Re: [kbuild-all] Re: include/linux/compiler_types.h:315:38: error: call to '__compiletime_assert_536' declared with attribute error: BUILD_BUG_ON failed: offsetof(struct can_frame, len) != offsetof(struct canfd_frame, len) || offsetof(struct can_frame, data) != offsetof(struc...

[Rasmus Villemoes <linux@rasmusvillemoes.dk>]

On 23/03/2021 19.59, Oliver Hartkopp wrote:
> 
> 
> On 23.03.21 15:00, Rasmus Villemoes wrote:

>> Now what CONFIG_* knobs are responsible for putting -mabi=apcs-gnu in
>> CFLAGS is left as an exercise for the reader. Regardless, it is not a
>> bug in the compiler. The error is the assumption that this language
>>
>> "Aggregates and Unions
>>
>> Structures and unions assume the alignment of their most strictly
>> aligned component.
> 
> (parse error in sentence)

It was a direct quote, but I can try to paraphrase with an example. If
you have a struct foo { T1 m1; T2 m2; T3 m3; }, then alignof(struct foo)
= max(alignof(T1), alignof(T2), alignof(T3)). Same for a "union foo".

But this is specifically for x86-64; for (some flavors of) ARM, other
rules apply - namely, alignof(T) is 4 unless T is char or short (or
(un)signed variants), ignoring bitfields which have their own rules.
Note that while

union u {char a; char b;}

has alignment 4 on ARM and 1 on x86-64, other types are less strictly
aligned on ARM; e.g. s64 aka long long is 8-byte aligned on x86-64 but
(still) just 4-byte aligned on ARM. And again, this is just for specific
-mabi= options.

>> Each member is assigned to the lowest available offset with the
>> appropriate
>> alignment. The size of any object is always a multiple of the object‘s
>> alignment."
>>
>> from the x86-64 ABI applies on all other architectures/ABIs.
>>
>>> I'm not a compiler expert but this does not seem to be consistent.
>>>
>>> Especially as we only have byte sizes (inside and outside of the union)
>>> and "A field with a char type is aligned to the next available byte."
>>
>> Yes, and that's exactly what you got before the anon union was
>> introduced.
> 
> Before(!) the union there is nothing to pad.

Just to be clear, my "before" was in the temporal sense, i.e. "prior to
commit ea7800565a128", all the u8s in struct can_frame were placed one
after the other. But after that commit, struct can_frame has a new
member replacing can_dlc which happens to occupy 4 bytes (for some
ABIs), pushing the subsequent members __pad, __res0 and len8_dlc
(formerly known as __res1) ahead.

>>> The union is indeed aligned to the word boundary - but the following
>>> byte is not aligned to the next available byte.
>>
>> Yes it is, because the union occupies 4 bytes. The first byte is shared
>> by the two char members, the remaining three bytes are padding.
> 
> But why is the union 4 bytes long here and adds a padding of three bytes
> at the end?

Essentially, because arrays. It's true for _any_ type T that sizeof(T)
must be a multiple of alignof(T). Take an array "T x[9]". If x[0] is
4-byte aligned, then in order for x[1] to be 4-byte aligned as well,
x[0] must occupy a multiple of 4 bytes.

It doesn't matter at all that this happens to be an anonymous union.
Layout-wise, you could as well have a definition

union uuu { __u8 len; __u8 can_dlc; }

and made struct can_frame

struct can_frame {
   canid_t can_id;
   union uuu u;
   __u8 __pad;
   ...
};

(you lose the anonymity trick so you'd have to do frame->u.can_dlc
instead of just frame->can_dlc). You have a member with alignof()==4 and
 sizeof()==4; that sizeof() cannot magically become 1 just because that
particular instance of the type is not part of an array. Imagine what
would happen if the compiler pulled subsequent char members into
trailing padding of a previous compound member. E.g. consider

struct a { int x; char y; } // alignof==4, sizeof==8, offsetof(y)==4
struct b { struct a a; char z; }

If I have a "struct b *b", I'm allowed to do "&b->a" and get a "pointer
to struct a". Then I can do memset(&b->a, 0, sizeof(struct a)). Clearly,
z must not have been placed inside the trailing padding of struct a.

Rasmus