Re: 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.