http://gcc.gnu.org/bugzilla/show_bug.cgi?id=54005
Hans-Peter Nilsson <hp at gcc dot gnu.org> changed: What |Removed |Added ---------------------------------------------------------------------------- CC| |crowl at gcc dot gnu.org --- Comment #6 from Hans-Peter Nilsson <hp at gcc dot gnu.org> 2012-08-14 03:52:07 UTC --- (In reply to comment #4) > A GCC port is incorrect if it is issuing any kind of lock. GCC is only > allowed > to issue a lock free sequence of some sort. If a lock is required, then a > call > to libatomic must be made. > > So if there is a type which is sometimes lock free and sometimes not due to > alignment or anything else, __atomic_always_lock_free should return FALSE > for > that type. Yes... > I think your current problem is that the infrastructure change for > __atomic_always_lock_free to handle alignment issues is not present yet. So it > erroneously says TRUE for this type when in fact it shouldn't. This PR is about the libstdc++ library is_lock_free function using the *per-object* builtin query when it should use the *per-type* query as per the referenced discussion. And that's been corrected. It's separate to my other target woes; an incidental observation. I agree the function has a name intuitively leading to thinking it should be per-object, but it isn't. > Its important that the libstdc++ routines call the __atomic_is_lock_free() so > that the answer can be determined at runtime if libatomic is utilized for any > atomic sequences. Maybe elsewhere, but not this particular code. If your argument is that is_lock_free should be per-object, then you'll have to argue with the standard-guys; I'm just quoting the reference above. > __atomic_always_lock_free() is *always* resolved to a 1 or a 0 at compile > time. > The compiler is only capable of answering the question, "Does the compiler > always generate a lock free sequence". Which is exactly the kind of answer sought here. Is the n2992 reference incorrect or am I (and bkoz) misinterpreting it? Are things about to change again standard-wise? Let's try and ask mr. Crowl.