magically a static member on init?

[Adam D. Ruppe]

On Saturday, 14 November 2020 at 23:20:55 UTC, Martin wrote:
> Is this intentional?

In the current language design, yes. For the many users who ask 
this, no.

All static initializers are run at compile time and refer to the 
static data segment - this is consistent across the language.

static x = y; // y is run at compile time

struct A {
     int[] a = [1,2,3]; // array built t compile time
}

so.....

void main() {
    A a;
    a.a[0] = 5;
    A b;
    b.a[0] == 5; // true!!!
}


Because both actually share the same initial array reference.


And with classes again, same deal. *Constructors* are run for 
each instance. But the rest of it is part of the static 
initializer that is shared....

> IMO this feels not consistent and its weird when a reference 
> leaks into another instance without having declared it static 
> member.


Yeah, it is weird and a frequent mistake people make, I almost 
wish it had to be more explicit (I'd be sad if it was removed 
though, I actually use this in places!).

But once you understand it it kinda makes sense.

Note btw that the reference itself is NOT static... just the 
object it refers to. So if you do

obj.a  = new Thing

then it doesn't affect other objects, just this one. But they all 
start off pointing to the same thing.