Can I add a dynamic method to an interface in c#?
So the interface method doesn't exist but exists on the underlying class and I want to call it via an interface.
thanks
That kind of defeats the point of an interface. If you want it on some of the classes and not others then you can implement it that way, but you will have to cast it to the known class type first.
You could probably use some reflection to check in the under-laying class type has the method you want to call if you want to be really fussy about it (and avoid knowing the type to cast)
Try making a new interface that implements the original interface as a sort of "middle man" between your class and the underlying interface
public interface IFoo
{
string PropA { get; set; }
}
public interface IFooExtended
{
void MyMethod();
}
public class ConcreteFoo : IFooExtended
{
// implementation...
}
It's not "dynamic" but this may help.
No - you could add an extension method to the interface to achieve a similar objective.
I think the right practice to follow that is as close as possible to what you're trying to achieve is create a new interface that inherits the old one.
Something like:
interface IMyInterface : IExistingInterface
{
void MyCustomMethod();
}
And you will address your new interface in your code. The classes implementing it will have your new method as well as the methods of the inherited interface.
You can't alter an existing interface, that would defeat the purpose of an interface itself.
If you're asking if you want to use a dynamic interface (e.g., strong-typed duck typing like in Visual Basic), then no, C# doesn't natively support this. But you can check out the Code Project article, Dynamic interfaces in any .NET language.
Related
In C# 8.0 interfaces can have default implementations for some members. These implementations are explicit meaning that they are inaccessible through the class instance and we must cast to the interface type to call them. Is there a way to expose this implementation on the class instance with the same name? It is possible to expose it under another name by casting the this pointer to the interface type and calling the method like:
void A()
{
((ISomething)this).B();
}
However I can't seem to find a way to expose the implementation with the original name of B because if I declare a method B it counts as the one implementing the interface which causes infinite recursion. Is there some way to expose the implementation without copying it or there is something I am missing?
To clarify, I am looking for a way to achieve traits-like functionality i.e. being able to import the implementation from the interface directly into the class's public API without changing the method name (presumably the name in the interface was the best one). The question is not about how to call the method as a user of the class but how to make it part of the public API of the class.
Extension methods are one solution but default interface implementations have access to protected members and can be overloaded.
One possibility is to provide properties for them. This is what I do when I need to refer to explicit implementations from within the class.
class Something : IInterfaceA, IInterfaceB {
public IInterfaceA A => this;
public IInterfaceB B => this;
}
...
something.A.AMethod();
something.B.BMethod();
You might also consider using extension methods instead of default implementations. They are somewhat similar anyway.
interface IInterface {
}
static class IInterfaceExtensions {
public static void DoSomething(
this IInterface me
) {
// do something
}
}
There's no simple way to do it; various ugly workarounds are possible that all boil down to delegating to some other method (a static method or relying on a helper class that doesn't itself implemented the method) to access it, and even then you need to pass in state somehow. There's a proposal for a base(T) syntax, draft specification here, which should allow base(ISomething).B() to refer to the implementation without causing a cyclic reference. This was originally slated to be part of C# 8, but this proved to be too ambitious and it was cut. As of writing, it's a candidate for inclusion in C# 9.
I'm looking for a way to get wildcards to work in .NET generics.
My code is as follows:
private class Rule<TSource, TSelected> where TSource : class where TSelected : class
{
// stuff in here
}
I want to be able to create a List<> of Rules where the TSource will be the same but the TSelected may be different.
You need to make a contravariant generic interface IRule<TSource, in TSelected> and make a list of that, where in addition TSelected is going to be constrained to some meaningful class. Constraining to any reference type as in your existing code will compile, but you won't be able to do anything meaningful with anything that has to do with TSelected.
At this time there is no other way to use variant generics (unless of course you go into reflection mode with List<dynamic> or something equivalent), so if this solution does not work for you you will need to redesign.
If I read your question right, I think you'd have to do this:
public interface ISelected
{
// ISelected interface
}
// A TSelected implementation
public class Implementation1: ISelected { }
// Another
public class Implementation2 : ISelected { }
// our Rule
private class Rule<TSource, TSelected> where TSource : class where TSelected ISelected
{
}
If the TSelected classes has the same super-class, you can just make a list of Rule<TSource, TSelectedSuperClass>. I believe you can use typeof (http://msdn.microsoft.com/en-us/library/58918ffs(v=vs.71).aspx) to get the exact subclass after reading the TSelected object again.
Alternatively you can make a container class to contain both and also store the exact types.
An interface could do it instead of a super class. If the Selected share implementation however, I prefer an abstract class.
Suppose I have an empty interface class IBaseInterface which is used only to "label" implementing classes as being interfaces themselves.
