https://gcc.gnu.org/bugzilla/show_bug.cgi?id=64798
--- Comment #8 from rguenther at suse dot de <rguenther at suse dot de> --- On Tue, 27 Jan 2015, jakub at gcc dot gnu.org wrote: > https://gcc.gnu.org/bugzilla/show_bug.cgi?id=64798 > > Jakub Jelinek <jakub at gcc dot gnu.org> changed: > > What |Removed |Added > ---------------------------------------------------------------------------- > CC| |jakub at gcc dot gnu.org > > --- Comment #6 from Jakub Jelinek <jakub at gcc dot gnu.org> --- > __BIGGEST_ALIGNMENT__ is unnecessarily large, e.g. on i?86-linux, it is 16, > while you only need to guarantee 8-byte alignment. > Isn't the bug just in the badalloc1.C testcase, if it provides its own malloc, > it should IMHO guarantee the system malloc alignments (generally, something > like alignment of > union U > { > long long ll; > double d; > long double ld; > void *p; > }; > . no, badalloc1.C provides __BIGGEST_ALIGNMENT__ aligned memory. The bug is in the EH allocator which allocates the extra size_t entry aligned but the EH object itself only size_t aligned. It's documented that __attribute__((aligned)) aligns to the biggest type so using __BIGGEST_ALIGNMENT__ looks correct here - this is also what malloc () guarantees for alignment, no? Do we have any other means of getting the same alignment as malloc () provides? Btw, it still wastes a lot less space than the previous allocator. Bug in the patch anyway and the following hunk is also needed: @@ -185,7 +186,7 @@ namespace { __gnu_cxx::__scoped_lock sentry(emergency_mutex); allocated_entry *e = reinterpret_cast <allocated_entry *> - (reinterpret_cast <char *> (data) - sizeof (std::size_t)); + (reinterpret_cast <char *> (data) - offsetof (allocated_entry, data)); std::size_t sz = e->size; if (!first_free_entry) {
