Issues with union and memory aligment

Question

i'm currently working on a Port of Embedded Code (on a Freescale S12) so GNU and i hava a issue with unions. i have the following union

typedef signed short        sint16;
typedef signed long         sint32;

typedef union
{
    sint32 Akku;
    sint16 AkkuHigh;
    sint16 AkkuLow;
} akku_type;

and i want to access the highest 2 byte of the union. The Problem is, that both AkkuHigh and AkkuLow have the same starting adress as Akku. It seems to be compiler specific. My Questions are: Is the there a Compiler Flag which changes the behaviour of the union? Can atribute((align (2))) help me?

Thank you in Advance

Pack the `high` and `low` fields in an inner anonymous `struct`. Strictly speaking it's not portable, but it should work. — user703016, Dec 03 '14 at 17:44
@ParkYoung-Bae: C allows it, and it's a common C++ extension. Wonder why the committee didn't cave in and legitimize it... — Deduplicator, Dec 03 '14 at 17:50
@ParkYoung-Bae: Padding at the *beginning* of a struct? Where did you get that standards-breaking compiler? (And non-neccessary padding *after* a member is just weird, and not yet found in the wild.) — Deduplicator, Dec 03 '14 at 17:55

score 3 · Answer 1 · answered Dec 03 '14 at 17:44

3

Yes, all of Akku, AkkuHigh, AkkuLow have the same address. This is how unions work in C. By the look of it, you intended to make an union with a 32-bit member and a member that is a struct of two 16-bit members instead. What you wrote is not the way to achieve it. Try instead:

typedef union
{
    sint32 Akku;
    struct s {
      sint16 AkkuHigh;
      sint16 AkkuLow;
    } representation;
} akku_type;

answered Dec 03 '14 at 17:44

Pascal Cuoq

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Is it guaranteed that `AkkuHigh` will be the high byte and `AkkuLow` will be the low byte? Couldn't the compiler decide to switch them around? – Unapiedra Dec 03 '14 at 17:45
@Unapiedra The guarantee is that the addresses of the members of a struct are in order. What these addresses correspond to is implementation-dependant (typically, this depends on the endianness of the target architecture. The struct as written works for most big-endian architectures). – Pascal Cuoq Dec 03 '14 at 17:48
@Unapiedra: Switching them around is not allowed, but adding arbitrary (implementation-defined) amounts of padding after `struct`-members is (*not* before them). – Deduplicator Dec 03 '14 at 17:57

score 0 · Answer 2 · edited May 23 '17 at 12:13

0

The correct definition of the union can be found in this answer.

atribute(align(2)) will definitely help you if you compile this on a 32bit or 64bit architecture. Also, on 64bit sizeof(sint32) is 8 (64 bits).

Depending on the endian-ness of the architecture, you may need to swap AkkuHigh and AkkuLow.

edited May 23 '17 at 12:13

Community

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answered Dec 03 '14 at 17:51

axiac

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Just mention that's valid C but not C++, though a common extension in C++. – Deduplicator Dec 03 '14 at 17:53
What exactly is not valid C++? The unnamed struct? – axiac Dec 03 '14 at 17:54
Yes. The unnamed member of anonymous type. Ping me when you corrected it. – Deduplicator Dec 03 '14 at 17:58
Got it. Removed the definition of the union entirely, it is described well in the other answer. – axiac Dec 03 '14 at 18:09
I would keep it and just add that disclaimer. It's the better solution in C, and arguably also in C++, considering how common that extension is. – Deduplicator Dec 03 '14 at 18:12

Issues with union and memory aligment

2 Answers2