Is there any way to do something like this?
For example:
public class MyClass : T where T : IBaseInterface
{
}
No, you can't do that, since the compiler has to know which interface the class implements when you declare the class. You can have generic parameters to the interface, but the actual interface has to be specified.
Not like that, there isn't. I would strongly recommend using a composition pattern to try and achieve whatever you're trying. As an alternative, you might find DynamicProxy (or some other proxy solution) is what you're going for.
The type you're declaring isn't even generic. Something like this:
class MyClass<T> : T where T : IBaseInterface
could work under some circumstances (for example, if C++ templates were used instead of .Net generics), but it's simply not valid C# code.
I'm not sure what are the “labels” used for, but an interface with a property
ClassType ClassType { get; }
where
ClassType is an enum could work.
I have a generic interface:
public interface IUnauthorizedRequestRespondable<out T> where T:class
{
T GetResponseForUnauthorizedRequest();
}
(I'm not sure why Resharper recommended T is "out", but that's not the question).
In my scenario, the object returned by GetResponseForUnauthorizedRequest is always of the type that implements the interface.
So all the interface's implementations look like:
public class SignInReturnedObject : IUnauthorizedRequestRespondable<SignInReturnedObject>
(class name and the type in brackets are always the same).
Which seems a bit awkward - isn't there a neater way to tell the compiler that the interface's method returns the type it's a part of?
Thanks.
As far as I know, there is no way to do that currently in C#.
A theoretical side-note: The feature that would allow you to do this is called self types but that's not available in C#. The idea of the feature is that you have a special type that refers to the actual type of this, so if you had a special type named self, you could probably write something like:
public interface IUnauthorizedRequestRespondable {
self GetResponseForUnauthorizedRequest();
}
...and the actual type used in place of self when you have a class SignInReturnedObject implementing the interface would be SignInReturnedObject, but unfortunatelly, that's not
available in C# :-)
If the only way you want to use that template is in that manner, I would use:
public interface IUnauthorizedRequestRespondable<T> where T:IUnauthorizedRequestRespondable<T>
{
T GetResponseForUnauthorizedRequest();
}
This has the advantage of guaranteeing it isn't used in any other way.
The class declaration won't change, but I don't see anything as awkward in this myself. Considering that you are defining a relationship between the class and itself, any more concise form might be inadvisable.
Actually, that about sums it up. That is how the syntax works.
You can see it used in .NET itself with the IEquatable inteface--you are almost always comparing an object to itself, yet you always have to provide your own class name as a template parameter.
This is simply to provide flexibility such that you can compare to anything, not necessarily yourself.
You can create a nongeneric version and just use that but I think it is more trouble than it is worth
public interface ICastUnauthorizedRequestRespondable : IUnauthorizedRequestRespondable<SignInReturnedObject>
{
}
Since T can be any class (does not have to be the class you are implementing) you need to name your class.
public class SignInReturnedObject : IUnauthorizedRequestRespondable<ANYCLASS>
I have an object that implements an interface. I want to call on the object's method if it is implemented. What's the best way in doing this?
Update
A few of you mentioned that my question was vague. Sorry about that. When i said "if it is implemented" i meant "if it is callable". Thanks for your answers and effort guys (or girls!). I'm amazed how much developer support there is on this website.
If this really the way you need it to work, an interface is the wrong choice. Instead, you could have an abstract class from which your class derives with a virtual method. Virtual allows it to be overridden, but does not require it. Since a virtual method has an implementation, it cannot be part of an interface.
Not quite sure what you mean by "if it is implemented." If the method is in the interface and your object implements the interface it must implement the method.
If you want to test if an object implements the interface so you can call the method, you can do it like so:
interface IFoo { void Bar(); }
object o = GetObjectThatMayImplementIFoo();
IFoo foo = o as IFoo;
if (foo != null) {
foo.Bar();
}
I think that's what you were asking?
Create two interfaces, and inherit both interfaces where all methods are required. Inherit only one of the interfaces where the optional methods aren't required.
You can also create a base interface, from which all your interface will inherit, for OOP uses.
I think what you're really looking for is a partial method. These are new in .NET 3.5. You simply declare the method as "partial":
partial void OnLoaded();
The method can be called normally:
OnLoaded();
The neat thing is that if the method is not implemented anywhere, the compiler is smart enough not to generate the call.
This was implemented primarily for LINQ to SQL and for Entity Framework; this allows generated code (using partial classes) to define and call methods without knowing whether they are implemented.
Mixing partial methods with interfaces would be interesting (I haven't tried it), but my first try would be declaring a partial method in the interface.
