Jan Engelhardt <jeng...@medozas.de> writes:
> I noticed that __alignof__(uint64_t) will return 8, while
> __alignof__(struct { uint64_t x; }) will give only 4. This
> run on a typical 32-bit x86 CPU (GCC config below).
>
> What I am wondering about is why GCC was coded to give different
> alignments here. If aligning a single uint64_t to a boundary of 8 for
> whatever reason there may be (performance?), not doing so when it is
> inside a struct appears to be a discrepancy.
This question would have been more appropriate for the mailing list
gcc-h...@gcc.gnu.org rather than g...@gcc.gnu.org. Please take any
followups to gcc-help. Thanks.
The x86 ABI specifies the aligment of fields within a struct. gcc can
not change that alignment without breaking the ABI. This does not
matter much when the only field has type uint64_t, but it matters for a
case like "struct { int f1; uint64_t f2; };". For a variable of type
uint64_t, it's OK for gcc to increase the alignment, since it does not
affec the ABI (although the required alignment is only 4, any program
must work if all uint64_t variables happen to be aligned on an 8-byte
boundary).
Ian
