If I have
public class AImplementation:IAInterface
{
void IAInterface.AInterfaceMethod()
{
}
void AnotherMethod()
{
((IAInterface)this).AInterfaceMethod();
}
}
How to call AInterfaceMethod() from AnotherMethod() without explicit casting?
There are lots of ways of doing this without using the cast operator.
Technique #1: Use "as" operator instead of cast operator.
void AnotherMethod()
{
(this as IAInterface).AInterfaceMethod(); // no cast here
}
Technique #2: use an implicit conversion via a local variable.
void AnotherMethod()
{
IAInterface ia = this;
ia.AInterfaceMethod(); // no cast here either
}
Technique #3: write an extension method:
static class Extensions
{
public static void DoIt(this IAInterface ia)
{
ia.AInterfaceMethod(); // no cast here!
}
}
...
void AnotherMethod()
{
this.DoIt(); // no cast here either!
}
Technique #4: Introduce a helper:
private IAInterface AsIA => this;
void AnotherMethod()
{
this.AsIA.IAInterfaceMethod(); // no casts here!
}
as would be considered "explicit casting" (no idea if that's correct in terms of language spec definitions, or whether the question is even meaningless in that context).
A intends that a derived class must be able to call and/or override the base class implementation then the author of class A should make a protected virtual method rather than an explicit interface method. If the author of class A does not intend that, and the author of class B wishes that the author of class A had done so then the author of class B should either (1) find a different class to extend, perhaps by writing their own, or (2) negotiate with the author of class A. Inheritance must be a designed-in feature of a class.You can introduce a helper private property:
private IAInterface IAInterface => this;
void IAInterface.AInterfaceMethod()
{
}
void AnotherMethod()
{
IAInterface.AInterfaceMethod();
}
Tried this and it works...
public class AImplementation : IAInterface
{
IAInterface IAInterface;
public AImplementation() {
IAInterface = (IAInterface)this;
}
void IAInterface.AInterfaceMethod()
{
}
void AnotherMethod()
{
IAInterface.AInterfaceMethod();
}
}
And yet another way (which is a spin off of Eric's Technique #2 and also should give a compile time error if the interface is not implemented)
IAInterface AsIAInterface
{
get { return this; }
}
The reason that this is not possible without an explicit or implicit cast may is, that you could implement more interfaces within the same class and all interfaces may provide the same method (same name and signature). Think about the following example:
class MyClass : IAInterface1, IAInterface2
{
void IAInterface1.DoSomething() { }
void IAInterface2.DoSomething() { }
void AnotherMethod
{
this.DoSomething(); // ⚡ERROR
// how should the compiler know which method to link?
}
}
But the simples and in my opinion most elegant (and performant) solution for your use case is to provide two methods:
class MyClass : IAInterface
{
void IAInterface.DoSomething() => this.DoSomething();
private void DoSomething
{
}
void AnotherMethod
{
this.DoSomething();
}
}
You can't, but if you have to do it a lot you could define a convenience helper:
private IAInterface that { get { return (IAInterface)this; } }
Whenever you want to call an interface method that was implemented explicitly you can use that.method() instead of ((IAInterface)this).method().
this to IAInterface which is implements).Yet another way (not best):
(this ?? default(IAInterface)).AInterfaceMethod();
((IAInterface)this), with the added downside that if this is null, and isn't supposed to be, it will call AInterfaceMethod on a default instance of your class, rather than throwing an exception. Almost certainly not wanted behavior.Can't you just remove the "IAInterface." from the method signature?
public class AImplementation : IAInterface
{
public void AInterfaceMethod()
{
}
void AnotherMethod()
{
this.AInterfaceMethod();
}
}
public(and removing the explicitIAInterface.from declaration), which makes it more widely visible than you might want. So I'm not recommending this, just mentioning it as an alternative solution that is possible.