//DMD 2.073.1 and latest 2.075.0-master-972eaed
//Windows 7 32-bit

union Union1
{
  align(1):
  byte[5] bytes5;
  struct
  {
    align(1):
    char char1;
    uint int1;
  }
}

void main ()
{
  import std.stdio: writefln;
  writefln("Union1.sizeof=%d", Union1.sizeof);  //prints 8, not 5
}

I expect size of Union1 is 5 (5 bytes == char + uint == 5).
Is this my bug or DMD?
Can anybody reproduce?

On 03/25/2017 11:37 PM, zabruk70 wrote:
> union Union1
> {
>   align(1):
>   byte[5] bytes5;
>   struct
>   {
>     align(1):
>     char char1;
>     uint int1;
>   }
> }
>
> void main ()
> {
>   import std.stdio: writefln;
>   writefln("Union1.sizeof=%d", Union1.sizeof);  //prints 8, not 5
> }

I'm not sure how the align stuff is supposed to work exactly, but you get 5 when you add `align(1)` to the union itself, too:

----
align(1) union Union1
{
  align(1):
  byte[5] bytes5;
  struct
  {
    char char1;
    uint int1;
  }
}

pragma(msg, Union1.sizeof); /* "5LU" */
----

zabruk70 wrote:

> //DMD 2.073.1 and latest 2.075.0-master-972eaed
> //Windows 7 32-bit
>
> union Union1
> {
>    align(1):
>    byte[5] bytes5;
>    struct
>    {
>      align(1):
>      char char1;
>      uint int1;
>    }
> }
>
> void main ()
> {
>    import std.stdio: writefln;
>    writefln("Union1.sizeof=%d", Union1.sizeof);  //prints 8, not 5
> }
>
> I expect size of Union1 is 5 (5 bytes == char + uint == 5).
> Is this my bug or DMD?

`align(1) union Union1` will do the trick.

what you did is members packing. but the union itself is padded to integer size too. i.e. internal `align` will set aligning for *members*, and external `align` will change padding of the whole thing.

Thank you ag0aep6g and ketmar!!

I will use additional outside align.
I want packing inside, you are right.
But i check result size with assert() and failed.

But for clearness...
I was thinked, that align not changes SIZE, but changes LOCATION.
I was thinked, that "align(X) union Union1"
just force compiler to place Union1 on boundaries of X bytes...

On Saturday, 25 March 2017 at 22:54:30 UTC, zabruk70 wrote:
> But for clearness...
> I was thinked, that align not changes SIZE, but changes LOCATION.
> I was thinked, that "align(X) union Union1"
> just force compiler to place Union1 on boundaries of X bytes...

In order for all Union1 instances in an array to be aligned on boundaries of X bytes, the size of Union1 needs to be padded to a multiple of X in such a case.

On Saturday, 25 March 2017 at 22:45:22 UTC, ketmar wrote:
> zabruk70 wrote:
>
>> //DMD 2.073.1 and latest 2.075.0-master-972eaed
>> //Windows 7 32-bit
>>
>> union Union1
>> {
>>    align(1):
>>    byte[5] bytes5;
>>    struct
>>    {
>>      align(1):
>>      char char1;
>>      uint int1;
>>    }
>> }
>>
>> void main ()
>> {
>>    import std.stdio: writefln;
>>    writefln("Union1.sizeof=%d", Union1.sizeof);  //prints 8, not 5
>> }
>>
>> I expect size of Union1 is 5 (5 bytes == char + uint == 5).
>> Is this my bug or DMD?
>
> `align(1) union Union1` will do the trick.
>
> what you did is members packing. but the union itself is padded to integer size too. i.e. internal `align` will set aligning for *members*, and external `align` will change padding of the whole thing.

The union should have an implicit alignment of 1 already though, right? It's defined as the maximum of all member alignments, and both the bytes5 array and the anonymous struct members have an explicit alignment of 1. The alignment of the anonymous struct's int1 member (explicitly 1 too) shouldn't even matter.

On Saturday, 25 March 2017 at 23:36:07 UTC, kinke wrote:
> On Saturday, 25 March 2017 at 22:45:22 UTC, ketmar wrote:
>> zabruk70 wrote:
>>
>>> [...]
>>
>> `align(1) union Union1` will do the trick.
>>
>> what you did is members packing. but the union itself is padded to integer size too. i.e. internal `align` will set aligning for *members*, and external `align` will change padding of the whole thing.
>
> The union should have an implicit alignment of 1 already though, right? It's defined as the maximum of all member alignments, and both the bytes5 array and the anonymous struct members have an explicit alignment of 1. The alignment of the anonymous struct's int1 member (explicitly 1 too) shouldn't even matter.

Why should it have an implicit alignment of 1? The alignment completely depends on the definition unless specified AFAIK.

zabruk70 wrote:

> Thank you ag0aep6g and ketmar!!
>
> I will use additional outside align.
> I want packing inside, you are right.
> But i check result size with assert() and failed.
>
> But for clearness...
> I was thinked, that align not changes SIZE, but changes LOCATION.
> I was thinked, that "align(X) union Union1"
> just force compiler to place Union1 on boundaries of X bytes...

most of the time either location or padding will work the same. ;-)

but no, the idea of aligning is that structs/unions/etc. that comes together one after another are aligned from the starting offset of the first element, not from "address zero" for the whole app.

On Sunday, 26 March 2017 at 05:09:15 UTC, ketmar wrote:
> most of the time either location or padding will work the same.

hmm.. you ruined my expirence..

i made another experiment.
whould you please explain me S2 size 6?
thank you for you time.

https://dpaste.dzfl.pl/9a31b6e370a0

struct S1 //sizeof=6
{
  align(1):
  byte[3] b1; //offsetof=0, sizeof=3
  byte[3] b2; //offsetof=3, sizeof=3
}

struct S2 //sizeof must be 7, but DMD say 6
{
  align(4):
  byte[3] b1; //offsetof=0, sizeof=3
  byte[3] b2; //offsetof=4, sizeof=3
}

zabruk70 wrote:

> On Sunday, 26 March 2017 at 05:09:15 UTC, ketmar wrote:
>> most of the time either location or padding will work the same.
>
> hmm.. you ruined my expirence..
>
> i made another experiment.
> whould you please explain me S2 size 6?
> thank you for you time.

yes. you have a typo in second `writefln`: S1 instead of S2. ;-)