Shouldn't the object's class implement every method of the interface?
If the object's class inherits from an abstract class, it is possible that it might not override("implement") some methods. Perhaps you are mixing the two up in your mind.
As with the other answers, I'm not sure what you mean. The closest that a class implementing an interface can get to not implementing one of the interface methods is throwing a NotImplementedException. The way to handle this is to specifically catch that exception when calling the method. However, the whole point of an interface is to define a contract between classes, so maybe some clarification would help.
My first response is don't do this. It creates conditional logic around the possibility of a method being there, it goes against the statically typeness of C# and breaks a couple of the SOLID principles. My experience tells me this is the wrong path to walk down.
With that said it can be done via Reflection or using the 'is/as' solution wojo demonstrates.
This type of behavior might be better implemented in a dynamic language. It sounds similar to Duck typing. I'm not a dynamic language guy, but if you have unit tests, it may be alright.
You cannot really know if the method is actually implemented (or if the class just has a "dummy" implementation). Therefore, you may use a pattern such as one of the following to find out if a specific method is supported:
-> Have multiple interfaces and see if the class actually implements it; this is probably the cleanest way to deal with it, but it may leave you with a large number of different interfaces, which may not be desirable:
IIntfA = inst as IIntfA;
if (inst != null) {
// inst seems to be implemented
}
-> Use methods in the TryXxx style, which return true if they were successfull (like TryParse() etc.).
-> Use NotImplementedException - but note that catching those is very expensive and should only be used for calls which are performed rarely, or where a missing implementation is not expected. The Stream class works like this, for instance if it cannot be written to (but additionally there is a property telling what the class supports, e.g. IsWritable in the Stream class).
Hey guys, don't forget the "is" keyword :P
You can check if an object implements an interface like this too:
if (inst is IInterface)
{
// you can safely cast it
}
I prefer it that way but of course you could also use the "as" keyword
IInterface a = inst as IInterface;
if (a != null)
{
// use a, already casted
}
Depending on how you're referencing an object, certain members will be visible. An interface might be implicitly defined or explicitly defined, or might be implemented by a derived class and you're using a base class reference. In other words, it's not always immediately evident all the available members on an object.
So if you want to test for implementation of a certain interface (ISomething) by your object (yourObj), one choice is testing the data type, using reflection. Based on the result of this test, you can explicitly cast an implementing object into the interface Type and use its members...
if (yourObj is ISomething)
((ISomething)yourObj).DoSomething();
This is the same thing done another way (more "wordy" using method calls):
if (typeof(ISomething).IsAssignableFrom(yourObj.GetType()))
((ISomething)yourObj).DoSomething();
This sample assumes the ISomething interface is defined as:
public interface ISomething {
void DoSomething();
// other members ...
}
In summary, this code says: if the interface ISomething Is-Assignable-From your object of choice, then your object implements that interface and therefore has those public members.
I don't know if you might be looking for something like this. This uses an attribute that you can flag a method with whether or not it is implemented. Next I added an extension method to the interface to allow for checking if ithe method is implemented. Finally, the code will allow you to ask an object if the method is implemented. I don't like this but it might be what you are looking for.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Reflection;
namespace ConsoleApplication1
{
public static class Program
{
static void Main(string[] args)
{
EmployeeA empA = new EmployeeA();
if (empA.IsImplemented("TestMethod"))
empA.TestMethod();
EmployeeB empB = new EmployeeB();
if (empB.IsImplemented("TestMethod"))
empB.TestMethod();
Console.ReadLine();
}
public static bool IsImplemented(this IEmp emp, string methodName)
{
ImplementedAttribute impAtt;
MethodInfo info = emp.GetType().GetMethod(methodName);
impAtt = Attribute.GetCustomAttribute(info, typeof(ImplementedAttribute), false)
as ImplementedAttribute;
return (impAtt == null) ? true : impAtt.Implemented;
}
}
public class EmployeeA : IEmp
{
#region IEmp Members
[Implemented(false)]
public void TestMethod()
{
Console.WriteLine("Inside of EmployeeA");
}
#endregion
}
public class EmployeeB : IEmp
{
#region IEmp Members
[Implemented(true)]
public void TestMethod()
{
Console.WriteLine("Inside of EmployeeB");
}
#endregion
}
public class ImplementedAttribute : Attribute
{
public bool Implemented { get; set; }
public ImplementedAttribute():this(true)
{
}
public ImplementedAttribute(bool implemented)
{
Implemented = implemented;
}
}
public interface IEmp
{
void TestMethod();
}
}
EDIT: After original author reworded question, you definitely just want to implement the interface guranteeing the method does exist. I will leave above code for curiosity sake.