typedef behavior

[Simen Kjærås]

On Monday, 12 February 2018 at 09:58:13 UTC, Alex wrote:
> On Monday, 12 February 2018 at 09:37:56 UTC, Simen Kjærås wrote:
>>
>> Not really, since D doesn't have a concept of an address 
>> associated with a type, only with instances of it. So when you 
>> use a static array, the address is hard-coded.
>>
>> --
>>   Simen
>
> Ok... so the query on ptr on a static is not to be allowed? And 
> the fact I can perform it is a bug?

I'm sorry, I was apparently unclear. When I said 'static array' 
above, I meant 'static member'.

Since we've been using arrays in our examples, there could be 
conflation of ideas there. The fact that you can access (and even 
modify) the static member's .ptr property (as in S.arr.ptr = 
[1,2,3].ptr), doesn't mean you can change the address of the 
array itself (the static data in S). This can be shown by 
writeln(&S.arr), which will not change.

When you call a static method, as the one in this example:

struct S {
     static int[] arr;
     static void foo() { arr[0]++; }
}

unittest {
     S.foo();
}

No pointer is being passed to foo - it's exactly equivalent to 
this code:

module test;
int[] arr;
void foo() { arr[0]++; }

unittest {
     foo();
}

Likewise, when a non-static method modifies a static member, it 
doesn't need to look up the address of the static member - its 
address is hard-coded.

As an example, try this:

struct S {
    static int n1;
    int n2;
}

unittest {
     import std.stdio;
     S s1;
     S s2;
     // These should be equal:
     writeln(&s1.n1);
     writeln(&s2.n1);
     // These should be different:
     writeln(&s1.n2);
     writeln(&s2.n2);
}

--
   Simen