On Wed, Jul 29, 2026 at 01:53:26PM +0200, Stefan Schulze Frielinghaus wrote:
> > The ugly thing about the x86_64 (or s390x) decision that some people
> > don't like is that we have the __int128 type and that passing of
> > _BitInt(N) for N in [65, 128] may differ from __int128 (on x86_64
> > actually it doesn't), and more importantly that
> > alignof (__int128) != alignof (_BitInt(128)).
>
> Can you elaborate on that? On s390x, alignment/size of __int128 and
> _BitInt(128) are the same and both are passed via memory. Also
> _BitInt(N) with 64<N<128 should have same sizeof/alignof as __int128 and
> being passed via memory. Since those _BitInt(N) even require extension,
> there shouldn't be any difference compared to __int128.
>
> That being said, it was deliberate to make __int128 and _BitInt(128)
> behave the same. Otherwise we could have went for passing _BitInt(128)
> via vector registers for example.
Ah, sorry, indeed s390x has alignof (_BitInt(128)) == 8.
On x86_64 it stands though.
And, note, even for __int128 the alignof is doable without going to 128-bit
ABI limbs, by simply having special cases for sizes up to 128 bits
inclusive. Just alignof (_BitInt(129)) would be strangely smaller than
alignof (_BitInt(128)).
alignof sizeof multilimb
n <= 8 1 1 false
n <= 16 2 2 false
n <= 32 4 4 false
n <= 64 8 8 false
n <= 128 16 16 false
n > 128 8 (n+63)/64*8 true
Jakub
