This probably doesn't quite work for 32-bit big endian? At least, I used this:

static __inline uint128_t
to128(const void *p)
{
#if __SIZEOF_LONG__ == 4
	return (le64dec(p));
#else
	uint64_t lo, hi;

	lo = le64dec(p);
	hi = le64dec((const char *)p + 8);
	return ((uint128_t)hi << 64 | lo);
#endif
}

(I think it's a bit cleaner to use le64dec instead of assuming it is aligned). Also, if it weren't invasive to do the rename I would rather call this le128dec instead of to128

In D44651#1018919, @jhb wrote:

This probably doesn't quite work for 32-bit big endian? At least, I used this:

static __inline uint128_t
to128(const void *p)
{
#if __SIZEOF_LONG__ == 4
	return (le64dec(p));
#else
	uint64_t lo, hi;

	lo = le64dec(p);
	hi = le64dec((const char *)p + 8);
	return ((uint128_t)hi << 64 | lo);
#endif
}

(I think it's a bit cleaner to use le64dec instead of assuming it is aligned). Also, if it weren't invasive to do the rename I would rather call this le128dec instead of to128

edited There's reasons I was so lazy... but they don't matter.

Here's what I landed on...

#if __STDC_VERSION__ >= 202311L
typedef unsigned __BitInt(128) uint128_t;
#elif defined(__SIZEOF_INT128__)
typedef __uint128_t uint128_t;
#else
typedef uint64_t uint128_t;
#endif

static __inline uint128_t
to128(void *p)
{
#if __STDC_VERSION__ >= 202311L || defined(__SIZEOF_INT128__)
        uint64_t lo, hi;

        lo = le64dec(p);
        hi = le64dec((const char *)p + 8);
        return ((uint128_t)hi << 64 | lo);
#else
     	return (le64dec(p));
#endif
}

but I think you need gcc14 or clang 18 though... We'd have to add CSTD=c23 to the Makefile too...

ah, turns out

I do like the idea of using C23 for the future as a way to fix 32-bit platforms.