provide common functionality, logging, and exception handling in the base implementation class...
You cannot use the virtual modifier with the static, abstract, private or override modifiers.
default modifier is private
Related
I have a question while studying C#.
When I declare the interface, I usually declare it like the code below
public interface IRequireInitialization
{
void Init();
}
public class TESTManager : IRequireInitialization
{
public void Init(){ throw new NotImplementedException(); }
}
but my friend said that I should add abstract before the declaration of the interface member function.
I didn't feel any big difference in implementing the interface, what's the difference? And if there is a difference, when should I use it?
public interface IRequireInitialization
{
abstract void Init();
}
public class TESTManager : IRequireInitialization
{
public void Init(){ throw new NotImplementedException(); }
}
C# 8 introduced default interface methods. This allows to define a default implementation (body) for a method in the interface itself.
As part of this proposal they relaxed the syntax to allow modifiers for interface methods:
The syntax for an interface is relaxed to permit modifiers on its members. The following are permitted: private, protected, internal, public, virtual, abstract, sealed, static, extern, and partial.
A method in an interface is abstract by default and abstract modified just makes that explicit:
similarly, although abstract is the default on interface members without bodies, that modifier may be given explicitly.
I expose the following class in assembly A:
public abstract class ServiceDependencyHost
{
protected virtual T ReferenceService<T>() where T : ServiceBase
{
// Virtual implementation here...
}
}
I expose this derived class in a separate assembly (B):
public sealed class ProcessServiceOperation : ServiceDependencyHost
{
public override T ReferenceService<T>()
{
// Override implementation here...
return base.ReferenceService<T>();
}
}
Using the code as shown above, the compiler complains on this line:
return base.ReferenceService<T>();
The type 'T' cannot be used as type parameter 'T' in the generic type or method A.ReferenceService(). There is no boxing conversion or type parameter conversion from 'T' to 'System.ServiceProcess.ServiceBase'.
Naturally, I tried replicating the constraints in assembly B:
public override T ReferenceService<T>() where T : ServiceBase
But the compiler now warns on the line above...
Constraints for override and explicit interface implementation methods are inherited from the base method, so they cannot be specified directly.
This answer indicates that my solution should have worked. I want to avoid using reflection to expose this method publicly. It should be so simple!
Thanks in advance to anyone who can spot what mistake I am making.
The issue is not strictly due to generics. The cause of the issue is that the base class method is protected, whilst the derived class method is public.
An override declaration cannot change the accessibility of the virtual method. Both the override method and the virtual method must have the same access level modifier.
Consequently, the compiler assumes that the two methods are distinct, and the derived method fails to inherit the where T : ServiceBase generic constraint from the base method. Since the derived method knows nothing about T, it expectedly complains that T cannot be converted to ServiceBase when you attempt to pass it to the base method.
The error message seems to be misleading.
Note, that you overrode a protected virtual method with a public override. That's not going to work. Make the override protected too, and the code should compile just fine...
(I am not sure if the seemingly wrong error message is a glitch, or whether there is a reasonable explanation behind this particular behavior of the compiler.)
its because of the public => protected. This compiles fine
public abstract class ServiceDependencyHost
{
protected virtual T ReferenceService<T>() where T : new()
{
return new T();
}
}
public sealed class ProcessServiceOperation : ServiceDependencyHost
{
protected override T ReferenceService<T>()
{
// Override implementation here...
return base.ReferenceService<T>();
}
}
Here's the c# code I'm trying to compile
Fairly simple, testDMO that inherits from testDMOBase
Then test that inherits from testBase
public abstract class testDMOBase { }
public class testDMO : testDMOBase { }
public abstract class testBase
{
abstract protected void LoadCRUD(testDMOBase dmo);
}
public class test : testBase
{
override protected void LoadCRUD(testDMO dmo) { }
}
I'm getting the following errors:
'test' does not implement inherited abstract member
'testBase.LoadCRUD(testDMOBase)' 'test.LoadCRUD(testDMO)': no suitable
method found to override
Shouldn't the use of a subclass be ok on the override method?
Shouldn't the use of a subclass be ok on the override method?
No. Aside from anything else, which implementation would you expect to be called if the caller provided an instance other than your subclass?
testBase t = new test();
t.LoadCRUD(new SomeOtherDMO()); // What would be called here?
You might well argue that it would make sense to be able to override the base method with a subclass method which is more general (e.g. with a parameter which is a superclass of the original parameter type, or with a return type which is a subclass of the original return type) but .NET doesn't allow either of these anyway. The parameter and return types of the overriding method have to match the original method exactly, at least after generic type parameter substitution.
It sounds like you may want to make your base type generic:
public abstract class TestBase<T> where T : TestDmoBase
{
public abstract void LoadCrud(T dmo);
}
public class Test : TestBase<TestDmo>
{
public override void LoadCrud(TestDmo dmo)
{
...
}
}
Note that you should follow .NET naming conventions, too - even in sample code.
No, in this case you have to exactly follow the signature of the abstract method, in order to provide valid override.
So, you have to write:
public class test : testBase
{
override protected void LoadCRUD(testDMOBase dmo) //BASE CLASS
{ }
}
Shouldn't the use of a subclass be ok on the override method?
No, method overrides must use the same parameter types as their original declarations.
