在引用其实现的接口时，对象会存在吗？

Question

I have a class implementing an interface. I don't need a reference to objects of that class - only reference to their interfaces. It looks like:

interface A {}

class B : A {}

//in code:
A a = (A) new B();

My question is: Will instance of B to live (not collecting by GC) while I have a reference to A of that B?

Answer 1

是的，因为尽管您只能看到实现该接口A的部分，但是您仍然可以引用该new B() 。

Answer 2

The reference is the same actual value no matter whether your variable is typed as the class or the interface. So yes: it will stay alive.

Answer 3

是的，对象的实例是相同的，您可以将对象强制转换为其实现的任何接口，但实例是一个。

Answer 4

Yes, because a reference to an object through an interface is still a reference to that object.

Casting an object to an interface does not create a new object, it just alters the "portal" you use to talk to the object through.

You can easily test this in LINQPad :

void Main()
{
    A a = (A)new B();
    GC.Collect();
    GC.WaitForPendingFinalizers();
    GC.Collect();
    GC.KeepAlive(a);
    Debug.WriteLine("Got here");
}

public interface A
{
}

public class B : A
{
    ~B()
    {
        Debug.WriteLine("B was finalized");
    }
}

When executed, you'll get:

Got here

And then, optionally:

B was finalized

But notice that B survived the full GC cycle, even though you had a reference to it through A.