Say I have two structs, defined like this:

    struct A { /* could contain whatever */ }

    struct B { A a; }

My question is, is it now guaranteed that A.sizeof==B.sizeof, regardless of how A is defined (member variable types, alignment, etc.)?  More to the point, say I have a function foo() which looks like this:

    extern(C) void foo(A* ptr, size_t len);

Is it now guaranteed that I can safely pass it a pointer to an array of Bs?  That is,

    auto arr = new B[10];
    foo(cast(A*) arr.ptr, arr.length);

Thanks,
Lars

On 4/19/14, Lars T. Kyllingstad via Digitalmars-d-learn <digitalmars-d-learn@puremagic.com> wrote:
> Say I have two structs, defined like this:
>
>      struct A { /* could contain whatever */ }
>
>      struct B { A a; }
>
> My question is, is it now guaranteed that A.sizeof==B.sizeof?

The best thing to do is add a static assert and then you can relax:

-----
struct A { }
struct B { A a; }
static assert(A.sizeof == B.sizeof);
-----

As for the ABI, I don't think I've ever seen it mentioned anywhere.

On Saturday, 19 April 2014 at 12:26:16 UTC, Andrej Mitrovic via Digitalmars-d-learn wrote:
> On 4/19/14, Lars T. Kyllingstad via Digitalmars-d-learn
> <digitalmars-d-learn@puremagic.com> wrote:
>> Say I have two structs, defined like this:
>>
>>      struct A { /* could contain whatever */ }
>>
>>      struct B { A a; }
>>
>> My question is, is it now guaranteed that A.sizeof==B.sizeof?
>
> The best thing to do is add a static assert and then you can relax:

That's what I've done, but it would be nice to know the code won't break due to some combination of platform and/or compiler switches I didn't think to test.  Anyway, I've played around a bit, and found that a combination of struct and field alignment *can* break my assumption:

    align(1) struct A
    {
        char c;
        align(1) int i;
    }

    struct B { A a; }

Now, A.sizeof is 5, while B.sizeof is 8.  I'd have to add align(1) to the declaration of B to fix it.