How do I prevent a method from being overridden in a derived class?
In Java I could do this by using the final modifier on the method I wish to prevent from being overridden.
How do I achieve the same in C#?
I am aware of using sealed but apparently I can use it only with the override keyword?
class A
{
public void methodA()
{
// Code.
}
public virtual void methodB()
{
// Code.
}
}
class B : A
{
sealed override public void methodB()
{
// Code.
}
}
So in the above example I can prevent the methodB() from being overridden by any classes deriving from class B, but how do I prevent class B from overriding the methodB() in the first place?
Update: I missed the virtual keyword in the methodB() declaration on class A when i posted this question. Corrected it.
You don't need to do anything. The virtual modifier specifies that a method can be overridden. Omitting it means that the method is 'final'.
Specifically, a method must be virtual, abstract, or override for it to be overridden.
Using the new keyword will allow the base class method to be hidden, but it will still not override it i.e. when you call A.methodB() you will get the base class version, but if you call B.methodB() you will get the new version.
As you mentioned, you can prevent further overriding of MethodB in class B by using sealed with override
class B : A
{
public sealed override void methodB()
{
Console.WriteLine("Class C cannot override this method now");
}
}
Use of the sealed modifier along with override prevents a derived class from further overriding the method.
If you do not want methodB in class A to be overridden by any child classes, do not mark that method virtual. Simply remove it. virtual keyword enable the method to be overridden in child classes
public void methodA()
{
}
Use sealed keyword on your classes to prevent further overriding of the class
In C#, a function not marked virtual (which also includes overrides of virtual functions) is effectively sealed and cannot be overridden. So, your example code actually won't compile because the override keyword is not valid unless there is a method marked virtual with the same signature in a base class.
If A.methodB() were marked virtual, then you could override the method from A, but prevent it being further overridden in classes deriving more indirectly, using the sealed keyword exactly as you have shown.
One thing to keep in mind is that while method overriding can be prevented, method hiding cannot. Given your current definition of class A, the following definition of class B is legal and there's nothing you can do about it:
class B:A
{
public new void methodB()
{
//code
}
}
The new keyword basically "breaks" the inheritance/overriding hierarchy as it pertains to this one method; any reference to a class B, treated as a class B (or any further derived type) will use the implementation from class B and ignore the one from class A unless B's implementation specifically calls back to it. However, if you were to treat an instance of class B as a class A (by casting it or passing it as a parameter), then the "new" implementation is ignored.
This differs from overriding, where a class B that is being treated as a class A and truly overrides a virtual methodB would still use class B's override of the method. Also understand that method hiding is inferred (though you will get a compiler warning); if you declare a method with the same signature in a derived class and do not specify either new or override, the base class method will be hidden.
In a base class, the sealed keyword is only used to prevent a class from being derived, but in inherited classes it can be used to prevent another inherited class from overriding the method.
To prevent a base class method from being overridden, just do not specify it as virtual. In the example you provided, class B could not override methodB because methodB was not marked as virtual on the original class.
this will compile:
class A
{
public virtual void methodA()
{
//code
}
public virtual void methodB()
{
//code
}
}
class B:A
{
public override void methodB()
{
//code
}
}
this will not:
class A
{
public void methodA()
{
//code
}
public void methodB()
{
//code
}
}
class B:A
{
public override void methodB()
{
//code
}
}
EDITTED: clarified and corrected my original statement about the sealed keyword
I have a class which inherits an interface. An interface member method is implemented in my class without an access modifier (so, by default it's private ) .
I am getting the error "cannot implement an interface member because it is not public".
Why it is not allowed? Can't I override the accessibility?
Here's an example of why it doesn't make sense to be able to override the visibility:
interface someI
{
void doYourWork();
}
public class A : someI
{
public void doYourWork()
{
//...
}
}
public class B : someI
{
private void doYourWork()
{
//...
}
}
void Main()
{
List<someI> workers = getWorkers();
foreach(var worker in workers)
worker.doYourWork();
}
What happens when your worker is of type B? You're calling a method as if it were public, but it's a private method. If you want this functionality, then it's not really a private method is it?
If you only want it to be public when referenced through your interface, then you can define it as such:
public class B : someI
{
void someI.doYourWork()
{
//...
}
}
And you end up with this:
var b = new B();
b.doYourWork(); // Not accessible
((someI)b).doYourWork(); // Accessible
Methods have to be implemented public because they have to be callable through the interface, thus from where the interface is accessible as a type.
You have a few options here to "change" the visibility of that method. Given:
public interface IFoo
{
bool IsFoo();
}
A. Implement the method explicitly
public class Foo : IFoo
{
bool IFoo.IsFoo() { return true; }
}
The method will only be available through the interface (IFoo in this case.)
B. Change the visibility of the interface
Define the interface as internal instead of public. As a consequence, however, Foo will have to be internal too.
Requiring that an interface implementation be public is simply a logical requirement. When you implement an interface, you're telling the compiler "Hey I implement every method on this interface". So making the method private makes it no longer accessible - and logically not implemented. Interfaces serve as a contract to code that uses your object saying you can always call any method defined in the interface on my object. If the implementing method were private, that would not longer be true.
If you want to hide your implementation from, say Intellisense, then you can simply implement the method explicitly as #Bryan mentions.