Class member always has the same address

[szymski via Digitalmars-d-learn]

On Sunday, 20 March 2016 at 02:21:51 UTC, Mike Parker wrote:
> On Saturday, 19 March 2016 at 20:24:15 UTC, szymski wrote:
>
>>>class A {
>>>	B b = new B();	
>>>}
>
> This is *default* initialization, not per instance 
> initialization. The compiler will create one instance of B and 
> it will become the default initializer of b in *every* instance 
> of A. You can verify that with this:
>
> class B {}
> class A {
>     B b = new B;
> }
> void main() {
>     auto as = [new A, new A, new A];
>     assert(as[0].b is as[1].b);
>     assert(as[1].b is as[2].b);
>     assert(as[0].b is as[2].b);
> }
>
> Here, all of the asserts will pass. But add a constructor to A 
> that does this:
>
> this() { b = new B; }
>
> And now the first assert will fail. This is *per-instance* 
> initialization.

Ok, I understand now, thanks. I used C# a lot before and there 
default initialization worked like per instance initialization.